Set-theory-&-relation-1
Definition.
Let A and B be two non-empty sets, then every subset of A × B defines a relation from A to B and every
relation from A to B is a subset of A × B.
Let R ⊆ A × B and (a, b) ∈ R. Then we say that a is related to b by the relation R and write it as a R b . If
(a,b)∈ R , we write it as a R b .
Example: Let A = {1, 2, 5, 8, 9}, B = {1, 3} we set a relation from A to B as: a R b iff a ≤ b; a ∈ A, b ∈ B .
Then R = {(1, 1)}, (1, 3), (2, 3)} ⊂ A × B
(1) Total number of relations : Let A and B be two non-empty finite sets consisting of m and n elements
respectively. Then A × B consists of mn ordered pairs. So, total number of subset of A × B is 2mn. Since each
subset of A × B defines relation from A to B, so total number of relations from A to B is 2mn. Among these 2mn
relations the void relation φ and the universal relation A × B are trivial relations from A to B.
(2) Domain and range of a relation : Let R be a relation from a set A to a set B. Then the set of all first
components or coordinates of the ordered pairs belonging to R is called the domain of R, while the set of all second
components or coordinates of the ordered pairs in R is called the range of R.
Thus, Dom (R) = {a : (a, b) ∈ R} and Range (R) = {b : (a, b) ∈ R}.
It is evident from the definition that the domain of a relation from A to B is a subset of A and its range is a subset
of B.
(3) Relation on a set : Let A be a non-void set. Then, a relation from A to itself i.e. a subset of A × A is called
a relation on set A.
Example: 1 Let A = {1, 2, 3}. The total number of distinct relations that can be defined over A is
(a) 29 (b) 6 (c) 8 (d) None of these
Solution: (a) n(A × A) = n(A).n(A) = 32 = 9
Example: 2 So, the total number of subsets of A × A is 29 and a subset of A × A is a relation over the set A.
Let X = {1, 2, 3, 4,5} and Y = {1, 3,5,7,9} . Which of the following is/are relations from X to Y
(a) R1 = {(x, y)|y = 2 + x, x ∈ X, y ∈ Y} (b) R2 = {(1,1),(2,1),(3, 3),(4, 3),(5,5)}
(c) R3 = {(1,1),(1, 3)(3,5),(3,7),(5,7)} (d) R4 = {(1, 3),(2,5),(2, 4),(7,9)}
Solution: (a,b,c) R4 is not a relation from X to Y, because (7, 9) ∈ R4 but (7, 9) ∉ X × Y .
Example: 3 Given two finite sets A and B such that n(A) = 2, n(B) = 3. Then total number of relations from A to B is
(a) 4 (b) 8 (c) 64 (d) None of these
Solution: (c) Here n(A × B) = 2 × 3 = 6
Since every subset of A × B defines a relation from A to B, number of relation from A to B is equal to number of subsets
of A × B = 26 = 64, which is given in (c).
Example: 4 The relation R defined on the set of natural numbers as {(a, b) : a differs from b by 3}, is given by
(a) {(1, 4, (2, 5), (3, 6),.....} (b) {(4, 1), (5, 2), (6, 3),.....} (c) {(1, 3), (2, 6), (3, 9),..} (d) None of these
Solution: (b) R = {(a,b) : a,b ∈ N,a − b = 3} = {((n + 3), n) : n ∈ N} = {(4,1), (5, 2), (6, 3).....}
Inverse Relation.
Let A, B be two sets and let R be a relation from a set A to a set B. Then the inverse of R, denoted by R–1, is a
relation from B to A and is defined by R −1 = {(b, a) : (a, b) ∈ R}
Clearly (a, b) ∈ R ⇔ (b, a) ∈ R–1 . Also, Dom (R) = Range (R−1) and Range (R) = Dom (R−1)
Example : Let A = {a, b, c}, B = {1, 2, 3} and R = {(a, 1), (a, 3), (b, 3), (c, 3)}.
Then, (i) R–1 = {(1, a), (3, a), (3, b), (3, c)}
(ii) Dom (R) = {a, b, c} = Range (R−1)
(iii) Range (R) = {1, 3} = Dom (R−1)
Example: 5 Let A = {1, 2, 3}, B = {1, 3, 5}. A relation R : A → B is defined by R = {(1, 3), (1, 5), (2, 1)}. Then R−1 is defined by
Solution: (c) (a) {(1,2), (3,1), (1,3), (1,5)} (b) {(1, 2), (3, 1), (2, 1)} (c) {(1, 2), (5, 1), (3, 1)} (d) None of these
Example: 6
(x, y)∈ R ⇔ (y, x)∈ R−1 , ∴ R−1 = {(3,1),(5,1),(1, 2)} .
Solution: (b)
The relation R is defined on the set of natural numbers as {(a, b) : a = 2b}. Then R−1 is given by
(a) {(2, 1), (4, 2), (6, 3).....} (b) {(1, 2), (2, 4), (3, 6)....} (c) R−1 is not defined (d) None of these
R = {(2, 1), (4, 2), (6, 3),......} So, R−1 = {(1, 2), (2, 4), (3, 6),.....}.
Types of Relations.
(1) Reflexive relation : A relation R on a set A is said to be reflexive if every element of A is related to itself.
Thus, R is reflexive ⇔ (a, a) ∈ R for all a ∈ A.
A relation R on a set A is not reflexive if there exists an element a ∈ A such that (a, a) ∉ R.
Example: Let A = {1, 2, 3} and R = {(1, 1); (1, 3)}
Then R is not reflexive since 3 ∈ A but (3, 3) ∉ R
Note : The identity relation on a non-void set A is always reflexive relation on A. However, a reflexive
relation on A is not necessarily the identity relation on A.
The universal relation on a non-void set A is reflexive.
(2) Symmetric relation : A relation R on a set A is said to be a symmetric relation iff
(a, b) ∈ R ⇒ (b, a) ∈ R for all a, b ∈ A
i.e. aRb ⇒ bRa for all a, b ∈ A.
it should be noted that R is symmetric iff R−1 = R
Note : The identity and the universal relations on a non-void set are symmetric relations.
A relation R on a set A is not a symmetric relation if there are at least two elements a, b ∈ A such that
(a, b) ∈ R but (b, a) ∉ R.
A reflexive relation on a set A is not necessarily symmetric.
(3) Anti-symmetric relation : Let A be any set. A relation R on set A is said to be an anti-symmetric relation
iff (a, b) ∈ R and (b, a) ∈ R ⇒ a = b for all a, b ∈ A.
Thus, if a ≠ b then a may be related to b or b may be related to a, but never both.
Example: Let N be the set of natural numbers. A relation R ⊆ N × N is defined by xRy iff x divides y(i.e., x/y).
Then x R y, y R x ⇒ x divides y, y divides x ⇒ x = y
Note : The identity relation on a set A is an anti-symmetric relation.
The universal relation on a set A containing at least two elements is not anti-symmetric, because if
a ≠ b are in A, then a is related to b and b is related to a under the universal relation will imply that
a = b but a ≠ b.
The set {(a, a): a ∈ A} = D is called the diagonal line of A × A . Then “the relation R in A is
antisymmetric iff R ∩ R −1 ⊆ D ”.
(4) Transitive relation : Let A be any set. A relation R on set A is said to be a transitive relation iff
(a, b) ∈ R and (b, c) ∈ R ⇒ (a, c) ∈ R for all a, b, c ∈ A i.e., aRb and bRc ⇒ aRc for all a, b, c ∈ A.
In other words, if a is related to b, b is related to c, then a is related to c.
Transitivity fails only when there exists a, b, c such that a R b, b R c but a R c.
Example: Consider the set A = {1, 2, 3} and the relations
R1 = {(1, 2),(1, 3)} ; R2 = {(1, 2)}; R3 = {(1, 1)}; R4 = {(1, 2), (2, 1), (1, 1)}
Then R1 , R2 , R3 are transitive while R4 is not transitive since in R4 ,(2, 1)∈ R4 ;(1, 2)∈ R4 but (2, 2)∉ R4 .
Note : The identity and the universal relations on a non-void sets are transitive.
The relation ‘is congruent to’ on the set T of all triangles in a plane is a transitive relation.
(5) Identity relation : Let A be a set. Then the relation IA = {(a, a) : a ∈ A} on A is called the identity relation on A.
In other words, a relation IA on A is called the identity relation if every element of A is related to itself only. Every
identity relation will be reflexive, symmetric and transitive.
Example: On the set = {1, 2, 3}, R = {(1, 1), (2, 2), (3, 3)} is the identity relation on A .
Note : It is interesting to note that every identity relation is reflexive but every reflexive relation need not be an
identity relation.
Also, identity relation is reflexive, symmetric and transitive.
(6) Equivalence relation : A relation R on a set A is said to be an equivalence relation on A iff
(i) It is reflexive i.e. (a, a) ∈ R for all a ∈ A
(ii) It is symmetric i.e. (a, b) ∈ R ⇒ (b, a) ∈ R, for all a, b ∈ A
(iii) It is transitive i.e. (a, b) ∈ R and (b, c) ∈ R ⇒ (a, c) ∈ R for all a, b, c ∈ A.
Note : Congruence modulo (m) : Let m be an arbitrary but fixed integer. Two integers a and b are said to
be congruence modulo m if a − b is divisible by m and we write a ≡ b (mod m).
Thus a ≡ b (mod m) ⇔ a − b is divisible by m. For example, 18 ≡ 3 (mod 5) because 18 – 3 = 15
which is divisible by 5. Similarly, 3 ≡ 13 (mod 2) because 3 – 13 = –10 which is divisible by 2. But 25
≠ 2 (mod 4) because 4 is not a divisor of 25 – 3 = 22.
The relation “Congruence modulo m” is an equivalence relation.
Important Tips
• If R and S are two equivalence relations on a set A , then R ∩ S is also an equivalence relation on A.
• The union of two equivalence relations on a set is not necessarily an equivalence relation on the set.
• The inverse of an equivalence relation is an equivalence relation.
Equivalence Classes of an Equivalence Relation.
Let R be equivalence relation in A(≠ φ) . Let a ∈ A . Then the equivalence class of a, denoted by [a] or {a} is
defined as the set of all those points of A which are related to a under the relation R. Thus [a] = {x ∈ A : x R a}.
It is easy to see that
(1) b ∈ [a] ⇒ a ∈ [b] (2) b ∈ [a] ⇒ [a] = [b] (3) Two equivalence classes are either disjoint or identical.
As an example we consider a very important equivalence relation x ≡ y(mod n) iff n divides (x − y),n is a fixed
positive integer. Consider n = 5. Then
[0] = {x : x ≡ 0 (mod 5)} = {5p : p ∈ Z} = { 0,± 5,± 10,± 15,.....}
[1] = {x : x ≡ 1(mod 5)} = {x : x − 1 = 5k,k ∈ Z} = {5k + 1 : k ∈ Z} = {1,6,11,......., − 4,− 9,.....} .
One can easily see that there are only 5 distinct equivalence classes viz. [0], [1], [2], [3] and [4], when n = 5.
Example: 7 Given the relation R = {(1, 2), (2, 3)} on the set A = {1, 2, 3}, the minimum number of ordered pairs which when added
Solution: (c) to R make it an equivalence relation is
Example: 8 (a) 5 (b) 6 (c) 7 (d) 8
Solution: (a)
Example: 9 R is reflexive if it contains (1, 1), (2, 2), (3, 3)
Solution: (a)
(1, 2) ∈ R,(2, 3)∈ R
∴ R is symmetric if (2, 1), (3, 2) ∈ R. Now, R = {(1, 1), (2, 2), (3, 3), (2, 1), (3, 2), (2, 3), (1, 2)}
R will be transitive if (3, 1); (1, 3) ∈ R. Thus, R becomes an equivalence relation by adding (1, 1) (2, 2) (3, 3) (2, 1) (3,2)
(1, 3) (3, 1). Hence, the total number of ordered pairs is 7.
The relation R = {(1, 1), (2, 2), (3, 3), (1, 2), (2, 3), (1, 3)} on set A = {1, 2, 3} is
(a) Reflexive but not symmetric (b) Reflexive but not transitive
(c) Symmetric and Transitive (d) Neither symmetric nor transitive
Since (1, 1); (2, 2); (3, 3) ∈ R therefore R is reflexive. (1, 2) ∈ R but (2, 1) ∉ R, therefore R is not symmetric. It can be
easily seen that R is transitive.
Let R be the relation on the set R of all real numbers defined by a R b iff |a − b|≤ 1 . Then R is [Roorkee 1998]
(a) Reflexive and Symmetric (b) Symmetric only (c) Transitive only (d) Anti-symmetric only
|a − a|= 0 < 1 ∴a R a ∀ a ∈ R
∴ R is reflexive, Again a R b ⇒ |a − b|≤ 1 ⇒|b − a|≤ 1 ⇒ bRa
∴ R is symmetric, Again 1R 1 and 1 R1 but 1 ≠ 1
2 2 2
∴ R is not anti-symmetric
Further, 1 R 2 and 2 R 3 but 1 R 3
[|1 − 3|= 2 > 1 ]
∴ R is not transitive.
Example: 10 The relation "less than" in the set of natural numbers is [UPSEAT 1994, 98; AMU 1999]
Solution: (b)
(a) Only symmetric (b) Only transitive (c) Only reflexive (d) Equivalence relation
Example: 11
Solution: (d) Since x < y, y < z ⇒ x < z ∨ x, y, z ∈ N
Example: 12 ∴ x R y, yR z ⇒ x R z , ∴ Relation is transitive , ∴ x < y does not give y < x , ∴ Relation is not symmetric.
Solution: (c)
Since x < x does not hold, hence relation is not reflexive.
Example: 13
Solution: (b) With reference to a universal set, the inclusion of a subset in another, is relation, which is [Karnataka CET 1995]
Example: 14
(a) Symmetric only (b) Equivalence relation (c) Reflexive only (d) None of these
Solution: (d)
Example: 15 Since A ⊆ A ∴ relation ' ⊆' is reflexive.
Solution: (b)
Example: 16 Since A ⊆ B, B ⊆ C ⇒ A ⊆ C
Solution: (d) ∴ relation ' ⊆' is transitive.
But A ⊆ B, ⇒ B ⊆ A , ∴ Relation is not symmetric.
Let A = {2, 4,6,8} . A relation R on A is defined by R = {(2, 4),(4, 2),(4,6),(6, 4)} . Then R is [Karnataka CET 1995]
(a) Anti-symmetric (b) Reflexive (c) Symmetric (d) Transitive
Given A = {2, 4, 6, 8}
R = {(2, 4)(4, 2) (4, 6) (6, 4)}
(a, b) ∈ R ⇒ (b, a) ∈ R and also R−1 = R . Hence R is symmetric.
Let P = {(x, y)| x2 + y2 = 1, x, y ∈ R} . Then P is
(a) Reflexive (b) Symmetric (c) Transitive (d) Anti-symmetric
Obviously, the relation is not reflexive and transitive but it is symmetric, because x2 + y2 = 1 ⇒ y2 + x2 = 1 .
Let R be a relation on the set N of natural numbers defined by nRm ⇔ n is a factor of m (i.e., n|m). Then R is
(a) Reflexive and symmetric (b) Transitive and symmetric
(c) Equivalence (d) Reflexive, transitive but not symmetric
Since n | n for all n ∈ N , therefore R is reflexive. Since 2 | 6 but 6 | 2 , therefore R is not symmetric.
Let n R m and m R p ⇒ n|m and m|p ⇒ n|p ⇒ nRp. So R is transitive.
Let R be an equivalence relation on a finite set A having n elements. Then the number of ordered pairs in R is
(a) Less than n (b) Greater than or equal to n (c) Less than or equal to n (d) None of these
Since R is an equivalence relation on set A, therefore (a, a) ∈ R for all a ∈ A . Hence, R has at least n ordered pairs.
Let N denote the set of all natural numbers and R be the relation on N × N defined by (a, b) R (c, d) if
ad(b + c) = bc(a + d), then R is [Roorkee 1995]
(a) Symmetric only (b) Reflexive only (c) Transitive only (d) An equivalence relation
For (a, b), (c, d) ∈ N × N
(a,b)R(c, d) ⇒ ad(b + c) = bc(a + d)
Reflexive: Since ab(b + a) = ba(a + b)∀ab ∈ N ,
∴ (a,b)R(a,b) , ∴ R is reflexive.
Symmetric: For (a,b),(c, d)∈ N × N , let (a,b)R(c, d)
∴ ad(b + c) = bc(a + d) ⇒ bc(a + d) = ad(b + c) ⇒ cb(d + a) = da(c + b) ⇒ (c, d)R(a,b)
∴ R is symmetric
Transitive: For (a,b),(c, d),(e, f)∈ N × N, Let (a,b)R(c, d),(c, d)R(e, f)
∴ ad(b + c) = bc(a + d) , cf(d + e) = de(c + f)
⇒ adb + adc = bca + bcd .....(i) and cfd + cfe = dec + def .......(ii)
(i) × ef + (ii) × ab gives, adbef + adcef + cfdab + cfeab = bcaef + bcdef + decab + defab
⇒ adcf(b + e) = bcde(a + f) ⇒ af(b + e) = be(a + f) ⇒ (a,b)R(e, f) . ∴ R is transitive. Hence R is an equivalence relation.
Example: 17 For real numbers x and y, we write x Ry ⇔ x − y + 2 is an irrational number. Then the relation R is
Solution: (a)
(a) Reflexive (b) Symmetric (c) Transitive (d) None of these
Example: 18
Solution: (b) For any x ∈ R, we have x − x + 2 = 2 an irrational number.
Example: 19 ⇒ xRx for all x. So, R is reflexive.
Solution: (a) R is not symmetric, because 2R1 but 1 R 2 , R is not transitive also because 2 R 1 and 1R2 2 but 2 R 2 2 .
Example: 20 Let X be a family of sets and R be a relation on X defined by ‘A is disjoint from B’. Then R is
Solution: (c)
(a) Reflexive (b) Symmetric (c) Anti-symmetric (d) Transitive
Clearly, the relation is symmetric but it is neither reflexive nor transitive.
Let R and S be two non-void relations on a set A. Which of the following statements is false
(a) R and S are transitive ⇒ R ∪ S is transitive (b) R and S are transitive ⇒ R ∩ S is transitive
(c) R and S are symmetric ⇒ R ∪ S is symmetric (d) R and S are reflexive ⇒ R ∩ S is reflexive
Let A = {1, 2, 3} and R = {(1, 1), (1, 2)}, S = {(2, 2) (2, 3)} be transitive relations on A.
Then R ∪ S = {(1, 1); (1, 2); (2, 2); (2, 3)}
Obviously, R ∪ S is not transitive. Since (1, 2) ∈ R ∪ S and (2, 3)∈ R ∪ S but (1, 3) ∉ R ∪ S .
The solution set of 8x ≡ 6(mod 14), x ∈ Z , are
(a) [8] ∪ [6] (b) [8] ∪ [14] (c) [6] ∪ [13] (d) [8] ∪ [6] ∪ [13]
8x − 6 = 14P(P ∈ Z) ⇒ x = 1 [14 P + 6], x ∈ Z
8
⇒x = 1 (7P + 3) ⇒ x = 6, 13, 20, 27, 34, 41, 48,.......
4
∴ Solution set = {6, 20, 34, 48,.....} ∪ {13, 27, 41, ......} = [6] ∪ [13].
Where [6], [13] are equivalence classes of 6 and 13 respectively.
Composition of Relations.
Let R and S be two relations from sets A to B and B to C respectively. Then we can define a relation SoR from
A to C such that (a, c) ∈ SoR ⇔ ∃ b ∈ B such that (a, b) ∈ R and (b, c) ∈ S.
This relation is called the composition of R and S.
For example, if A = {1, 2, 3}, B = {a, b, c, d}, C = {p, q, r, s} be three sets such that R = {(1, a), (2, c), (1, c),
(2, d)} is a relation from A to B and S = {(a, s), (b, q), (c, r)} is a relation from B to C. Then SoR is a relation from
A to C given by SoR = {(1, s) (2, r) (1, r)}
In this case RoS does not exist.
In general RoS ≠ SoR. Also (SoR)–1 = R–1oS–1.
Example: 21 If R is a relation from a set A to a set B and S is a relation from B to a set C, then the relation SoR
Solution: (a)
Example: 22 (a) Is from A to C (b) Is from C to A (c) Does not exist (d) None of these
Solution: (b)
Example: 23 It is obvious.
Solution: (c) If R ⊂ A × B and S ⊂ B × C be two relations, then (SoR)−1 =
Example: 24 (a) S−1oR−1 (b) R−1oS−1 (c) SoR (d) RoS
Solution: (a) It is obvious.
If R be a relation < from A = {1,2, 3, 4} to B = {1, 3, 5} i.e., (a, b) ∈ R ⇔ a < b, then RoR−1 is
(a) {(1, 3), (1, 5), (2, 3), (2, 5), (3, 5), (4, 5)}
(b) {(3, 1) (5, 1), (3, 2), (5, 2), (5, 3), (5, 4)}
(c) {(3, 3), (3, 5), (5, 3), (5, 5)}
(d) {(3, 3) (3, 4), (4, 5)}
We have, R = {(1, 3); (1, 5); (2, 3); (2, 5); (3, 5); (4, 5)}
R−1 = {(3, 1), (5, 1), (3, 2), (5, 2); (5, 3); (5, 4)}
Hence RoR−1 = {(3, 3); (3, 5); (5, 3); (5, 5)}
Let a relation R be defined by R = {(4, 5); (1, 4); (4, 6); (7, 6); (3, 7)} then R−1oR is
(a) {(1, 1), (4, 4), (4, 7), (7, 4), (7, 7), (3, 3)} (b) {(1, 1), (4, 4), (7, 7), (3, 3)}
(c) {(1, 5), (1, 6), (3, 6)} (d) None of these
We first find R−1, we have R−1 = {(5, 4);(4,1);(6, 4);(6,7);(7, 3)} we now obtain the elements of R −1oR we first pick the
element of R and then of R−1 . Since (4, 5) ∈ R and (5, 4) ∈ R−1 , we have (4, 4)∈ R−1oR
Similarly, (1, 4)∈ R,(4,1)∈ R−1 ⇒(1,1)∈ R−1oR
(4, 6)∈ R,(6, 4)∈ R−1 ⇒(4, 4)∈ R−1oR, (4, 6)∈ R,(6,7)∈ R−1 ⇒ (4,7)∈ R−1oR
(7, 6)∈ R,(6, 4)∈ R−1 ⇒ (7, 4)∈ R−1oR, (7, 6) ∈ R,(6,7) ∈ R−1 ⇒ (7,7) ∈ R−1oR
(3,7)∈ R,(7, 3)∈ R−1 ⇒ (3, 3)∈ R−1oR,
Hence R−1oR = {(1, 1); (4, 4); (4, 7); (7, 4), (7, 7); (3, 3)}.
Axiomatic Definitions of the Set of Natural Numbers (Peano's Axioms).
The set N of natural numbers (N = {1, 2, 3, 4......}) is a set satisfying the following axioms (known as peano's
axioms)
(1) N is not empty.
(2) There exist an injective (one-one) map S : N → N given by S(n) = n+ , where n+ is the immediate
successor of n in N i.e., n + 1 = n+ .
(3) The successor mapping S is not surjective (onto).
(4) If M ⊆ N such that,
(i) M contains an element which is not the successor of any element in N, and
(ii) m ∈ M ⇒ m+ ∈ M , then M = N
This is called the axiom of induction. We denote the unique element which is not the successor of any element is
1. Also, we get 1+ = 2, 2+ = 3 .
Note : Addition in N is defined as,
n + 1 = n+
n + m+ = (n + m)+
Multiplication in N is defined by,
n.1 = n
n.m+ = n.m + n
Set-theory-&-relation-2
Definition.
Let A and B be two non-empty sets, then every subset of A × B defines a relation from A to B and every
relation from A to B is a subset of A × B.
Let R ⊆ A × B and (a, b) ∈ R. Then we say that a is related to b by the relation R and write it as a R b . If
(a,b)∈ R , we write it as a R b .
Example: Let A = {1, 2, 5, 8, 9}, B = {1, 3} we set a relation from A to B as: a R b iff a ≤ b; a ∈ A, b ∈ B .
Then R = {(1, 1)}, (1, 3), (2, 3)} ⊂ A × B
(1) Total number of relations : Let A and B be two non-empty finite sets consisting of m and n elements
respectively. Then A × B consists of mn ordered pairs. So, total number of subset of A × B is 2mn. Since each
subset of A × B defines relation from A to B, so total number of relations from A to B is 2mn. Among these 2mn
relations the void relation φ and the universal relation A × B are trivial relations from A to B.
(2) Domain and range of a relation : Let R be a relation from a set A to a set B. Then the set of all first
components or coordinates of the ordered pairs belonging to R is called the domain of R, while the set of all second
components or coordinates of the ordered pairs in R is called the range of R.
Thus, Dom (R) = {a : (a, b) ∈ R} and Range (R) = {b : (a, b) ∈ R}.
It is evident from the definition that the domain of a relation from A to B is a subset of A and its range is a subset
of B.
(3) Relation on a set : Let A be a non-void set. Then, a relation from A to itself i.e. a subset of A × A is called
a relation on set A.
Example: 1 Let A = {1, 2, 3}. The total number of distinct relations that can be defined over A is
(a) 29 (b) 6 (c) 8 (d) None of these
Solution: (a) n(A × A) = n(A).n(A) = 32 = 9
Example: 2 So, the total number of subsets of A × A is 29 and a subset of A × A is a relation over the set A.
Let X = {1, 2, 3, 4,5} and Y = {1, 3,5,7,9} . Which of the following is/are relations from X to Y
(a) R1 = {(x, y)|y = 2 + x, x ∈ X, y ∈ Y} (b) R2 = {(1,1),(2,1),(3, 3),(4, 3),(5,5)}
(c) R3 = {(1,1),(1, 3)(3,5),(3,7),(5,7)} (d) R4 = {(1, 3),(2,5),(2, 4),(7,9)}
Solution: (a,b,c) R4 is not a relation from X to Y, because (7, 9) ∈ R4 but (7, 9) ∉ X × Y .
Example: 3 Given two finite sets A and B such that n(A) = 2, n(B) = 3. Then total number of relations from A to B is
(a) 4 (b) 8 (c) 64 (d) None of these
Solution: (c) Here n(A × B) = 2 × 3 = 6
Since every subset of A × B defines a relation from A to B, number of relation from A to B is equal to number of subsets
of A × B = 26 = 64, which is given in (c).
Example: 4 The relation R defined on the set of natural numbers as {(a, b) : a differs from b by 3}, is given by
(a) {(1, 4, (2, 5), (3, 6),.....} (b) {(4, 1), (5, 2), (6, 3),.....} (c) {(1, 3), (2, 6), (3, 9),..} (d) None of these
Solution: (b) R = {(a,b) : a,b ∈ N,a − b = 3} = {((n + 3), n) : n ∈ N} = {(4,1), (5, 2), (6, 3).....}
Inverse Relation.
Let A, B be two sets and let R be a relation from a set A to a set B. Then the inverse of R, denoted by R–1, is a
relation from B to A and is defined by R −1 = {(b, a) : (a, b) ∈ R}
Clearly (a, b) ∈ R ⇔ (b, a) ∈ R–1 . Also, Dom (R) = Range (R−1) and Range (R) = Dom (R−1)
Example : Let A = {a, b, c}, B = {1, 2, 3} and R = {(a, 1), (a, 3), (b, 3), (c, 3)}.
Then, (i) R–1 = {(1, a), (3, a), (3, b), (3, c)}
(ii) Dom (R) = {a, b, c} = Range (R−1)
(iii) Range (R) = {1, 3} = Dom (R−1)
Example: 5 Let A = {1, 2, 3}, B = {1, 3, 5}. A relation R : A → B is defined by R = {(1, 3), (1, 5), (2, 1)}. Then R−1 is defined by
Solution: (c) (a) {(1,2), (3,1), (1,3), (1,5)} (b) {(1, 2), (3, 1), (2, 1)} (c) {(1, 2), (5, 1), (3, 1)} (d) None of these
Example: 6
(x, y)∈ R ⇔ (y, x)∈ R−1 , ∴ R−1 = {(3,1),(5,1),(1, 2)} .
Solution: (b)
The relation R is defined on the set of natural numbers as {(a, b) : a = 2b}. Then R−1 is given by
(a) {(2, 1), (4, 2), (6, 3).....} (b) {(1, 2), (2, 4), (3, 6)....} (c) R−1 is not defined (d) None of these
R = {(2, 1), (4, 2), (6, 3),......} So, R−1 = {(1, 2), (2, 4), (3, 6),.....}.
Types of Relations.
(1) Reflexive relation : A relation R on a set A is said to be reflexive if every element of A is related to itself.
Thus, R is reflexive ⇔ (a, a) ∈ R for all a ∈ A.
A relation R on a set A is not reflexive if there exists an element a ∈ A such that (a, a) ∉ R.
Example: Let A = {1, 2, 3} and R = {(1, 1); (1, 3)}
Then R is not reflexive since 3 ∈ A but (3, 3) ∉ R
Note : The identity relation on a non-void set A is always reflexive relation on A. However, a reflexive relation
on A is not necessarily the identity relation on A.
The universal relation on a non-void set A is reflexive.
(2) Symmetric relation : A relation R on a set A is said to be a symmetric relation iff
(a, b) ∈ R ⇒ (b, a) ∈ R for all a, b ∈ A
i.e. aRb ⇒ bRa for all a, b ∈ A.
it should be noted that R is symmetric iff R−1 = R
Note : The identity and the universal relations on a non-void set are symmetric relations.
A relation R on a set A is not a symmetric relation if there are at least two elements a, b ∈ A such that
(a, b) ∈ R but (b, a) ∉ R.
A reflexive relation on a set A is not necessarily symmetric.
(3) Anti-symmetric relation : Let A be any set. A relation R on set A is said to be an anti-symmetric relation
iff (a, b) ∈ R and (b, a) ∈ R ⇒ a = b for all a, b ∈ A.
Thus, if a ≠ b then a may be related to b or b may be related to a, but never both.
Example: Let N be the set of natural numbers. A relation R ⊆ N × N is defined by xRy iff x divides y(i.e., x/y).
Then x R y, y R x ⇒ x divides y, y divides x ⇒ x = y
Note : The identity relation on a set A is an anti-symmetric relation.
The universal relation on a set A containing at least two elements is not anti-symmetric, because if
a ≠ b are in A, then a is related to b and b is related to a under the universal relation will imply that
a = b but a ≠ b.
The set {(a, a): a ∈ A} = D is called the diagonal line of A × A . Then “the relation R in A is
antisymmetric iff R ∩ R −1 ⊆ D ”.
(4) Transitive relation : Let A be any set. A relation R on set A is said to be a transitive relation iff
(a, b) ∈ R and (b, c) ∈ R ⇒ (a, c) ∈ R for all a, b, c ∈ A i.e., aRb and bRc ⇒ aRc for all a, b, c ∈ A.
In other words, if a is related to b, b is related to c, then a is related to c.
Transitivity fails only when there exists a, b, c such that a R b, b R c but a R c.
Example: Consider the set A = {1, 2, 3} and the relations
R1 = {(1, 2),(1, 3)} ; R2 = {(1, 2)}; R3 = {(1, 1)}; R4 = {(1, 2), (2, 1), (1, 1)}
Then R1 , R2 , R3 are transitive while R4 is not transitive since in R4 ,(2, 1)∈ R4 ;(1, 2)∈ R4 but (2, 2)∉ R4 .
Note : The identity and the universal relations on a non-void sets are transitive.
The relation ‘is congruent to’ on the set T of all triangles in a plane is a transitive relation.
(5) Identity relation : Let A be a set. Then the relation IA = {(a, a) : a ∈ A} on A is called the identity relation on A.
In other words, a relation IA on A is called the identity relation if every element of A is related to itself only. Every
identity relation will be reflexive, symmetric and transitive.
Example: On the set = {1, 2, 3}, R = {(1, 1), (2, 2), (3, 3)} is the identity relation on A .
Note : It is interesting to note that every identity relation is reflexive but every reflexive relation need not be an
identity relation.
Also, identity relation is reflexive, symmetric and transitive.
(6) Equivalence relation : A relation R on a set A is said to be an equivalence relation on A iff
(i) It is reflexive i.e. (a, a) ∈ R for all a ∈ A
(ii) It is symmetric i.e. (a, b) ∈ R ⇒ (b, a) ∈ R, for all a, b ∈ A
(iii) It is transitive i.e. (a, b) ∈ R and (b, c) ∈ R ⇒ (a, c) ∈ R for all a, b, c ∈ A.
Note : Congruence modulo (m) : Let m be an arbitrary but fixed integer. Two integers a and b are said to
be congruence modulo m if a − b is divisible by m and we write a ≡ b (mod m).
Thus a ≡ b (mod m) ⇔ a − b is divisible by m. For example, 18 ≡ 3 (mod 5) because 18 – 3 = 15
which is divisible by 5. Similarly, 3 ≡ 13 (mod 2) because 3 – 13 = –10 which is divisible by 2. But 25
≠ 2 (mod 4) because 4 is not a divisor of 25 – 3 = 22.
The relation “Congruence modulo m” is an equivalence relation.
Important Tips
• If R and S are two equivalence relations on a set A , then R ∩ S is also an equivalence relation on A.
• The union of two equivalence relations on a set is not necessarily an equivalence relation on the set.
• The inverse of an equivalence relation is an equivalence relation.
Equivalence Classes of an Equivalence Relation.
Let R be equivalence relation in A(≠ φ) . Let a ∈ A . Then the equivalence class of a, denoted by [a] or {a} is
defined as the set of all those points of A which are related to a under the relation R. Thus [a] = {x ∈ A : x R a}.
It is easy to see that
(1) b ∈ [a] ⇒ a ∈ [b] (2) b ∈ [a] ⇒ [a] = [b] (3) Two equivalence classes are either disjoint or identical.
As an example we consider a very important equivalence relation x ≡ y(mod n) iff n divides (x − y),n is a fixed
positive integer. Consider n = 5. Then
[0] = {x : x ≡ 0 (mod 5)} = {5p : p ∈ Z} = { 0,± 5,± 10,± 15,.....}
[1] = {x : x ≡ 1(mod 5)} = {x : x − 1 = 5k,k ∈ Z} = {5k + 1 : k ∈ Z} = {1,6,11,......., − 4,− 9,.....} .
One can easily see that there are only 5 distinct equivalence classes viz. [0], [1], [2], [3] and [4], when n = 5.
Example: 7 Given the relation R = {(1, 2), (2, 3)} on the set A = {1, 2, 3}, the minimum number of ordered pairs which when added
Solution: (c) to R make it an equivalence relation is
Example: 8 (a) 5 (b) 6 (c) 7 (d) 8
Solution: (a)
Example: 9 R is reflexive if it contains (1, 1), (2, 2), (3, 3)
Solution: (a)
(1, 2) ∈ R,(2, 3)∈ R
∴ R is symmetric if (2, 1), (3, 2) ∈ R. Now, R = {(1, 1), (2, 2), (3, 3), (2, 1), (3, 2), (2, 3), (1, 2)}
R will be transitive if (3, 1); (1, 3) ∈ R. Thus, R becomes an equivalence relation by adding (1, 1) (2, 2) (3, 3) (2, 1) (3,2)
(1, 3) (3, 1). Hence, the total number of ordered pairs is 7.
The relation R = {(1, 1), (2, 2), (3, 3), (1, 2), (2, 3), (1, 3)} on set A = {1, 2, 3} is
(a) Reflexive but not symmetric (b) Reflexive but not transitive
(c) Symmetric and Transitive (d) Neither symmetric nor transitive
Since (1, 1); (2, 2); (3, 3) ∈ R therefore R is reflexive. (1, 2) ∈ R but (2, 1) ∉ R, therefore R is not symmetric. It can be
easily seen that R is transitive.
Let R be the relation on the set R of all real numbers defined by a R b iff |a − b|≤ 1 . Then R is [Roorkee 1998]
(a) Reflexive and Symmetric (b) Symmetric only (c) Transitive only (d) Anti-symmetric only
|a − a|= 0 < 1 ∴a R a ∀ a ∈ R
∴ R is reflexive, Again a R b ⇒ |a − b|≤ 1 ⇒|b − a|≤ 1 ⇒ bRa
∴ R is symmetric, Again 1R 1 and 1 R1 but 1 ≠ 1
2 2 2
∴ R is not anti-symmetric
Further, 1 R 2 and 2 R 3 but 1 R 3
[|1 − 3|= 2 > 1 ]
∴ R is not transitive.
Example: 10 The relation "less than" in the set of natural numbers is [UPSEAT 1994, 98; AMU 1999]
Solution: (b)
(a) Only symmetric (b) Only transitive (c) Only reflexive (d) Equivalence relation
Example: 11
Solution: (d) Since x < y, y < z ⇒ x < z ∨ x, y, z ∈ N
Example: 12 ∴ x R y, yR z ⇒ x R z , ∴ Relation is transitive , ∴ x < y does not give y < x , ∴ Relation is not symmetric.
Solution: (c)
Since x < x does not hold, hence relation is not reflexive.
Example: 13
Solution: (b) With reference to a universal set, the inclusion of a subset in another, is relation, which is [Karnataka CET 1995]
Example: 14
(a) Symmetric only (b) Equivalence relation (c) Reflexive only (d) None of these
Solution: (d)
Example: 15 Since A ⊆ A ∴ relation ' ⊆' is reflexive.
Solution: (b)
Example: 16 Since A ⊆ B, B ⊆ C ⇒ A ⊆ C
Solution: (d) ∴ relation ' ⊆' is transitive.
But A ⊆ B, ⇒ B ⊆ A , ∴ Relation is not symmetric.
Let A = {2, 4,6,8} . A relation R on A is defined by R = {(2, 4),(4, 2),(4,6),(6, 4)} . Then R is [Karnataka CET 1995]
(a) Anti-symmetric (b) Reflexive (c) Symmetric (d) Transitive
Given A = {2, 4, 6, 8}
R = {(2, 4)(4, 2) (4, 6) (6, 4)}
(a, b) ∈ R ⇒ (b, a) ∈ R and also R−1 = R . Hence R is symmetric.
Let P = {(x, y)| x2 + y2 = 1, x, y ∈ R} . Then P is
(a) Reflexive (b) Symmetric (c) Transitive (d) Anti-symmetric
Obviously, the relation is not reflexive and transitive but it is symmetric, because x2 + y2 = 1 ⇒ y2 + x2 = 1 .
Let R be a relation on the set N of natural numbers defined by nRm ⇔ n is a factor of m (i.e., n|m). Then R is
(a) Reflexive and symmetric (b) Transitive and symmetric
(c) Equivalence (d) Reflexive, transitive but not symmetric
Since n | n for all n ∈ N , therefore R is reflexive. Since 2 | 6 but 6 | 2 , therefore R is not symmetric.
Let n R m and m R p ⇒ n|m and m|p ⇒ n|p ⇒ nRp. So R is transitive.
Let R be an equivalence relation on a finite set A having n elements. Then the number of ordered pairs in R is
(a) Less than n (b) Greater than or equal to n (c) Less than or equal to n (d) None of these
Since R is an equivalence relation on set A, therefore (a, a) ∈ R for all a ∈ A . Hence, R has at least n ordered pairs.
Let N denote the set of all natural numbers and R be the relation on N × N defined by (a, b) R (c, d) if
ad(b + c) = bc(a + d), then R is [Roorkee 1995]
(a) Symmetric only (b) Reflexive only (c) Transitive only (d) An equivalence relation
For (a, b), (c, d) ∈ N × N
(a,b)R(c, d) ⇒ ad(b + c) = bc(a + d)
Reflexive: Since ab(b + a) = ba(a + b)∀ab ∈ N ,
∴ (a,b)R(a,b) , ∴ R is reflexive.
Symmetric: For (a,b),(c, d)∈ N × N , let (a,b)R(c, d)
∴ ad(b + c) = bc(a + d) ⇒ bc(a + d) = ad(b + c) ⇒ cb(d + a) = da(c + b) ⇒ (c, d)R(a,b)
∴ R is symmetric
Transitive: For (a,b),(c, d),(e, f)∈ N × N, Let (a,b)R(c, d),(c, d)R(e, f)
∴ ad(b + c) = bc(a + d) , cf(d + e) = de(c + f)
⇒ adb + adc = bca + bcd .....(i) and cfd + cfe = dec + def .......(ii)
(i) × ef + (ii) × ab gives, adbef + adcef + cfdab + cfeab = bcaef + bcdef + decab + defab
⇒ adcf(b + e) = bcde(a + f) ⇒ af(b + e) = be(a + f) ⇒ (a,b)R(e, f) . ∴ R is transitive. Hence R is an equivalence relation.
Example: 17 For real numbers x and y, we write x Ry ⇔ x − y + 2 is an irrational number. Then the relation R is
Solution: (a)
(a) Reflexive (b) Symmetric (c) Transitive (d) None of these
Example: 18
Solution: (b) For any x ∈ R, we have x − x + 2 = 2 an irrational number.
Example: 19 ⇒ xRx for all x. So, R is reflexive.
Solution: (a) R is not symmetric, because 2R1 but 1 R 2 , R is not transitive also because 2 R 1 and 1R2 2 but 2 R 2 2 .
Example: 20 Let X be a family of sets and R be a relation on X defined by ‘A is disjoint from B’. Then R is
Solution: (c)
(a) Reflexive (b) Symmetric (c) Anti-symmetric (d) Transitive
Clearly, the relation is symmetric but it is neither reflexive nor transitive.
Let R and S be two non-void relations on a set A. Which of the following statements is false
(a) R and S are transitive ⇒ R ∪ S is transitive (b) R and S are transitive ⇒ R ∩ S is transitive
(c) R and S are symmetric ⇒ R ∪ S is symmetric (d) R and S are reflexive ⇒ R ∩ S is reflexive
Let A = {1, 2, 3} and R = {(1, 1), (1, 2)}, S = {(2, 2) (2, 3)} be transitive relations on A.
Then R ∪ S = {(1, 1); (1, 2); (2, 2); (2, 3)}
Obviously, R ∪ S is not transitive. Since (1, 2) ∈ R ∪ S and (2, 3)∈ R ∪ S but (1, 3) ∉ R ∪ S .
The solution set of 8x ≡ 6(mod 14), x ∈ Z , are
(a) [8] ∪ [6] (b) [8] ∪ [14] (c) [6] ∪ [13] (d) [8] ∪ [6] ∪ [13]
8x − 6 = 14P(P ∈ Z) ⇒ x = 1 [14 P + 6], x ∈ Z
8
⇒x = 1 (7P + 3) ⇒ x = 6, 13, 20, 27, 34, 41, 48,.......
4
∴ Solution set = {6, 20, 34, 48,.....} ∪ {13, 27, 41, ......} = [6] ∪ [13].
Where [6], [13] are equivalence classes of 6 and 13 respectively.
Composition of Relations.
Let R and S be two relations from sets A to B and B to C respectively. Then we can define a relation SoR from
A to C such that (a, c) ∈ SoR ⇔ ∃ b ∈ B such that (a, b) ∈ R and (b, c) ∈ S.
This relation is called the composition of R and S.
For example, if A = {1, 2, 3}, B = {a, b, c, d}, C = {p, q, r, s} be three sets such that R = {(1, a), (2, c), (1, c),
(2, d)} is a relation from A to B and S = {(a, s), (b, q), (c, r)} is a relation from B to C. Then SoR is a relation from
A to C given by SoR = {(1, s) (2, r) (1, r)}
In this case RoS does not exist.
In general RoS ≠ SoR. Also (SoR)–1 = R–1oS–1.
Example: 21 If R is a relation from a set A to a set B and S is a relation from B to a set C, then the relation SoR
Solution: (a)
Example: 22 (a) Is from A to C (b) Is from C to A (c) Does not exist (d) None of these
Solution: (b)
Example: 23 It is obvious.
Solution: (c) If R ⊂ A × B and S ⊂ B × C be two relations, then (SoR)−1 =
Example: 24 (a) S−1oR−1 (b) R−1oS−1 (c) SoR (d) RoS
Solution: (a) It is obvious.
If R be a relation < from A = {1,2, 3, 4} to B = {1, 3, 5} i.e., (a, b) ∈ R ⇔ a < b, then RoR−1 is
(a) {(1, 3), (1, 5), (2, 3), (2, 5), (3, 5), (4, 5)}
(b) {(3, 1) (5, 1), (3, 2), (5, 2), (5, 3), (5, 4)}
(c) {(3, 3), (3, 5), (5, 3), (5, 5)}
(d) {(3, 3) (3, 4), (4, 5)}
We have, R = {(1, 3); (1, 5); (2, 3); (2, 5); (3, 5); (4, 5)}
R−1 = {(3, 1), (5, 1), (3, 2), (5, 2); (5, 3); (5, 4)}
Hence RoR−1 = {(3, 3); (3, 5); (5, 3); (5, 5)}
Let a relation R be defined by R = {(4, 5); (1, 4); (4, 6); (7, 6); (3, 7)} then R−1oR is
(a) {(1, 1), (4, 4), (4, 7), (7, 4), (7, 7), (3, 3)} (b) {(1, 1), (4, 4), (7, 7), (3, 3)}
(c) {(1, 5), (1, 6), (3, 6)} (d) None of these
We first find R−1, we have R−1 = {(5, 4);(4,1);(6, 4);(6,7);(7, 3)} we now obtain the elements of R −1oR we first pick the
element of R and then of R−1 . Since (4, 5) ∈ R and (5, 4) ∈ R−1 , we have (4, 4)∈ R−1oR
Similarly, (1, 4)∈ R,(4,1)∈ R−1 ⇒(1,1)∈ R−1oR
(4, 6)∈ R,(6, 4)∈ R−1 ⇒(4, 4)∈ R−1oR, (4, 6)∈ R,(6,7)∈ R−1 ⇒ (4,7)∈ R−1oR
(7, 6)∈ R,(6, 4)∈ R−1 ⇒ (7, 4)∈ R−1oR, (7, 6) ∈ R,(6,7) ∈ R−1 ⇒ (7,7) ∈ R−1oR
(3,7)∈ R,(7, 3)∈ R−1 ⇒ (3, 3)∈ R−1oR,
Hence R−1oR = {(1, 1); (4, 4); (4, 7); (7, 4), (7, 7); (3, 3)}.
Axiomatic Definitions of the Set of Natural Numbers (Peano's Axioms).
The set N of natural numbers (N = {1, 2, 3, 4......}) is a set satisfying the following axioms (known as peano's
axioms)
(1) N is not empty.
(2) There exist an injective (one-one) map S : N → N given by S(n) = n+ , where n+ is the immediate
successor of n in N i.e., n + 1 = n+ .
(3) The successor mapping S is not surjective (onto).
(4) If M ⊆ N such that,
(i) M contains an element which is not the successor of any element in N, and
(ii) m ∈ M ⇒ m+ ∈ M , then M = N
This is called the axiom of induction. We denote the unique element which is not the successor of any element is
1. Also, we get 1+ = 2, 2+ = 3 .
Note : Addition in N is defined as,
n + 1 = n+
n + m+ = (n + m)+
Multiplication in N is defined by,
n.1 = n
n.m+ = n.m + n
Trigonometrical equations, properties of triangles and heights & distances
Relation between Sides and Angles.
A triangle has six components, three sides and three angles. The three angles of a ∆ABC are denoted by letters
A, B, C and the sides opposite to these angles by letters a, b and c respectively. Following are some well known
relations for a triangle (say ∆ABC ) a A
• A + B + C = 180o (or π)
• a+b>c, b+c >a, c +a>b c bC B
• |a − b|< c,|b − c |< a,|c − a|< b
Generally, the relations involving the sides and angles of a triangle are cyclic in nature, e.g. to obtain the
second similar relation to a + b > c , we simply replace a by b, b by c and c by a. So, to write all the relations, follow
the cycles given.
(1) The law of sines or sine rule : The sides of a triangle are proportional to the sines of the angles opposite
to them i.e. , a = b = c = k (say) A
sinA sinB sinC A
More generally, if R be the radius of the circumcircle of the triangle FE c b
C
ABC, a = b = c = 2R B O C
sinA sinB sinC B
D
Note : The above rule may also be expressed as a
sin A sin B sin C
a =b= c
The sine rule is very useful tool to express sides of a triangle in terms of sines of angle and vice-versa in the
following manner a = b = c = K (Let) ⇒ a= K sin A, b = K sin B, c = K sin C .
sin A sin B sin C
Similarly, sin A = sin B = sin C =λ (Let) ⇒ sin A = λa, sin B = λb, sin C = λc.
a b c
Example: 1 If the angles of a triangle are in the ratio 4 : 1 : 1, then the ratio of the longest side to the perimeter is [IIT Screening 2003]
Solution: (a)
(a) 3 : (2 + 3) (b) 1 : 6 (c) 1 : (2 + 3) (d) 2 : 3
Example: 2
4 x + x + x = 180 ⇒ 6x = 180 ⇒ x = 30o
sin 120o sin 30o sin 30o
a =b=c
∴ a : (a + b + c) = (sin 120o ) : (sin 120o + sin 30o + sin 30o ) = 3 : 3+ 2 = 3: 3 + 2.
2 2
In a triangle ABC, ∠B = π and ∠C = π and D divides BC internally in the ratio 1 : 3. Then sin ∠BAD is equal to
3 4 sin ∠CAD
[UPSEAT 2003, 2001; IIT 1995]
(a) 1 (b) 1 (c) 1 (d) 2
3 3 6 3
Solution: (c) Let ∠BAD = α, ∠CAD = β A
αβ
Example: 3 In ∆ADB , applying sine formulae, we get x = AD ........(i)
Solution: (c) sinα sin π ..........(ii) D
Example: 4 3
Solution: (c)
In ∆ADC, applying sine formulae, we get, 3x = AD
sin β
sin π
4 π/3 π/4
Bx 3x
Dividing (i) by (ii), we get, C
⇒ x × sin β = AD × sin π ⇒ sin β = 1 2 ⇒ sin β =3 2 = 6
sinα 3x 4 3 sinα 3 sinα 3
2=
sin π AD 3
3 2
∴ sin ∠BAD = sin α = 1.
sin ∠CAD sin β 6
In a ∆ABC, A : B : C = 3 : 5 : 4 . Then [a + b + c 2] is equal to [DCE 2001]
(a) 2b (b) 2c (c) 3b (d) 3a
A : B : C = 3 : 5 : 4 ⇒ A + B + C = 12x = 180o ⇒ x = 15o
∴ A = 45o , B = 75o , C = 60o
a = b = c =k (say)
sin 45o sin 75o sin 60o
i.e., a = 1 K, b = 3 + 1 K, c= 3 K . Hence [a + b + c 2] = 3b .
2 22 2
In any triangle ABC if 2 cos B = a , then the triangle is
c
(a) Right angled (b) Equilateral (c) Isosceles (d) None of these
2 cos B = a = k sin A = sin A ⇒ 2 cos B sin C = sin A ⇒ sin(B + C) − sin(B − C) = sin A
c k sin C sin C
⇒ sin(180o − A) − sin(B − C) = sinA ⇒ sin A − sin(B − C) = sin A ⇒ sin(B − C) = 0 ⇒ B − C = 0 ⇒ B = C
∴ Triangle is isosceles.
The Law of Cosines or Cosine Rule.
In any triangle ABC, the square of any side is equal to the sum of the squares of the other two sides diminished
by twice the product of these sides and the cosine of their included angle, that is for a triangle ABC,
(1) a2 = b2 + c2 − 2bc cos A ⇒ cos A = b2 + c2 − a2
2bc
(2) b2 = c2 + a2 − 2ca cos B ⇒ cos B = c2 + a2 − b2
2ca
(3) c2 = a2 + b2 − 2ab cos C ⇒ cos C = a2 + b2 − c2
2ab
Combining with sin A = a , sin B = b , sin C = c
2R 2R 2R
We have by division, tan A = R(b2 abc − a2), tan B = R(c 2 abc − b2) , tan C = abc
+ c2 + a2 R(a 2 + b 2 − c 2 )
Where R, be the radius of the circum-circle of the triangle ABC.
Example: 5 The smallest angle of the ∆ABC, when a = 7, b = 4 3 and c = 13, is [MP PET 2003]
Solution: (a)
Example: 6 (a) 30o (b) 15o (c) 45o (d) None of these
Solution: (b)
Smallest angle is opposite to smaller side
Example: 7
Solution: (b) ∴ cos C = b2 + a2 − c2 = 49 + 48 − 13 = 3 = 3 ⇒ ∠C = 30o .
Example: 8 2ab 2×7×4 3 23 2
Solution: (b) In a ∆ABC, if b+c = c+a = a+b , then cos C = [Karnataka CET 2003]
11 12 13
Example: 9
Solution: (a) (a) 7 (b) 5 (c) 17 (d) 16
5 7 36 17
b+c = c+a = a+b = λ (Let)
11 12 13
∴ b + c = 11λ ......(i)
.......(iii)
c + a = 12λ .......(ii) and a + b = 13λ .......(iv)
From (i) + (ii) + (iii), 2(a + b + c) = 36λ ∴ a + b + c = 18λ
Now subtract (i), (ii) and (iii) from (iv), a = 7λ, b = 6λ, c = 5λ .
Now cos C = a2 + b2 − c2 = (7λ)2 + (6λ)2 − (5λ)2 = 49λ2 + 36λ2 − 25λ2 = 60λ2 = 5 .
2ab 2.7λ.6λ 84λ2 84λ2 7
In a ∆ABC, 2ac sin A − B + C is equal to [IIT Screening 2000]
2
(a) a2 + b2 − c2 (b) c2 + a2 − b2 (c) b2 − c2 − a2 (d) c2 − a2 − b2
2ac sin A−B+C = 2ac sin π − 2B = 2ac cos B = 2ac c 2 + a2 − b2 = c2 + a2 − b2 .
2 2 2ca
In ambiguous case if a, b and A are given and if there are two possible values of third side, are c1 and c 2 , then
[UPSEAT 1999]
(a) c1 − c2 = 2 (a2 + b2 sin2 A) (b) c1 − c2 = 2 (a2 − b2 sin2 A)
(c) c1 − c2 = 4 (a2 + b2 sin2 A) (d) c1 − c2 = 3 (a2 − b2 sin2 A)
cos A = b2 + c2 − a2 or c2 − (2b cos A)c + (b2 − a2) = 0
2bc
Which is quadratic equation in c. Let there be two roots, c1 and c 2 of above quadratic equation then c1 + c2 = 2b cos A
and c1c2 = b2 − a2
∴ c1 − c2 = [(c1 + c2)2 − 4c1c2] = [(2b cos A)2 − 4(b2 − a2) = [4a2 − 4b2(1 − cos2 A)] = 2 (a2 − b2 sin2 A) .
In a ∆ABC, if sin A = sin(A − B) , then a2, b2, c2 are in [Karnataka CET 1999]
sin C sin(B − C)
(a) A.P. (b) G.P. (c) H.P. (d) None of these
sin A = sin A cos B − cos A sin B ⇒ a = a cos B − b cos A (Using sine formula)
sin C sin B cos C − cos B sin C c b cos C − c cos B
⇒ ab cos C − ac cos B = ac cos B − bc cos A ⇒ ab cos C + bc cos A = 2ac cos B
⇒ a2 + b2 − c2 + b2 + c2 − a2 = c2 + a2 − b2 ⇒ b2 = c2 + a2 − b2 ⇒ b2 = c2 + a2 ⇒ a2, b2, c2 are in A.P.
2 2 1 2
Example: 10 In a triangle ABC, AD is altitude from A. Given b > c, ∠C = 23o and AD = abc , then ∠B equal to [IIT 1994]
Solution: (c) b2 − c2
(a) 67o (b) 44 o (c) 113o (d) None of these
cos B = a2 + c2 − b2 = a2 − (b2 − c2)
2ac 2ac
Now, AD abc ; cos B = a2 − abc
b2 − c2 AD
= ∴ 2ac
Also AD = b sin 23o ; a − c
sin 23o
∴ cos B = 2c
sin(B + 23o ) − 1
sin 23o
By sine formulae ⇒ a = sin(B + 23o) ; ∴ cos B = sin 23o
c sin 23o
2
⇒ sin(23o − B) = −1 = sin(−90o) ; therefore 23o − B = −90o or B = 113o.
Projection Formulae. A
In any triangle ABC, b cos C + c cos B = k sin B cos C + k sin C cos B (from sine rule)
= k[sin(B + C)] = k sin(π − A) = k sin A = a
Similarly, we can deduct other projection formulae from sine rule. cb
(i) a = b cos c + c cos B (ii) b = c cos A + a cos C (iii) c = a cos B + b cos A
i.e., any side of a triangle is equal to the sum of the projection of other two sides on it. B aC
Example: 11 In a ∆ABC, cos C + cos A + cos B is equal to [EAMCET 2001]
c + a b
Solution: (b)
(a) 1 (b) 1 (c) 1 (d) c+a
Example: 12 a b c b
Solution: (c) cos C + cos A + cos B = (b cos C + b cos A) + (c cos B + a cos B)
Example: 13 c + a b b(c + a)
Solution: (d)
= (b cos C + c cos B) + (b cos A + a cos B) = a+c (Using projection formulae) = 1 .
b(c + a) b(c + a) b
If k be the perimeter of the ∆ABC , then b cos2 C + c cos2 B is equal to
2 2
(a) k (b) 2k (c) k (d) None of these
2
b cos2 C +c cos2 B = b (1 + cos C) + c (1 + cos B) = b + c + 1 (b cos C + c cos B) = a+b+c = k .
2 2 2 2 2 2 2 2 2
(a + b + c)(cos A + cos B + cos C) =
(a) ∑ a sin2 A (b) ∑ a cos2 A (c) 2 ∑ a sin2 A (d) 2 ∑ a cos2 A
2 2 2 2
(a + b + c)(cos A + cos B + cos C) = 9 terms.
= ∑ a cos A+ ∑(b cos C + c cos B) = a cos A + b cos B + c cos C + (a + b + c) = ∑ a(1 + cos A) = 2 ∑ a cos2 A .
2
(3) (6)
Theorem of the Medians: (Apollonius Theorem) c A C
In every triangle the sum of the squares of any two sides is equal to twice Bh
b
the square on half the third side together with twice the square on the median m
that bisects the third side. αβ
Dh
For any triangle ABC, b2 + c2 = 2(h2 + m2) = 2{m2 + (a / 2)2} by use of
cosine rule.
If ∆ be right angled, the mid point of hypotenuse is equidistant from the
three vertices so that DA = DB = DC
∴b2 + c2 = a2 which is pythagoras theorem. This theorem is very useful for solving problems of height and
distance.
Example: 14 If AD, BE and CF are the medians of a ∆ABC then (AD2 + BE2 + CF 2) : (BC2 + CA2 + AB2) is equal to
(a) 4 : 2 (b) 3 : 2 (c) 3 : 4 (d) 2 : 3
Solution: (c) We have, AB2 + AC2 = 2(AD2 + BD2) ⇒ c2 + b2 a2 = AD2 ......(i)
Example: 15 2 −4
Solution: (a)
a2 + c2 b2 = BE2 ......(ii) and a2 + b2 − c2 = CF 2 .........(iii)
2 −4 2 4
a2 + b2 + c2 − a2 + b2 + c2 = AD2 + BE2 + CF 2
4
Adding (i), (ii) and (iii) we get, (AD2 + BE2 + CF 2):(a2 + b2 + c2) = 3 : 4 .
AD is a median of the ∆ABC, if AE and AF are medians of the triangles ABD and ADC respectively and AD = m1,
AE = m2 , AF = m3 , then a2 is equal to
8
(a) m22 + m32 − 2m12 (b) m12 + m22 − 2m32 (c) m22 + m32 − m12 (d) None of these
In ∆ABC, AD2 = m12 = c2 + b2 − a2
2 4
a 2 A
2
In ∆ABD, AE 2 = m22 = c2 + AD2 −
2
4
In ∆ADC, AF 2 = m32 = AD2 + b2 − a 2
2 2
4
∴ m22 + m32 = AD2 + b2 + c2 − a2 = m12 + m12 + a2 − a2 BE DF C
2 8 4 8
m22 + m32 = 2m12 + a2 ⇒ a2 = m22 + m32 − 2m12 .
8 8
Napier's Analogy (Law of Tangents)
For any triangle ABC,
(1) tan A− B = a − b cot C (2) tan B − C = b − c cot A (3) tan C − A = c − a cot B
2 a + b 2 2 b + c 2 2 c + a 2
Note : Mollweide's formula: For any triangle, a+b cos 1 ( A − B) , a − b sin 1 (A − B) .
c 2 c 2 C
= 1 = 1
2 2
sin C cos
Example: 16 If tan B−C =x cot A , then x equal to [MP PET 1992, 2002]
Solution: (c) 2 2
Example: 17 (d) None of these
Solution:(b,c) (a) c−a (b) a−b (c) b−c
c+a a+b b+c [Roorkee 1997]
Example: 18
Solution:(b,c) We know, tan B −C = b − c cot A ⇒ x = b − c . (d) None of these
Example: 19 2 b + c 2 b + c
Solution: (b) (d) C = 60o − α
If in a ∆ABC a = 6, b =3 and cos(A − B) = 4 , then
5 (d) None of these
(a) C = π (b) A = sin−1 2 (c) ar(∆ABC) = 9
4 5
tan A − B = a − b cot C = 1 cot C
2 a + b 2 3 2
∴ 4 = cos(A − B) = 1− tan2 A−B = 1− 1 cot 2 C
5 1+ tan2 2 1+ 9 cot 2 2
1 C
A−B 9 2
2
∴ tan2 C =1 ⇒ C= π ∴ ar (∆ABC) = 1 ab = 1 .6.3 = 9
2 2 2 2
Also, sin A = 6 = 2 .
32 + 62 5
In a ∆ABC, A = π and b : c = 2 : 3. If tanα = 3 ,0 <α < π , then
3 5 2
(a) B = 60o + α (b) C = 60o + α (c) B = 60o − α
tan C − B = c − b cot A ⇒ tan C − B = 1 cot 30o = 3 = tan α
2 c + b 2 2 5 5
∴ C − B = 2α and C + B = 180o − 60o = 120o i.e., B = 60o − α , C = 60o + α .
In a ∆ABC, a = 2b and |A− B |= π . The measure of ∠C is
3
(a) π (b) π (c) π
4 3 6
Clearly, A > B ( a > b)
Now tan A − B = a − b cot C ⇒ tan 30o = 1 cot C
2 a + b 2 3 2
∴ 3 = cot C or C = π ⇒ C = π .
2 2 6 3
Area of Triangle.
Let three angles of ∆ABC are denoted by A, B, C and the sides opposite to these angles by letters a, b, c
respectively.
(1) When two sides and the included angle be given: The area of triangle ABC is given by,
∆ = 1 bc sin A = 1 ca sin B = 1 ab sin C i.e., ∆ = 1 (Product of two sides) × sine of included angle
2 2 2 2
(2) When three sides are given: Area of ∆ABC = ∆ = s(s − a)(s − b)(s − c)
where semiperimeter of triangle s = a+b+c A
2
(3) When three sides and the circum-radius be given: Area of triangle ∆ = abc , c b
4R
where R be the circum-radius of the triangle. B aC
(4) When two angles and included side be given :
∆ = 1 a2 sin B sin C = 1 b2 sin A sin C = 1 c2 sin A sin B
2 sin(B + C) 2 sin(A + C) 2 sin(A + B)
Example: 20 In a ∆ABC if a = 2x, b = 2y and ∠C = 120o , then the area of the triangle is [MP PET 1986, 2002]
Solution: (b)
Example: 21 (a) xy (b) xy 3 (c) 3xy (d) 2xy
Solution: (d)
Example: 22 ∆ = 1 ab sin C = 1 .2x.2y sin 120o = 3xy .
Solution: (b) 2 2
Example: 23 In a ∆ABC, cos A cos B cos C and a = 2, then the area of a triangle is [MP PET 2000; IIT 1993]
Solution: (c) a =b= c
(a) 1 (b) 2 (c) 3 (d) 3
2
By sine rule, tan A = tan B = tan C ; ∴ Triangle is equilateral .
Hence, ∆= 1 .a.a. sin 60o = 1 .2.2. 3 3.
2 2 2=
In a triangle ABC, a, b, A are given and c1, c2 are two values of third side c. The sum of the areas of triangles with sides
a, b, c1 and a, b, c2 is
(a) 1 a2 sin 2A (b) 1 b 2 sin 2 A (c) b2 sin 2A (d) a2 sin 2A
2 2
Let the triangles be ∆1 = ABC1 and ∆2 = ABC2 A, b, a are given and c has two values. Hence we apply cosine formulae
cos A = b2 + c2 − a2 or c2 − 2bc cos A + b2 − a2 = 0 .
2bc
Above is quadratic in c If c1, c2 be the two values of c, then c1 + c2 = 2b cos A, c1c2 = b2 − a2
∆1 = 1 ab sin C1 , ∆2 = 1 ab sin C2
2 2
∴ ∆1 + ∆2 = 1 ab(sin C1 + sin C2) = 1 abk(2b cos A) = b2ak cos A = b2 sin A cos A = 1 b2 sin 2A .
2 2 2
If ∆ stands for the area of a triangle ABC, then a2 sin 2B + b2 sin 2A = [WB JEE 1988]
(a) 3∆ (b) 2∆ (c) 4∆ (d) −4∆
Use sine rule, ∆ = 1 ab sin C
2
L.H.S. = k2(sin2 A.2 sin B cos B + sin2 B.2 sin A cos A) = k2[2 sin A. sin B. sin(A + B)] = 2ab sin C = 4∆ .
Example: 24 In a ∆ABC, A = 2π ,b − c = 3 3 and ar (∆ABC) = 93 cm2. Then a is
Solution: (b) 3 2
Example: 25
Solution: (b) (a) 6 3cm. (b) 9cm. (c) 18 cm. (d) None of these
1 bc sin 2π 93 or 1 . 3 .bc = 9 3 ⇒ bc =18
2 3 =2 2 2 2
Also, cos 2π = b2 + c2 − a2 ⇒ 1 = (b − c)2 + 2bc − a2 or (b − c)2 + 3bc − a2 = 0 or 27 + 54 = a2 ⇒ a = 9 .
3 2bc −2 2bc
If p1, p2, p3 are altitudes of a triangle ABC from the vertices A, B, C and ∆ , the area of the triangle, then p1−2 + p2−2 + p3−2
is equal to
(a) a + b + c (b) a2 + b2 + c2 (c) a2 + b2 + c2 (d) None of these
∆ 4∆2 ∆2
We have 1 ap1 = ∆, 1 bp2 = ∆, 1 cp3 = ∆ ⇒ p1 = 2∆ , p2 = 2∆ , p3 = 2∆
2 2 2 a b c
∴ 1 + 1 + 1 = a2 + b2 + c2 .
p12 p22 p32 4∆2
Half Angle Formulae.
If 2s shows the perimeter of a triangle ABC then, i.e., 2s = a + b + c, then
(1) Formulae for sin A , sin B , sin C :
2 2 2
(i) sin A = (s − b)(s − c) (ii) sin B = (s − a)(s − c) (iii) sin C = (s − a)(s − b)
2 bc 2 ca 2 ab
(2) Formulae for cos A , cos B , cos C :
2 2 2
(i) cos A = s(s − a) (ii) cos B = s(s − b) (iii) cos C = s(s − c)
2 bc 2 ca 2 ab
(3) Formulae for tan A , tan B , tan C :
2 2 2
(i) tan A = (s − b)(s − c) (ii) tan B = (s − c)(s − a) (iii) tan C = (s − a)(s − b)
2 s(s − a) 2 s(s − b) 2 s(s − c)
Note : sin A = 2 sin A cos A =2 (s − b)(s − c) . s(s − a) = 2 s(s − a)(s − b)(s − c) = 2∆
2 2 bc bc bc bc
Similarly sin B = 2∆ , sin C = 2∆
ca ab
tan A = (s − b)(s − c) , tan B = (s − c)(s − a) , tan C = (s − a)(s − b)
2 ∆ 2 ∆ 2 ∆
Important Tips
• tan A tan B = (s − b)(s − c) . (s − c)(s − a)1/ 2 = s −c ⇒ cot A cot B = s
2 2 s(s − a) s 2 2 s−c
s(s − b)
• tan A + tan B = s − c s −b + s − a = s − c s−b+s − a = c s(s − c) 1/ 2 = c cot C
2 2 s s −a s − b s (s − a)(s − b) s (s − a)(s − s 2
b)
Another form: c.(s − c) = c (s − c)
{s(s − a)(s − b)(s − c)}1/ 2 s
• tan A − tan B = a −b (s − c)
2 2 s
cot A + cot B tan A + tan B c cot C .
2 2 2 2 −c 2
• = A B = s
2 2
tan tan
Example: 26 If in any ∆ABC ; cot A , cot B , cot C are in A.P., then [MP PET 2003]
Solution: (b) 2 2 2
Example: 27
Solution: (b) (a) cot A cot B = 4 (b) cot A cot C = 3 (c) cot B cot C =1 (d) cot B tan C = 0
Example: 28 2 2 2 2 2 2 2 2
Solution: (d)
Example: 29 Trick: Take A = B = C = 60o, then cot A , cot B and cot C are in A.P. with common difference zero. Now option (b)
Solution: (b) 2 2 2
Example: 30 satisfies.
In a ∆ABC , if 3a = b + c ,then the value of cot B cot C is [Roorkee 1986; MP PET 1990, 97, 98]
2 2
(a) 1 (b) 2 (c) 3 (d) 2
cot B . cot C = (s s(s − b) c) . (s s(s − c) = s s
2 2 − a)(s − − a)(s − b) −a
Given 3a = b + c ⇒ a + b + c = 4a ⇒ ∴ cot B .cot C = 2a = 2.
2 2 a
In a triangle ABC, tan A = 5 and tan C = 2 , then [EAMCET 1994]
2 6 2 5
[Roorkee 1993]
(a) a, b, c are in A.P. (b) cos A, cos B, cos C are in A.P.
(c) sin A, sin B, sin C are in A.P. (d) Both (a) and (c)
tan A . tan C = (s − b) ⇒ 5 . 2 = s − b ⇒ 2s = 3b ⇒ a + b + c = 3b ⇒ a + c = 2b .
2 2 s 6 5 s
If the sides of triangle a, b, c be in A.P. then tan A + tan C equal to
2 2
(a) 2 cot A (b) 2 cot B (c) 2 cot C (d) None of these
3 2 3 2 3 2
tan A + tan C = (s − b)(s − c) + (s − a)(s − b) = b cot B = 2b cot B
2 2 s(s − a) s(s − c) s 2 2s 2
a, b, c in A.P.
∴ a + c = 2b ⇒ 2s = 3b = 2b cot B
3b 2
Hence, tan A + tan C = 2 cot B .
2 2 3 2
In ∆ABC , cot A + cot B a sin 2 B + b sin2 A equal to [Roorkee 1988]
2 2 2 2
(a) cot C (b) c cot C (c) cot C (d) c cot C
2 2
Solution: (d) cot A B a sin 2 B + b sin 2 A cos C a sin 2 B 2 A cos C sin B sin A
2 2 2 2 2 a sin 2 sin 2
+ cot = A B + b sin = A + b B
2 2 2 2 2 2 2
sin sin
(s − a)(s − c) (s − b)(s − c)
a (s − bc − c)
= s(s − c) ac +b a)(s = s(s − c) s − a ab + s − b ab
ab (s − b)(s − c) s − b s − a
ab
bc ac
= s(s − c) s−a+s − b = s(s − c) 2s − a − b = c s(s − c) = c cot C .
(s − a)(s − b) − a)(s − 2
(s − a)(s − b) (s b)
Trick : Such type of unconditional problems can be checked by putting the particular values for a = 1 , b = 3 , c = 2
and A = 30o , B = 60o , C = 90o , Here expression is equal to 2 which is given by (d).
Circle Connected with Triangle.
(1) Circumcircle of a triangle and its radius
(i) Circumcircle : The circle which passes through the angular points of a triangle is called its circumcircle.
The centre of this circle is the point of intersection of perpendicular bisectors of the sides and is called the
circumcentre. Its radius is always denoted by R. The circumcentre may lie within, outside or upon one of the sides
of the triangle.
(ii) Circum-radius : The circum-radius of a ∆ABC is given by A
(a) 2 a A = b B = c C = R (b) R = abc [∆ = area of ∆ABC ] FE
sin 2 sin 2 sin 4∆ O
B DC
(2) Inscribed circle or incircle of a triangle and its radius
(i) In-circle or inscribed circle : The circle which can be inscribed within a triangle so as to touch each of
its sides is called its inscribed circle or in circle. The centre of this circle is the point of intersection of the bisectors of
the angles of the triangle. The radius of this circle is always denoted by r and is equal to the length of the
perpendicular from its centre to any one of the sides of triangle.
(ii) In-radius : The radius r of the inscribed circle of a triangle ABC is given by A
∆ A B C A/
s 2 2 2
(a) r = (b) r = 4R sin sin sin E
r
A B C Fr
2 2 2
(c) r = (s − a) tan , r = (s − b) tan , r = (s − c) tan Ir
B BD
B C A C B A C/ C
2 2 2 2 2 2
(d) r= a sin sin ,r b sin sin ,r c sin sin 2
A = B = C
2 2 2
cos cos cos
(e) cos A + cos B + cos C =1 + r
R
(3) Escribed circles of a triangle and their radii
(i) Escribed circle : The circle which touches the side BC and two sides AB and AC produced of a triangle
ABC is called the escribed circle opposite to the angle A. Its radius is denoted by r1 . Similarly, r2 and r3 denote the
radii of the escribed circles opposite to the angles B and C respectively.
The centres of the escribed circles are called the ex-centres. The centre of the escribed circle opposite to the
angle A is the point of intersection of the external bisectors of angles B and C. The internal bisectors of angle A also
passes through the same point. The centre is generally denoted by I1 A
(ii) Radii of ex-circles : In any ∆ABC , we have
(a) r1 = s ∆ , r2 = s ∆ , r3 = s ∆ (b) r1 = s tan A , r2 = s tan B , r3 = s tan C A A
−a −b −c 2 2 2 2 2
(c) r1 a cos B cos C , r2 b cos C cos A , r3 c cos A cos B B DC
2 2 2 2 2 2
= A = B = C
2 2 2
cos cos cos F E
r1 I1 r1
1 1 1 1
(d) r1 + r2 + r3 − r = 4R (e) r1 + r2 + r3 =r
(f) 1 + 1 + 1 + 1 = a2 + b2 + c2 (g) 1 + 1 + 1 = 1
r2 r12 r22 r32 ∆2 bc ca ab 2Rr
(h) r1r2 + r2r3 + r3r1 = s2 (i) ∆ = 2R2 sin A.sin B.sin C = 4Rr cos A . cos B . cos C
2 2 2
(j) r1 = 4R sin A cos B cos C ; r2 = cos A . sin B . cos C ; r3 = 4R cos A cos B sin C
2 2 2 2 2 2 2 2 2
(4) Centroid (G) : Common point of intersection of medians of a triangle. Divides every median in the ratio
2:1. Always lies inside the triangle.
A
G
B DC
(5) Orthocentre of a triangle : The point of intersection of perpendicular drawn from the vertices on the
opposite sides of a triangle is called its orthocentre.
Pedal Triangle. F A
Let the perpendiculars AD, BE and CF from the vertices A, B and C on the opposite B
E
sides BC, CA and AB of ∆ABC respectively, meet at O. Then O is the orthocentre of O
the ∆ABC . The triangle DEF is called the pedal triangle of the ∆ABC . O
DC
Othocentre of the triangle is the incentre of the pedal triangle.
If O is the orthocentre and DEF the pedal triangle of the ∆ABC , where AD, BE, CF
are the perpendiculars drawn from A, B, C on the opposite sides BC, CA, AB
respectively, then
(i) OA = 2R cos A, OB = 2R cos B and OC = 2R cos C
(ii) OD = 2R cos B cos C, OE = 2R cos C cos A and OF = 2R cos A cos B
(1) Sides and angles of a pedal triangle: The angles of pedal triangle DEF are:
180 − 2A,180 − 2B,180 − 2C and sides of pedal triangle are:
EF = a cos A or R sin 2A ; FD = b cos B or R sin 2B ; DE = c cos C or R sin 2C A
If given ∆ABC is obtuse, then angles are been represented by 2A, 2B, 2C − 180o b cos BF a cos B c cos B
and the sides are a cos A, b cos B, − c cos C . OO E
(2) Area and circum-radius and in-radius of pedal triangle
Area of pedal triangle = 1 (Product of the sides) × (sine of included angle) B 180 – 2A D C
2
∆ = 1 R2.sin 2A.sin 2B. sin 2C
2
Circum-radius of pedal triangle = EF = R sin 2A R .
2 sin FDE 2 sin(180o − 2A) = 2
area of ∆DEF 1 R 2 sin 2 A. sin 2B. sin 2C
semi - perimeter of ∆DEF 2
In-radius of pedal triangle = = 2R sin A.sin B.sin C = 2R cos A.cos B.cos C .
Important Tips
• Circum-centre, Centroid and Orthocentre are collinear.
• In any right angled triangle, the orthocentre coincides with the vertex containing the right angled.
• The mid-point of the hypotenuse of a right angled triangle is equidistant from the three vertices of the triangle.
• The mid-point of the hypotenuse of a right angled triangle is the circumcentre of the triangle. A
• The length of the medians AD, BE, CF of ∆ABC are given by
AD = 1 2b2 + 2c2 − a2 = 1 b2 + c2 + 2bc cos A , BE = 1 2c 2 + 2a2 − b2 = 1 c2 + a2 + 2ca.cos B
2 2 2 2
F E
CF 1 2a 2 + 2b2 − c2 1 a2 + b2 + 2ab.cos c G
= 2 = 2
• The distance between the circumcentre O and centroid G of ∆ABC is given by BDC
OG = 1 OH = 1 R 1 − 8 cos A. cos B. cos C , Where H is the orthocentre of ∆ABC .
3 3
• The distance between the orthocentre H and centroid G of EMBED Equation.3 is given by EMBED Equation.3 .
• The distance between the circumcentre O and the incentre I of ∆ABC given by OI = R 1 − 8 sin A . sin B . sin C
2 2 2
• If I1 is the centre of the escribed circle opposite to the angle B, then OI1 = R 1 + 8 sin A . cos B . cos C
2 2 2
Similarly, OI2 = R 1 + 8 cos A . sin B . cos C , OI3 = R 1 + 8 cos A . cos B . sin C
2 2 2 2 2 2
• Circle circumscribing the pedal triangle of a given triangle bisects the sides of the given triangle and also the lines joining the vertices
of the given triangle to the orthocentre of the given triangle. This circle is known as "Nine point circle".
• Circumcentre of the pedal triangle of a given triangle bisects the line joining the circum-centre of the triangle to the orthocentre.
Ex-central Triangle.
Let ABC be a triangle and I be the centre of incircle. Let I1 , I2 and I3 be the centres of the escribed circles
which are opposite to A, B, C respectively then I1I2I3 is called the I3 I2
Ex-central triangle of ∆ABC .
90o – C/2 I 90o – B/2
I1I2I3 is a triangle, thus the triangle ABC is the pedal triangle of
its ex-central triangle I1I2I3 . The angles of ex-central triangle I1I2I3 BC
are 90o − A , 90o − B ,90o − C
2 2 2
and sides are I1I3 = 4R cos B ; I1 I 2 = 4 R cos C ; I2I3 = 4R cos A I1 90o – A/2
2 2 2
Area and circum-radius of the ex-central triangle
Area of triangle = 1 (Product of two sides) × (sine of included angles)
2
∆ = 1 4R cos B . 4R cos C × sin 90o − A
2 2 2 2
∆ = 8R2 cos A . cos B . cos C
2 2 2
I2I3 4R cos A
sin I 2 I1 I 3 2
Circum-radius = = = 2R .
2 2 sin 90 o A
− 2
Example: 31 In a ∆ABC r1 < r2 < r3 , then [EAMCET 2003]
Solution: (a)
Example: 32 (a) a < b < c (b) a > b > c (c) b < a < c (d) a < c < b
Solution: (d)
In a ∆ABC, r1 < r2 < r3
⇒ 11 1 ⇒ s−a > s−b > s−c ⇒ (s − a) > (s − b) > (s − c) ⇒ −a > −b > −c ⇒ a<b<c.
r1 > r2 > r3 ∆∆∆
Which of the following pieces of data does NOT uniquely determine an acute-angled triangle ABC (R being the radius of
the circum-circle) [IIT Screening 2002]
(a) a, sin A, sin B (b) a,b,c (d) a, sin B, R (d) a, sin A, R
a A = R and b = 2R sin B . So two sides and two angles are known. So ∠C is known. Therefore, two sides and
sin
included angle is known. So, ∆ is uniquely known in case (a).
If a, b, c are known the ∆ is uniquely known in case (b). b = 2R sin B, sin A = a . So, sides a, b and angle A, B are
2R
known. So ∠C is known. Therefore two sides and included angle is known. So, ∆ is uniquely known in case (c).
a = R. So, only a side and an angle is known. So, ∆ is not uniquely known in case (d).
sin A
Example: 33 In a triangle ABC, let ∠C = π . If r is the in radius and R is the circum-radius of the triangle, then 2(r + R) is equal to
Solution: (a) 2
Example: 34 (a) a + b (b) b + c (c) c + a [IIT Screening 2000]
Solution: (c) ∴ c = 2R sin 90o = 2R
Example: 35 (d) a + b + c
Solution: (b)
Example: 36 c = 2R ,
sin C
Solution: (a)
Example: 37 Also r = (s − c) tan C = (s − c) [ tan 45o = 1]
Solution: (a) 2
Example: 38 2r = 2s − 2c = a + b − c = a + b − 2R , 2(r + R) = a + b .
Solution: (c)
Example: 39 If R is the radius of the circumcircle of the ∆ABC and ∆ is its area, then [Karnataka CET 2000]
(a) R = a + b + c (b) R = a +b+ c (c) R = abc (d) R = abc
∆ 4∆ 4∆ ∆
Area of the triangle ABC(∆) = bc sin A . From the sine formula, a = 2R sin A or sin A = a ⇒ ∆ = 1 bc. a = abc or
2 2R 2 2R 4R
R = abc .
4∆
If the length of the sides of a triangle are 3, 4 and 5 units, then R (the circum-radius) is [UPSEAT 2000]
(a) 2.0 unit (b) 2.5 unit (c) 3.0 unit (d) 3.5 unit
Given, Sides are 3, 4, 5.
Since 32 + 42 = 52 . So, triangle is a right angle triangle. R= 5 = 2.5 unit .
2
If x, y, z are perpendicular drawn from the vertices of triangle having sides a, b, and c, then the value of bx + cy + az will
c a b
be [UPSEAT 1999]
(a) a2 + b2 + c2 (b) a2 + b2 + c2 (c) a2 + b2 + c2 (d) 2(a 2 + b 2 + c 2 )
2R R2 4R R
Let area of triangle be ∆ , then according to question, ∆ = 1 ax = 1 by = 1 cz
2 2 2
∴ bx + cy + az = b 2∆ + c 2∆ + a 2∆ = 2∆(b2 + c2 + a2) = 2(a2 + b2 + c2) . abc = a2 + b2 + c2 .
c a b c a a b b c abc abc 4R 2R
If r1, r2 , r3 in a triangle be in H.P., then the sides are in [EAMCET 1993]
(a) A.P. (b) G.P. (c) H.P. (d) None of these
11 11 1 − 1 = 1 − 1 ⇒ s−b−s+a = s−c−s+b ⇒ a−b = b−c
r2 − r1 = r3 − r2 ⇒ ∆ ∆ ∆ ∆ ∆ ∆ ∆∆
s−b s−a s−c s−b
⇒ b − a = c − b ⇒ 2b = a + c .
∆∆
If the sides of a triangle are 13,14 15, then the radius of its incircle is [EAMCET 1987]
(a) 67 (b) 65 (c) 4 (d) 24
8 4
r = ∆ and s = a+b+c = 13 + 14 + 15 = 21
s 2 2
⇒ ∆= 21 × 8 × 7 × 6 = 84 ⇒ r = 84 = 4 .
21
In an equilateral triangle the inradius and the circumradius are connected by [EAMCET 1983]
(a) r = 4R (b) r = R (c) r = R (d) None of these
2 3
Solution: (b) r = 4R sin A . sin B . sin C
2 2 2
For an equilateral triangle, A = B = C = 60o
∴ r = 4R sin 30o. sin 30o. sin 30o = 4 R. 1 . 1 . 1 = R .
2 2 2 2
Cyclic Quadrilateral .
A quadrilateral PQRS is said to be cyclic quadrilateral if there exists a circle passing through all its four vertices
P, Q, R and S.
Let a cyclic quadrilateral be such that PQ = a,QR = b, RS = c and SP = d . Then ∠Q + ∠S = 180o and
∠A + ∠C = 180o . Let 2s = a + b + c + d S
d
and = Area of cyclic quadrilateral PQRS
= Area of ∆PQR + Area of ∆PRS = 1 ab sin Q 1 cd sin S c
2 2
+
= 1 ab sin Q + 1 cd sin(π − Q) = 1 (ab + cd) sin Q P R
2 2 2 a b
= 1 (ab + cd) sin Q ......(i) Q
2
In ∆PQR and ∆PRS ,
From cosine rule, PR2 = PQ2 + QR2 − 2PQ.QR cos Q = a2 + b2 − 2ab cos Q ......(ii)
and PR2 = PS2 + RS2 − 2PS.RS cos S
PR2 = d2 + c2 − 2cd cos(π − Q)
PR2 = d2 + c2 + 2cd cos Q ........(iii)
From (ii) and (iii) we have, = (s − a)(s − b)(s − c)(s − d) ........(iv)
Therefore, (1) Area of cyclic quadrilateral = 1 (ab + cd) sin Q
2
(2) Area of cyclic quadrilateral = (s − a)(s − b)(s − c)(s − d) , where 2s = a + b + c + d
(3) cos Q = a2 + b2 − c2 − d2
2(ab + cd)
(4) Circumradius of cyclic quadrilateral : Circum circle of quadrilateral PQRS is also the circumcircle of
∆PQR . Hence circumradius of cyclic quadrilateral PQRS = R = circumradius of ∆PQR = PR = PR(ab + cd)
2 sin B 4∆
But PR = (ac + bd)(ad + bc)
(ab + cd)
Hence R= 1 (ac + bd)(ad + bc)(ab + cd) = 1 (ac + bd)(ad + bc)(ab + cd)
4∆ 4 (s − a)(s − b)(s − c)(s − d)
(5) Ptolemy's theorem : In a cyclic quadrilateral PQRS, the product of diagonals is equal to the sum of the
products of the length of the opposite sides i.e., According to Ptolemy's theorem, for a cyclic quadrilateral PQRS
PR.QS = PQ.RS + RQ.PS. A
B
C
D
Example: 40 A cyclic quadrilateral ABCD of area 33 is inscribed in a unit circle. If one of its sides AB = 1 and the diagonal BD =
Solution: (a) 4
Example: 41 3 then the lengths of the other sides are [Roorkee 1995]
Solution: (a)
(a) 2, 1, 1 (b) 2, 1, 2 (c) 3, 1, 2 (d) None of these
By sine formula in ∆ABC , 3 = 2R ⇒ 3 = 2 ⇒ sin A = 3 ⇒ A = π
sin A sin A 2 3
Now, AB = x = 1
By cosine formula in ∆ABD cos π = x2 + y2 − 3 ⇒ 1 = 1+ y2 − 3 ⇒ y = y2 − 2
3 2xy 2 2y
⇒ y2 − y − 2 = 0 ⇒ (y − 2)(y + 1) = 0 ⇒ y = 2 [ y ≠ −1 ] A
∴ x y
p 3D
Since ∠A = 60o ∴ ∠C = 120o
q
In ∆BDC, 3 = p2 + q2 − 2pq cos 120o ⇒ 3 = p2 + q2 + pq .........(i) B
Also area of quadrilateral ABCD = 33
4
∴ 33 = ∆ABD + ∆BCD = 1 .1.2 sin 60o + 1 p.q. sin 120o = 3 + 3 pq C
4 2 2 2 4
⇒ 3 pq = 3 3 3 3 3−2 3 3 ⇒ pq = 1
− ==
4 42 4 4
∴ (i) gives, 3 = p2 + q2 + 1 ⇒ p2 + q2 = 2 , [p, q > 0]
∴ p2 + 1 =2 ⇒ p4 − 2p2 + 1 = 0 ⇒ (p2 − 1)2 = 0 ⇒ p 2 = 1, ⇒ ∴ p = 1, q = 1
p2
∴ AB = 1, AD = 2, BC = CD = 1 .
The two adjacent sides of a cyclic quadrilateral are 2 and 5 and the angle between them is 60o. If the third side is 3, the
remaining fourth side is [UPSEAT 1994]
(a) 2 (b) 3 (c) 4 (d) 5
Since ABCD is cyclic quadrilateral and ∠ABC = 60o D
A 120o 3
∴ ∠ADC = 180o − 60o = 120o
Let AB = 2, BC = 5 and CD = 3
In ∆ABC, AB2 + AC2 + 2AB.BC cos 60o = AC2 or 4 + 25 − 2 × 2 × 5 × 1 = AC2 C
2 5
2
∴ AC2 = 19 ; In ∆ABC AD2 + CD2 + 2AD.CD cos 60o = AC2 60o
B
or AD2 + 9 + 2 AD.3. 1 = 19 or AD2 + 3AD − 10 = 0 or AD2 + 5 AD − 2AD − 10 = 0
2
or AD(AD + 5) − 2(AD + 5) = 0 or (AD − 2)(AD + 5) = 0 . Therefore, fourth side is AD = 2 .
Example: 42 Two adjacent sides of a cyclic quadrilateral are 2 and 5 and the angle between them is 60o. If the area of the quadrilateral
Solution: (a)
is 4 3 , then the remaining two sides are [Roorkee 1991]
(a) 2, 3 (b) 3, 4 (c) 4, 5 (d) 5, 6
Let a = PQ = 2 , b = QR = 5 , RS = c , SP = d S
Given, Area of quadrilateral PQRS = 4 3 120o c
d
Area of ( ∆PQR + ∆PRS) = 1 [2 5 sin 60o + c × d sin 120o]
⇒ 2 × P R
a
⇒ 4 3 = 53 + 3 cd ⇒ cd = 6 ........(i) 60o b
2 4
Q
Now by cosine formula RS2 + SP 2 − 2RS.SP. cos 120o = PR2 = PQ2 + QR2 − 2PQ.QR cos 60o
⇒ c2 + d2 − 2cd − 1 = 4 + 25 − 2(2)(5) 1 ⇒ c2 + d2 + cd = 19 ⇒ c2 + d2 = 19 − 6 = 13 ……..(ii)
2 2
Solving (i) and (ii) we get c = 2 and d = 3 .
Regular Polygon.
A regular polygon is a polygon which has all its sides equal and all its angles equal.
(1) Each interior angle of a regular polygon of n sides is 2n − 4 × right angles = 2n − 4 × π radians.
n n 2
(2) The circle passing through all the vertices of a regular polygon is called its circumscribed circle.
If a is the length of each side of a regular polygon of n sides, then the radius R of the circumscribed circle, is
given by R = a .cosec π
2 n
(3) The circle which can be inscribed within the regular polygon so as to touch all O C
its sides is called its inscribed circle. F R π/n
rR
Again if a is the length of each side of a regular polygon of n sides, then the
radius r of the inscribed circle is given by r = a .cot π AB
2 n
(4) The area of a regular polygon is given by ∆ = n × area of triangle OAB
= 1 n a2cot π (in terms of side)
4 n
= nr 2 .tan π (in terms of in-radius)
n
= n .R 2 sin 2π (in terms of circum-radius)
2 n
Example: 43 The sum of the radii of inscribed and circumscribed circles for an n sided regular polygon of side a, is [AIEEE 2003]
Solution: (c)
(a) a cot π (b) a cot π (c) a cot π (d) a cot π
n 2 2n 2n 4 2n
tan π = a and sin π = a ⇒ r + R = a cot π + cosec π = a cot π .
n 2r n 2R 2 n n 2 2n
Example: 44 The area of the circle and the area of a regular polygon of n sides and its perimeter equal to that of the circle are in the
Solution: (a)
ratio of [Roorkee 1992]
Example: 45
Solution: (a) (a) tan π : π (b) cos π : π (c) sin π : π (d) cot π : π
Example: 46 n n n n n n n n
Solution: (b)
Let r be the radius of the circle and A1 be its area ∴ A1 = πr 2
Since the perimeter of the circle is the same as the perimeter of a regular polygon of n sides ∴ 2πr = na , where 'a' is the
length of one side of the regular polygon, ∴ a = 2πr
n
Let A2 be the area of the polygon, then
A2 = 1 na2. cot π = 1 n. 4π 2r 2 cot π = πr 2 . π . cot π
4 n 4 n2 n n n
∴ A1 : A2 = πr 2 : πr 2. π . cot π = 1: π cot π = tan π : π .
n n n n n n
A regular polygon of nine sides, each of length 2 is inscribed in a circle. The radius of the circle is [IIT 1994]
(a) cosec π (b) cosec π (c) cot π (d) tan π
9 3 9 9
We know that radius of the circumcircle is given by R= a cosec π ; Here, a = 2, n = 9
2 n
∴ Required radius = 2 cosec π = cosec π .
2 9 9
If the number of sides of two regular polygons having the same perimeter be n and 2n respectively, their areas are in the
ratio
2 cos π 2 cos π cos π
n n n
(a) (b) (c) (d) None of these
cos π cos π sin π
2n 1 + n n
Let s be the perimeter of both the polygons. Then the length of each side of the first polygon is s and that of second
n
polygon is s .
2n
If A1 , A2 denote their areas, then A1 = n s 2 cot π and A2 = 1 .(2n) s 2. cot π
4 n n 4 2n 2n
A1 2 cot π 2 cos π sin π 2 cos π sin π A1 2 cos π
A2 n n 2n n 2n A2 n
= = = ⇒ = .
cot π sin π cos π sin π cos π cos π cos π
2n n 2n 2 2n 2n 2n 1 + n
Solutions of Triangles.
Different formulae will be used in different cases and sometimes the same problem may be solved in different
ways by different formulae. We should, therefore, look for that formula which will suit the problem best.
(1) Solution of a right angled triangle
(2) Solution of a triangle in general
(1) Solution of a right angled triangle
(i) When two sides are given: Let the triangle be right angled at C. Then we can determine the remaining
elements as given in the following table
Given Required
a,b tan A = a , B = 90o − A, c = a
b sin A
a, c
sin A = a , b = c cos A, B = 90o − A
c
(ii) When a side and an acute angle are given : In this case, we can determine the remaining elements as
given in the following table
Given Required
a, A B = 90o − A, b = a cot A, c = a
sin A
c, A
B = 90o − A, a = c sin A,b = c cos A
(2) Solution of a triangle in general
(i) When three sides a, b and c are given in this case, the remaining elements are determined by using the
following formulae, ∆ = s(s − a)(s − b)(s − c) , where 2s = a + b + c = perimeter of triangle
sin A = 2∆ , sin B = 2∆ , sin C = 2∆
bc ac ab
tan A = ∆ , tan B = ∆ b) , tan C = ∆
2 s(s − a) 2 s(s − 2 s(s − c)
(ii) When two sides a, b and the included angle C are given : In this case, we use the following formulae
∆ = 1 ab sin C ; tan A − B = a − b cot C ; A + B = 90o − C and c = a sin C
2 2 a + b 2 2 2 sin A
(iii) When one sides a and two angles A and B are given : In this case, we use the following formulae to
determine the remaining elements A + B + C = 180o ⇒ C = 180o − A − B
b = a sin B and c = a sin C ⇒ ∆= 1 ca sin B
sin A sin A 2
(iv) When two sides a, b and the angle A opposite to one side is given : In this case, we use the
following formulae sin B = b sin A ..........(i)
a
C = 180 o − (A + B), c = a sin C
sin A
Special Cases
Case I : When a is an acute angle C
(a) If a < b sin A, there is no triangle. When a < b sin A, from (i), b a b sin A
sin B > 1, which is impossible. B X
A
From the following figure, If AC = b,∠CAX = A , then perpendicular A N
CN = b sin A. Now taking C as centre, If we draw an arc of radius a then it
will never intersect the line AX and hence no triangle ABC can be
constructed in this case.
(b) If a = b sin A , then only one triangle is possible which is right angled C
at B. When a = b sin A, then from sine rule. sin B = 1, ∴ ∠B = 90o
from fig. It is clear that CB = a = b sin A
Thus, in this case, only one triangle is possible which is right angled at B. b a = b sinA
C (c) If a > b sin A, then three possibilities
b will arise: A A 90o
A
a (i) a = b In this case, from sine rule B X
A b sinA
X sin B = sin A ; ∴ B = A or B = 180o − A
NB
But B = 180o − A ⇒ A + B = 180o , which is not possible in a triangle.
∴ In this case, we get ∠A = ∠B . C
Hence, if b = a > b sin A then only one isosceles triangle ABC is possible in ab a
which ∠A = ∠B .
A b sinA X
(ii) a > b In the following figure, Let AC = b,∠CAX = A, and a > b, also
a > b sin A . B′ A N B
Now taking C as centre, if we draw an arc of radius a, it will intersect AX at one point B and hence only one
∆ABC is constructed. Also this arc will intersect XA produced at B′ and ∆AB′C is also formed but this ∆ is
inadmissible (because ∠CAB′ is an obtuse angle in this triangle)
Hence, if a > b and a > b sin A, then only one triangle is possible.
(iii) b > a (i.e., b > a > b sin A) C2 C C1
In fig. let AC = b,∠CAX = A . Now taking C as centre, if we draw an arc of a
b
a
radius a, then it will intersect AX at two points B1 and B2 . Hence if A b sinA X
N B1
b > a > sin A, then there are two triangles. A B2
Case II : When A is an obtuse angle: In this case, there is only one triangle, if a > b
C
b sinA ba
A
NAB X
Case III: b > c and B = 90o
Again the circle with A as centre and b as radius will cut the line only in one point. So, only one triangle is
possible. C
Bc A
b
Case IV: b ≤ c and B = 90o
The circle with A as centre and b as radius will not cut the line in any point. So, no triangle is possible.
This is, sometimes called an ambiguous case.
Alternative method: By applying cosine rule, we have cos B = a2 + c2 − b2
2ac
⇒ a2 − (2c cos B)a + (c 2 − b2) = 0
⇒ a = c cos B ± (c cos B)2 − (c 2 − b2) Bc A
⇒ a = c cos B ± b2 − (c sin B)2
b
This equation leads to following cases:
Case I : If b < c sin B ,no such triangle is possible.
Case II : Let b = c sin B, there are further following case.
(a) B is an obtuse angle ⇒ cos B is negative. There exists no such triangle.
(b) B is an acute angle ⇒ cos B is positive. There exists only one such triangle.
Case III : Let b > c sin B . There are further following cases :
(a) B is an acute angle ⇒ cosB is positive.
In this case two values of a will exists if and only if c cos B > b2 − (c sin B)2 or c > b
Two such triangle is possible. If c < b, only one such triangle is possible.
(b) B is an obtuse angle ⇒ cosB is negative. In this case triangle will exist if and only if b2 − (c sin B)2 >|c cos B|
⇒ b > c . So, in this case only one such triangle is possible. If b < c there exists no such triangle.
Note : If one side a and two angles B and C are given, then A = 180o − (B + C) and
b = a sin B ,c = a sin C
sin A sin A
If the three angles A, B, C are given. We can only find the ratios of the sides a, b, c by using sine rule
(since there are infinite similar triangle possible).
Important Tips
A triangle which does not contain a right angle is called an oblique triangle.
If triangle is an acute angled triangle means every angle is less than 90o.
If triangle is an obtuse angled triangle means one of its angle is greater than 90o.
If A and B are complementry angles then A + B = 90o.
If A and B are supplementary angle, then A + B = 180o.
Σ(p − q) = (p − q) + (q − r) + (r − p) = 0 , Σp(q − r) = p(q − r) + q(r − p) + r(p − q) = 0 .
Σ(p + a)(q − r) = Σp(q − r) + aΣ(q − r) = 0 .
Example: 47 If in a right angled triangle the hypotenuse is four times as long as the perpendicular drawn to it from opposite vertex, then
Solution: (a)
one of its acute angle is [MP PET 1998]
Example: 48
Solution: (a) (a) 15o (b) 30o (c) 45o (d) None of these
Example: 49
If x is length of perpendicular drawn to it from opposite vertex of a right angled triangle, so, length of diagonal
AB = y1 + y2 ........(i)
From ∆OCB, y2 = x cot θ and from ∆OCA, y1 = x tanθ
B
Put the values in equation (i), then AB = x(tanθ + cot θ ) .......(ii) (90o – θ)
c
Since, length of hypotenuse = 4 (Length of perpendicular) xθ
C
∴ x(tanθ + cot θ ) = 4 x ⇒ sin2 θ + cos2 θ =4 θ
sinθ . cosθ A
⇒ sin 2θ = 1 ⇒ 2θ = 30o ⇒θ = 15o .
2
Trick : Length of Length of hypotenuse vertex to hypotenuse = 2
perpendicular drawn from opposite sin 2θ
⇒ 4 = 2 ⇒ sin 2θ = 1 ⇒ sin 2θ = sin 30o ⇒ θ = 15o .
sin 2θ 2
The number of triangles ABC that can be formed with a = 3, b = 8 and sin A = 5 is [Roorkee 1998]
3
(a) 0 (b) 1 (c) 2 (d) 3
Given, a = 3, b = 8 and sin A = 5 .
13
∴b sin A = 8 × 5 = 40 > a(= 3)
13 13
Thus in this case no triangle is possible.
In any triangle AB = 2, BC = 4, CA = 3 and D is mid point of BC, then [Roorkee 1995]
(a) cos B = 11 (b) cos B = 7 (c) AD = 2.4 (d) AD2 = 2.5
6 8
Solution: (d) From ∆ABC , cos B = 22 + 42 − 32 = 11
2× 2× 4 16
A
From ∆ABD , cos B = 22 + 22 − AD 2 = 11 2 3
2× 2× 2 16
∴ 11 22 + 22 − AD2
16 = 2× 2× 2
B2 D2 C
⇒ AD2 = 2.5 .
Example: 50 If b= 3, c = 4 and B = π , then the number of triangles that can be constructed is [Roorkee 1992]
Solution: (d) 3
Example: 51
Solution: (b) (a) Infinite (b) Two (c) One (d) Nil
Example: 52 Here, c sin B = 4 sin π = 2 3 > b(= 3)
Solution: (a) 3
Thus, we have b < c sin B
Hence no triangle is possible, i.e., the number of triangles that can be constructed is nil.
If two sides of a triangle are 2 3 and 2 2 the angle opposite the shorter side is 45o . The maximum value of the third
side is
(a) 2 + 6 (b) 2 + 6 (c) 6 − 2 (d) None of these
Let a = 2 3, b = 2 2 ∴ B = 45o
∴ a b c 23 22 c .......(i)
sin A = sin B = sin C ⇒ sin A = sin 45o = sin C
∴ sin A = 3 ⇒ A = 60o .
2
∴ C = 180o − A − B = 75o
From (i) c = 4 sin C = 4 sin(45o + 30o ) = 2 + 6 .
In an ambiguous case, If the remaining angles of the triangles formed with a, b and A be B1, C1 and B2, C2 then
sin C1 + sin C2 =
sin B1 sin B2
(a) 2 cos A (b) cos A (c) 2 sin A (d) sinA
In ∆' s ACB1 and ACB2
sin C1 = AB1 = c1 and sin C2 = c2 C2 C
sin B1 AC b sin B2 b b C1
a
∴ sin C1 + sin C2 = c1 + c2 A
sin B1 sin B2 b A B2 b sinA
N B1 X
cos A = b2 + c2 − a2 or c 2 − (2b cos A)c + (b2 − a2) = 0
2bc
∴ c1 + c2 = 2b cos A
sin C1 + sin C2 = 2b cos A = 2cos A .
sin B1 sin B2 b
Trigonometrical-equations-properties-of-triangles-and-heights-and-distances-3
This chapter deals with the applications of trigonometry to practical situations concerning measurement of
heights and distances which are otherwise not directly measurable By the use of trigonometry we can measure the
following :
(i) Height of tower or temple (ii) Breadth of river
(iii) Distance between inaccessible points (iv) Angle of vision etc.
We need to first define certain terms and state some properties before applying the principles of trigonometry.
Some Terminology Related to Heights and Distances .
(1) Angle of elevation and depression: Let O and P be two points such that P is at higher level than O. Let
PQ, OX be horizontal lines through P and O, respectively. If an observer (or eye) is at Q Horizontal line P
O and the object is at P, then ∠XOP is called the angle of elevation of P as seen from Angle of depression
O. This angle is also called the angular height of P from O.
If an observer (or eye) is at P and the object is at O, then ∠QPO is called the O Angle of elevation
angle of depression of O as seen from P. Horizontal line X
(2) Method of solving a problem of heights and distances
(i) Draw the figure neatly showing all angles and distances as far as possible.
(ii) Always remember that if a line is perpendicular to a plane then it is perpendicular to every line in that plane.
(iii) In the problems of heights and distances we come across a right angled triangle in which one (acute) angle
and a side is given. Then to find the remaining sides, use trigonometrical ratios in which known (given) side is used,
i.e., use the formula.
(iv) In any triangle other than right angled triangle, we can use 'the sine rule'.
i.e., formula, a = b = c , or cosine formula i.e., cos A = b2 +c2 − a2 etc.
sin A sin B sin C 2bc
(v) Find the length of a particular side from two different triangles containing that side common and then
equate the two values thus obtained.
(3) Geometrical properties and formulae for a triangle
(i) In a triangle the internal bisector of an angle divides the opposite side in the ratio of the arms of the angle.
BD = c . A
DC b
A
(ii) In an isosceles triangle the median is perpendicular to the
base i.e., AD ⊥ BC . c A/2 b
(iii) In similar triangles the corresponding sides are proportional. A/2
(iv) The exterior angle is equal to sum of interior opposite angles.
B C CB 90o C
Bc Db D
(4) North-east: North-east means equally inclined to north and east, south-east means equally inclined to
south and east. ENE means equally inclined to east and north-east.
N
P1 NE
45o P2 ENE
45o
22.5o
W O 45o E
P3
S
(5) Bearing : In the figure, if the observer and the object i.e., O and P be on the same level then bearing is
defined. To measure the ‘Bearing’, the four standard directions East, West, North and South are taken as the
cardinal directions. N
Angle between the line of observation i.e., OP and any one standard direction– P
east, west, north or south is measured.
Thus, ∠POE = θ is called the bearing of point P with respect to O measured W θ E
from east to north. In other words the bearing of P as seen from O is the direction O
in which P is seen from O.
S
(6) Problem on two dimensions : If the actual figure is located in one plane, the problem is of two
dimensions. For direction in two dimensional figures, cross vertically as shown in the figure.
NW PN
W
90o
90o A O
ES E
S B
(7) Problems on three dimensions : If total actual figure is located in more than one plane, the problem will
be of three dimensions. For direction in three dimensional figures, cross obliquely N
as shown. Clearly this oblique cross represents the horizontal plane.
If OP be a vertical tower perpendicular to the plane then it will be represented W E
like the figure, clearly ∠POA = 90o . If the observer at A moves in east direction.
We draw a line AB parallel to east to represent this movement. Clearly
∠OAB = 90o (angle between north and east).
S
(8) m-n cot theorem of trigonometry: (m + n) cot θ
= m cot α − n cot β = n cot A − m cot B (θ on the right)
C
αβ
I II
Aθ n BB
A mD
I II
Note : If θ is on the left then angle in the right is π −θ and cot(π − θ ) = − cotθ . Hence in this case m- n
theorem becomes − (m + n) cotθ = m cot α − n cot β = n cot A − m cot B (θ on the left).
Some Properties Related to Circle.
(1) Angles in the same segment of a circle are equal i.e., ∠APB = ∠AQB = ∠ARB .
PQ
R
AB
(2) Angles in the alternate segments of a circle are equal. A
α
A
α
B
C α B O T
O α
C
(3) If the line joining two points A and B subtends the greatest angle α at a point P then the circle, will touch the
straight line XX’ at the point P. A
B
α P
α
X′ P X
2α
(4) The angle subtended by any chord at the centre is twice the angle subtended AB
by the same on any point on the circumference of the circle.
Some Important Results. (2)
(1)
h
αβ H
ad x
β
a = h (cot α − cot β ) = h sin(β − α) α
sinα. sin β
H = x cotα tan(α + β )
∴h = a sinα sin β cosec (β − α) and
d = hcot β = a sinα.cos β.cosec(β − α)
(3) (4)
P
α h β B h H
A Qa d α β
a = h(cotα + cot β ) , where by H = hcot β
h = a sinα. sin β.cosec(α + β ) and cot α
d = hcot β = a sinα.cos β.cosec (α + β )
(6)
(5)
α
α H aβ H
h H
β
h= H sin(β − α) or H = h cot α H = a sin(α + β )
cosα sin β cotα − cot β sin(β − α)
(7) A (8) OP – Tower P
y A – South h N
D B – East
O βB W E
α
dS
α β A
Cx B
h= d
AB = CD . Then, x = y tan α +β cot 2 β + cot 2 α
2
(10)
(9) P P
O h
α β
α
A Q
β aγ
B A B
h = AB h = AP sinα = a sinα. sin γ .cosec (β − γ ) and
cot 2 β − cot 2 α
if AQ = d , then d = AP cosα = a cosα. sin γ .cosec (β − γ )
(11) AP = a sin γ .cosec(α − γ ) P
AQ = a sin δ .cosec(β − δ ) Q
and apply,
PQ 2 = AP 2 + AQ 2 − 2AP.AQ cosθ α δγ
β aB
A
Important Tips
• In the application of sine rule, the following point be noted. We are given one side a and some other side x is to be found. Both these are
in different triangles. We choose a common side y of these triangles. Then apply sine rule for a A
and y in one triangle and for x and y for the other triangle and eliminate y. Thus, we will get a
unknown side x in terms of a. In the adjoining figure a is known side of ∆ ABC and x is unknown
θ
is side of triangle ACD. The common side of these triangle is AC = y (say) Now apply sine rule Bβ
a y xy y γD
sinα sin β sinθ = sinγ x
∴ = …….. (i) and ……..(ii)
Dividing (ii) by (i) we get, x sinα = sin β ; ∴ x = a sin β sin θ α
a sinθ sin γ sin α sin γ C
Miscellaneous Examples.
Example: 1 The angle of elevation of a tower at a point distant d metres from its base is 30o . If the tower is 20 meters high, then the
value of d is [MP PET 1982, 1988]
(a) 10 3 m (b) 20 m (c) 20 3 m (d) 10 m
3
Solution: (c) d = cot 30o
20
d = 20 cot 30o = 20 3m . 20
30°
d
Example: 2 The angle of elevation of the top of a tower from a point 20 meters away from its base is 45o . The height of the tower is
Solution: (b)
Example: 3 [MP PET 1984, 1989]
Solution: (b)
Example: 4 (a) 10 m (b) 20 m (c) 40 m (d) 20 3 m
Solution: (b) Let height of the tower be h.
h = tan 45o h
20
h = 20m .
45°
20
If the angle of elevation of the top of a tower at a distance 500 m from its foot is 30o , then height of the tower is
[Kerala (Engg.) 2002]
(a) 1 (b) 500 (c) 3 (d) 1
3 3 500
Let the height be h
∴ tan 30o = h ⇒h = 500 . h
500 3
30°
500 m
A person standing on the bank of a river finds that the angle of elevation of the top of a tower on the opposite bank is
45o . Then which of the following statements is correct [MP PET 1994]
(a) Breadth of the river is twice the height of the tower
(b) Breadth of the river and the height of the tower are the same A
(c) Breadth of the river is half of the height of the tower
(d) None of these Tower
AB is tower and BC is river.
From ∆ABC, AB = tan 45o or AB = BC C 45° B
BC River
Height of tower = Breadth of river.
Example: 5 A ladder 5 metre long leans against a vertical wall. The bottom of the ladder is 3 metre from the wall. If the bottom of the
Solution: (a)
ladder is pulled 1 metre farther from the wall, how much does the top of the ladder slide down the wall [AMU 2000]
(a) 1 m A A
(b) 7 m 5m D
(c) 2 m
(d) None of these 5m
AB = 4m ⇒ BD = 3m
∴ AD = 4 − 3 = 1m . B 3m C B 4m C
Example: 6 From the top of a light house 60 metre high with its base at the sea level the angle of depression of a boat is 15o. The
Solution: (b)
Example: 7 distance of the boat from the foot of the light house is [MP PET 2001, 1994; IIT 1983; UPSEAT 2000, 1988]
Solution: (b)
(a) 3 − 1 60 metre (b) 3 + 1 60 metre (c) 3 + 1 metre (d) None of these
Example: 8 3 + 1 3 − 1 3 − 1
Solution: (d) 15°
Required distance = 60 cot 15o
= 60 3 + 1 metre. 60
3 −
1 15°
x
A person observes the angle of deviation of a building as 30o . The person proceeds towards the building with a speed of
25( 3 − 1)m / hour . After 2 hours, he observes the angle of elevation as 45o . The height of the building (in metres) is
[UPSEAT 2003]
(a) 100 (b) 50 (c) 50( 3 + 1) (d) 50( 3 − 1)
In ∆PQR, tan 30o = PQ 30° P
QR h
1h
⇒ 3 = 50( 3 − 1) + h
⇒ 3h = 50( 3 − 1) + h R 30° 45° h Q
⇒ ( 3 − 1) h = 50( 3 − 1) ⇒ h = 50 metre. S
50(√3–1)
The shadow of a tower standing on a level ground is found to be 60 m longer when the sun's altitude is 30o than when it
is 45o . The height of the tower is
(a) 60 m (b) 30 m (c) 60 3m (d) 30( 3 + 1)m
AB = AM − BM ⇒ AB AM BM P
h= h−h
AB = cot 30o − cot 45o ⇒ h= 60 = ( 60( 3 + 1) h
h 3 −1 3 − 1)( 3 + 1)
⇒ h = 60( 3 + 1) ⇒ h = 30( 3 + 1)m . A 30° 45° M
3 −1 60m B
Example: 9 A person is standing on a tower of height 15( 3 + 1)m and observing a car coming towards the tower. He observed that
Solution: (a)
angle of depression changes from 30o to 45o in 3 sec. What is the speed of the car [Karnataka CET 1998]
(a) 36 km/hr (b) 72 km/hr (c) 18 km/hr (d) 30 km/hr
AB = OP[cot α − cot β ] , where α = 30o, β = 45o P
⇒ AB = 15( 3 + 1) ( 3 − 1) = 15(3 − 1) = 30 metre.
Speed = Distance = 30 = 10m/sec 15( 3 + 1)
time 3
= 10 × 18 km/hr 30o 45o
5 B A
= 36 km/hr. O
Example: 10 The angle of elevation of the top of a pillar at any point A on the ground is 15o. On walking 40 metre towards the pillar,
Solution: (b)
the angle becomes 30o. The height of the pillar is [MP PET 2001; EAMCET 1984; MNR 1975; IIT 1967]
(a) 40 metre (b) 20 metre (c) 20 3 metre (d) 40 3 metre
3
AB = OP[cot α − cot β ] ⇒ 40 = h[cot 15o − cot 30o ] P
h = 40
cot15o − cot 30o
h= 40 = 40(2 − 3) h
1 4−2 3 O
3
−
2− 3
15o 30o
h = 20 metre . A B
Example: 11 A man from the top of a 100 metre high tower looks a car moving towards the tower at an angle of depression of 30o. After some
Solution: (b) time, the angle of depression becomes 60o. The distance (in metre) travelled by the car during this time is [IIT Screening 2001]
(a) 100 3 (b) 200 3 (c) 100 3 (d) 200 3
3 3
d = AB = OA − OB P
= 100[cot 30o − cot 60o]
= 100 3− 1 = 200 3 metre. 100
3 3 O
30o 60o
A B
Example: 12 A tower is situated on horizontal plane. From two points, the line joining these points passes through the base and which
are a and b distance from the base. The angle of elevation of the top are α and 90o − α and θ is that angle which two
points joining the line makes at the top, the height of tower will be [UPSEAT 1999]
(a) a+b (b) a−b (c) ab 1
a−b a+b
(d) (ab)3
Solution: (c) Let there are two points C and D on horizontal line passing from point B of the base of the tower AB. The distance of these
Example: 13 points are b and a from B respectively i.e., BD = a and BC = b A
Solution: (b)
∴Line CD, on the top of tower A subtends an angle θ , hence ∠CAD = θ
Example: 14
Solution: (b) According to question, on point C and D, the elevation of top are θ
Example: 15
α and 90o − α .
∴ ∠BCA = α and ∠BDA = 90o − α .........(i) D 90 – α α
In ∆ABC, AB = BC tan α = b tan α .........(ii) Cb
and in ∆ABD, AB = BD tan(90o − α) = a cot α B
a
Multiplying equation (i) and (ii) (AB)2 = (b tan α)(a cot α) = ab, ⇒ AB = (ab) .
A tower of height b subtends an angle at a point O on the level of the foot of the tower and at a distance a from the foot of
the tower. If a pole mounted on the tower also subtends an equal angle at O, the height of the pole is [MP PET 1993]
(a) b a2 − b2 (b) b a2 + b2 (c) a a2 − b2 (d) a a2 + b2
a2 + b2 a2 − b2 a2 + b2 a2 − b2
Let AB is tower and AC is pole of height h.
From ∆ABO, b = tanα ......(i)
a
C
From ∆CBO, b + h = tan 2α or b+h = 2 tanα A
a a 1 − tan2 α B
or b+h= 2a b (Put value of tan α from (i))
a
b2 α
a2
1 − A α
a
or h= b(a2 + b2) .
a2 − b2
Remember the result h = b(a2 + b2) in which b= height of tower, h = height of pole, a = distance of observation point
a2 − b2
from the tower.
A vertical pole consists of two parts, the lower part being one third of the whole. At a point in the horizontal plane through
the base of the pole and distance 20 metres from it, the upper part of the pole subtends an angle whose tangent is 1 . The
2
possible heights of the pole are [IIT 1964]
(a) 20m and 20 3 m (b) 20 m and 60 m (c) 16 m and 48 m (d) None of these
H cot α =d and H cot β =d or H = tan α and H = tan β
3 3d d
HH
1 1 d − 3d H2 4H 2h/3
⇒ tan(β − α) = 2 ⇒ 2 = H2 ⇒ 1+ 3d 2 = d h/3
3d 2
1+ β
θα
⇒ H 2 − 4dH + 3d2 = 0
d = 20m
⇒ H 2 − 80H + 3(400) = 0 ⇒ H = 20 or 60 m.
A vertical pole (more than 100 m high) consists of two portions the lower being 1 rd of the whole. If the upper portion
3
subtends an angle tan−1 1 at a point in a horizontal plane through the foot of the pole and distance 40 ft. from it, then the
2
height of the pole is [AMU 1981]
(a) 100 ft. (b) 120 ft. (c) 150 ft. (d) None of these
Solution: (b) Obviously from figure, tanα = h ........(i)
120
Example: 16
Solution: (b) tan β = 3h , ........(ii)
Example: 17 120
Solution: (b) 2h/3
Therefore, tanθ = tan(β − α) h/3
3h h θα β
1 120 − 120
⇒ 2 = ⇒ h = 120, 40 40
1 + 3h2
14400
But h = 40 can not be taken according to the condition, therefore h = 120 ft. .
20 metre high flag pole is fixed on a 80 metre high pillar, 50 metre away from it, on a point on the base of pillar the flag
pole makes an angle α, then the value of tan α is [MP PET 2003]
(a) 2 (b) 2 (c) 21 (d) 21
11 21 2 4
Let ∠BAC = β ∴ tan β = 80 D
50
Now tan(α + β) = 100 20m
50 C
80m
tan α + tan β 2 tan α + 8 2 tan α 2 . α
1 − tan α. tan β 5 21 B
⇒ = ⇒ 8 = ⇒ = β
5 A 50m
1 tan α
−
The top of a hill observed from the top and bottom of a building of height h is at the angle of elevation p and q
respectively. The height of the hill is [UPSEAT 2001]
(a) h cot q (b) h cot p (c) h tan p (d) None of these
cot q − cot p cot p − cot q tan p − tan q
Let AD be the building of height h and BP be the hill, then P
tan q = h+ x and tan p = x
y y
⇒ tan q = h+ x ⇒ x cot p = (h + x) cot q x
x cot p
C
⇒ x = h cot q q ⇒ h + x = h cot p q p
cot p − cot cot p − cot D h
q y B
A
Example: 18 The angular depressions of the top and foot of a chimney as seen from the top of a second chimney, which is 150 m high
and standing on the same level as the first are θ and φ respectively, then the distance between their tops when
tanθ = 4 and tan φ = 5 is [IIT 1965]
3 2
(a) 150 metres (b) 100 3metres (c) 150 metres (d) 100metres
3
Solution: (d) d = 150 cot φ = 60m
Also, h = 60 tanθ = 80m x
h
Hence, x = 802 + 602 = 100 m.
150
θ
φ
d
Example: 19 The angle of elevation of a cliff at a point A on the ground and a point B, 100 m vertically at A are α and β respectively.
Solution: (c)
The height of the cliff is [EAMCET 1986]
Example: 20
Solution: (a) (a) 100 cot α (b) 100 cot β (c) 100 cot β (d) 100 cot β
cot α − cot β cot α − cot β cot β − cot α cot β + cot α
If OP = h, then CP = h − 100 P
Now, equate the values of OA and BC
h cot α = (h − 100) cot β h – 100
∴ h= 100 cot β . Bβ C
cot β − cot α 100 100
O
α y
A
For a man, the angle of elevation of the highest point of the temple situated east of him is 60o . On walking 240 metres to
north, the angle of elevation is reduced to 30o, then the height of the temple is [MP PET 2003]
(a) 60 6m (b) 60m (c) 50 3 m (d) 30 6m
Total distance from temple = x2 + (240)2 N
where x= h = h 240° 30°
tan 60o 3 h
So distance = h2 + (240)2 , but h1 ⇒ h2 1 W 60° E
3 = = x
3 h2 3
h2 + (240)2 3 + (240)2
3
After solving h = 60 6 metre. S
Example: 21 Two men are on the opposite side of a tower. They measure the angles of elevation of the top of the tower 45o and 30o
respectively. If the height of the tower is 40m, find the distance between the men [Karnataka CET 1998]
(a) 40 m (b) 40 3m (c) 68.280 m (d) 109.28 m
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