mathsvol1ans

51
√ π π π π π
11. Prove that 32( 3) sin cos cos cos cos = 3.
48 48 24 12 6
Exercise 3.6:

1. Express each of the following as a sum or difference
◦
◦
(i) sin 35 cos 28 (ii) sin 4x cos 2x (iii) 2 sin 10θ cos 2θ (iv) cos 5θ cos 2θ (v) sin 5θ sin 4θ.
Solution:
1 1
◦
◦
◦
◦
◦
(i) sin 35 cos 28 ◦ = {sin(35 + 28) + sin(35 − 28) } = {sin(63) + sin(7) }
2 2
1 1
(ii) sin 4x cos 2x = {sin(4x + 2x) + sin(4x − 2x)} = {sin 6x + sin 2x}
2 2
(iii) 2 sin 10θ cos 2θ = sin(10θ + 2θ) + sin(10θ − 2θ) = sin 12θ + sin 8θ
1 1
(iv) cos 5θ cos 2θ = {cos(5θ + 2θ) + cos(5θ − 2θ)} = {cos 7θ + cos 3θ}
2 2
1 1
(v) sin 5θ sin 4θ = {cos(5θ − 4θ) + cos(5θ + 4θ)} = − {cos 9θ − cos θ}
2 2
2. Express each of the following as a product
◦
◦
◦
◦
◦
◦
(i) sin 75 − sin 35 ◦ (ii) cos 65 + cos 15 (iii) sin 50 + sin 40 ◦ (iv) cos 35 − cos 75 .
Solution:

75 − 35 75 + 35
◦
◦
(i) sin 75 − sin 35 ◦ = 2 sin cos = 2 sin 20 cos 55 ◦
2 2

65 + 15 65 − 15
◦
◦
◦
(ii) cos 65 + cos 15 = 2 cos cos = 2 cos 40 cos 25 ◦
2 2

50 + 40 50 − 40
◦
◦
(iii) sin 50 + sin 40 ◦ = 2 sin cos = 2 sin 45 cos 5 ◦
2 2

75 + 35 75 − 35
◦
◦
◦
(iv) cos 35 − cos 75 = 2 sin sin = 2 sin 55 sin 20 ◦
2 2
1
◦
◦
◦
3. Show that sin 12 sin 48 sin 54 = .
8
Solution:

1
sin 12 (sin 48 cos 36) = (cos 36 − cos 60) cos 36
2
1 1

2
= cos 36 − cos 36
2 4
√ √
   
! 2
1  5 + 1 1 5 + 1 1
=   −  =
4 4 4 8
2 
π 2π 3π 4π 5π 6π 7π 1
4. Show that cos cos cos cos cos cos cos = .
15 15 15 15 15 15 15 128
Solution:

52

π
cos π cos 2π cos 3π cos 4π cos 5π cos 6π cos 7π = cos π cos 2π cos 3π cos 4π cos cos 6π cos 7π
15 15 15 15 15 15 15 15 15 15 15 3 15 15
= cos π cos 2π cos 3π cos 4π 1 cos 6π cos π − 8π
15 15 15 15 2 15 15
1
= − cos π cos 2π cos 4π cos 8π cos 3π cos 6π
2 15 15 15 15 15 15
= − 1 sin 2π sin 4π sin 8π sin 16π 3π cos 6π
15
15
15
15
8π cos
2 2 sin π 2 sin 2π 2 sin 4π 2 sin 15 15
15 15 15 15
= − 1 sin 16π π cos 3π cos 2 3π
15
4
2 2 sin 15 15
15
π
sin(π+ ) sin 6π sin 12π
1 15 15 15
= − 5 π 3π 6π
2 sin 2 sin 2 sin
15 15 15
sin(π− )
3π
= 1 15
2 7 sin 3π
15
= 1
128
sin 8x cos x − sin 6x cos 3x
5. Show that = tan 2x.
cos 2x cos x − sin 3x sin 4x
Solution:
sin 8x cos x − sin 6x cos 3x 2 sin 8x cos x − 2 sin 6x cos 3x
=
cos 2x cos x − sin 3x sin 4x 2 cos 2x cos x − 2 sin 3x sin 4x
sin 9x + sin 7x − sin 9x − sin 3x
=
cos 3x + cos x − cos x + cos 7x
sin 7x − sin 3x
=
cos 3x + cos 7x
2 sin 2x cos 5x
=
2 cos 5x cos 2x
= tan 2x
(cos θ − cos 3θ) (sin 8θ + sin 2θ)
6. Show that = 1.
(sin 5θ − sin θ) (cos 4θ − cos 6θ)
Solution:
(cos θ − cos 3θ) (sin 8θ + sin 2θ) (2 sin 2θ sin θ) (2 sin 5θ cos 3θ)
= = 1
(sin 5θ − sin θ) (cos 4θ − cos 6θ) (2 sin 2θ cos 3θ) (2 sin 5θ sin θ)
7. Prove that sin x + sin 2x + sin 3x = sin 2x (1 + 2 cos x).

Solution:
sin 2x + sin x + sin 3x = sin 2x + 2 sin 2x cos x = sin 2x (1 + 2 cos x)


sin 4x + sin 2x
8. Prove that = tan 3x.
cos 4x + cos 2x

Solution:
sin 4x + sin 2x 2 sin 3x cos x
= = tan 3x
cos 4x + cos 2x 2 cos 3x cos x


9. Prove that 1 + cos 2x + cos 4x + cos 6x = 4 cos x cos 2x cos 3x.

Solution:

53

2
1 + cos 6x + cos 2x + cos 4x = 2 cos 3x + 2 cos 3x cos x
= 2 cos 3x (cos 3x + cos x)

= 2 cos 3x (2 cos 2x cos x)
= 4 cos x cos 2x cos 3x

θ 7θ 3θ 11θ
10. Prove that sin sin + sin sin = sin 2θ sin 5θ.
2 2 2 2
Solution:

θ 7θ 3θ 11θ
2 sin sin + 2 sin sin = (cos 3θ − cos 4θ) + (cos 4θ − cos 7θ)
2 2 2 2
= cos 3θ − cos 7θ

= 2 sin 5θ sin 2θ


θ 7θ 3θ 11θ
Hence sin sin + sin sin = sin 2θ sin 5θ
2 2 2 2
1
◦
◦
◦
◦
11. Prove that cos (30 − A) cos (30 + A) + cos (45 − A) cos (45 + A) = cos 2A + .
4
Solution:
◦
◦
◦
◦
◦
◦
2 cos (30 − A) cos (30 + A) + 2 cos (45 − A) cos (45 + A) = (cos 60 + cos 2A) + (cos 90 + cos 2A)
1
= + 2 cos 2A
2
1
◦
◦
◦
◦
Hence cos (30 − A) cos (30 + A) + cos (45 − A) cos (45 + A) = + cos 2A
4
sin x + sin 3x + sin 5x + sin 7x
12. Prove that = tan 4x.
cos x + cos 3x + cos 5x + cos 7x
Solution:
sin x + sin 7x + sin 3x + sin 5x 2 sin 4x cos 3x + 2 sin 4x cos x
=
cos x + cos 7x + cos 3x + cos 5x 2 cos 4x cos 3x + 2 cos 4x cos x
2 sin 4x (cos 3x + cos x)
=
2 cos 4x (cos 3x + cos x)
= tan 4x
sin (4A − 2B) + sin (4B − 2A)
13. Prove that = tan (A + B).
cos (4A − 2B) + cos (4B − 2A)
Solution:
sin (4A − 2B) + sin (4B − 2A) 2 sin(A + B) cos(3A − 3B)
= = tan (A + B)
cos (4A − 2B) + cos (4B − 2A) 2 cos(A + B) cos(3A − 3B)
4 cos 2A
◦
◦
14. Show that cot (A + 15 ) − tan (A − 15 ) = .
1 + 2 sin 2A
Solution:

54

◦
◦
cos (A + 15 ) sin (A − 15 )
◦
◦
cot (A + 15 ) − tan (A − 15 ) = −
◦
◦
sin (A + 15 ) cos (A − 15 )
◦
◦
◦
◦
cos (A + 15 ) cos (A − 15 ) − sin (A − 15 ) sin (A + 15 )
=
◦
◦
sin (A + 15 ) cos (A − 15 )
cos 2A
=
◦
◦
sin (A + 15 ) cos (A − 15 )
2 cos 2A
=
◦
◦
2 sin (A + 15 ) cos (A − 15 )
2 cos 2A
=
sin 2A + sin 30 ◦
4 cos 2A
=
1 + 2 sin 2A
Exercise 3.7:
◦
1. If A + B + C = 180 , prove that
(i) sin 2A + sin 2B + sin 2C = 4 sin A sin B sin C
Solution:
sin 2A + sin 2B + sin 2C = sin 2A + 2 sin(B + C) cos(B − C)
= 2 sin A cos A + 2 sin A cos(B − C)
= 2 sin A (cos A + cos(B − C))

A + B − C A − B + C
= 4 sin A cos cos
2 2

π − 2C π − 2B
= 4 sin A cos cos
2 2
= 4 sin A sin C sin B
A B C
(ii) cos A + cos B − cos C = −1 + 4 cos cos sin
2 2 2
Solution:

A + B A − B
cos A + cos B − cos C = 2 cos cos − cos C
2 2

C A − B C
= 2 sin cos − 1 + 2 sin 2
2 2 2

C A − B C
= 2 sin cos + sin − 1
2 2 2

C A − B A + B
= 2 sin cos + cos − 1
2 2 2

C A B
= 2 sin 2 cos cos − 1
2 2 2
2
2
2
(iii) sin A + sin B + sin C = 2 + 2 cos A cos B cos C
Solution:

55

2
2
2
2 sin A + 2 sin B + 2 sin C = 3 − cos 2A − cos 2B − cos 2C
= 3 − cos 2A − (2 cos(B + C) cos(B − C))
2
= 3 − 2 cos A + 1 − 2 cos A cos(B − C)
= 4 − 2 cos A (cos A + cos(B − C))

= 4 − 2 cos A (cos(B + C) + cos(B − C))

= 4 + 2 cos A (2 cos B cos C)



2
2
2
(iv) sin A + sin B − sin C = 2 sin A sin B cos C
Solution:
2
2
2
2 sin A + 2 sin B − 2 sin C = 1 − cos 2A − cos 2B + cos 2C
= 1 − cos 2A + (2 sin(B + C) sin(B − C))
2
= 1 + 2 sin A − 1 + 2 sin A sin(B − C)

= 2 sin A (sin A + sin(B − C))

= 2 sin A (sin(B + C) + sin(B − C))

= 2 sin A (2 sin B cos C)



A B B C C A
(v) tan tan + tan tan + tan tan = 1
2 2 2 2 2 2
Solution:

A + B + C = π
A B C π
+ + =
2 2 2 2
A B π C
+ = −
2 2 2 2

A B π C
tan + = tan −
2 2 2 2
A B
tan + tan
2 2 = cot C
A B 2
1 − tan tan
2 2
C A C B A B
tan tan + tan tan = 1 − tan tan
2 2 2 2 2 2
A B B C C A
tan tan + tan tan + tan tan = 1
2 2 2 2 2 2
A B C
(vi) sin A + sin B + sin C = 4 cos cos cos
2 2 2
Solution:

56


B + C B − C
sin A + sin B + sin C = sin A + 2 sin cos
2 2
A A A B − C
= 2 sin cos + 2 cos cos
2 2 2 2

A A B − C
= 2 cos sin + cos
2 2 2

A B + C B − C
= 2 cos cos + cos
2 2 2

A B C
= 2 cos 2 cos cos
2 2 2
(vii) sin(B + C − A) + sin(C + A − B) + sin(A + B − C) = 4 sin A sin B sin C.

Solution:
sin(B + C − A) + sin(C + A − B) + sin(A + B − C) = sin(π − 2A) + sin(π − 2B) + sin(π − 2C)

= sin 2A + sin 2B + sin 2C

= 2 sin A cos A + 2 sin(B + C) cos(B − C)

= 2 sin A cos A + 2 sin(A) cos(B − C)

= 2 sin A (cos A + cos(B − C))
= 2 sin A (− cos(B + C) + cos(B − C))

= 2 sin A (2 sin B sin C)

= 4 sin A sin B sin C

2. If A + B + C = 2s, then prove that sin(s − A) sin(s − B) + sin s sin(s − C) = sin A sin B.


Solution:
2 sin(s − A) sin(s − B) + 2 sin s sin(s − C)

= (cos(B − A) − cos(C)) + (cos(C) − cos(A + B))

= cos(B − A) − cos(B + A)

= 2 sin B sin A
2x 2y 2z 2x 2y 2z
3. If x + y + z = xyz, then prove that + + = .
2
2
1 − x 2 1 − y 2 1 − z 2 1 − x 1 − y 1 − z 2
Solution: Let x = tan A, y = tan B, z = tan C.

57

tan A + tan B + tan C = tan A tan B tan C

tan A + tan B = − tan C{1 − tan A tan B}

tan A + tan B
= − tan C
1 − tan A tan B
tan (A + B) = tan (−C)

A + B + C = π

2A + 2B + 2C = 2π

tan 2A + tan 2B + tan 2C = tan 2A tan 2B tan 2C

2 tan A 2 tan B 2 tan C 2 tan A 2 tan B 2 tan C
+ + =
2
2
2
2
2
2
1 − tan A 1 − tan B 1 − tan C 1 − tan A 1 − tan B 1 − tan C
2x 2y 2z 2x 2y 2z
+ + =
1 − x 2 1 − y 2 1 − z 2 1 − x 1 − y 1 − z 2
2
2
π
4. If A + B + C = , prove the following
2
(i) sin 2A + sin 2B + sin 2C = 4 cos A cos B cos C
(ii) cos 2A + cos 2B + cos 2C = 1 + 4 sin A sin B cos C.
Solution:
sin 2A + sin 2B + sin 2C = sin 2A + 2 sin(B + C) cos(B − C)
π
= 2 sin A cos A + 2 sin( − A) cos(B − C)
2
= 2 sin A cos A + 2 cos A cos(B − C)
= 2 cos A (sin A + cos(B − C))

= 2 cos A (cos(B + C) + cos(B − C))

= 2 cos A (2 cos B cos C)

= 4 cos A cos B cos C

cos 2A + cos 2B + cos 2C = cos 2A + 2 cos(B + C) cos(B − C)

2
= 1 − 2 sin A + 2 sin A cos(B − C)
= 1 − 2 sin A (sin A − cos(B − C))

= 1 − 2 sin A (cos(B + C) − cos(B − C))

= 1 + 2 sin A (2 sin B sin C)

= 1 + 4 sin A sin B sin C

π
5. If 4ABC is a right triangle and if ∠A = , then prove that
2
2
2
(i) cos B + cos C = 1
Solution:

58

2
2
(ii) sin B + sin C = 1
√ B C
(iii) cos B − cos C = −1 + 2 2 cos sin .
2 2
Exercise 3.8:
1. Find the principal solution and general solutions of the following:
1 √ 1
(i) sin θ = −√ (ii) cot θ = 3 (iii) tan θ = −√ .
2 3

Solution:
1
(i) sin θ = −√ < 0 principal value lies in the IV quadrant
2
1 π n π
sin θ = −√ = − sin θ = nπ + (−1) − . n ∈ Z
2 4 4
√ 1
(ii) cot θ = 3 ⇒ tan θ = √ > 0 principal value lies in the I quadrant
3
1 π π
tan θ = √ = tan θ = nπ + n ∈ Z
3 6 6
1
(iii) tan θ = −√ < 0 principal value lies in the IV quadrant
3
1 π π
tan θ = −√ = − tan θ = nπ − n ∈ Z
3 6 6
◦
2. Solve the following equations for which solutions lies in the interval 0 ≤ θ < 360 ◦
4
2
2
2
4
2

(i) sin x = sin x sin x − sin x = 0 ⇒ sin x sin x − 1 = 0
sin (x) = 1, sin (x) = −1, sin (x) = 0
π 3π
sin (x) = 1 ⇒ x = : sin (x) = −1 ⇒ x = : sin (x) = 0 ⇒ x = 0 x = π
2 2
1
2
2
(ii) 2 cos x + 1 = −3 cos x 2 cos x + 3 cos x + 1 = 0 ⇒ cos (x) = − , cos (x) = −1
2
1 2π 4π
cos (x) = − ⇒ x = , : cos (x) = −1 ⇒ x = π
2 3 3
1
2
2
(iii) 2 sin x + 1 = 3 sin x 2 sin x − 3 sin x + 1 = 0 ⇒ sin (x) = 1, sin (x) =
2
π 1 π 5π
sin (x) = 1 ⇒ x = : sin (x) = ⇒ x = :
2 2 6 6
(iv) cos 2x = 1 − 3 sin x.

3
2
1 − 2 sin x = 1 − 3 sin x ⇒ sin x sin x − ⇒ sin x = 0 ⇒ x = 0, π
2
3. Solve the following equations:
(i) sin 5x − sin x = cos 3x
2 sin 2x cos 3x = cos 3x ⇒ cos 3x(2 sin 2x − 1) = 0
π π
When cos 3x = 0, 3x = (2n + 1) and hence x = (2n + 1)
2 6
1 π π π
when sin 2x = , 2x = (−1) n + nπ and hence x = (−1) n + n
2 6 12 2

59

2
(ii) 2 cos θ + 3 sin θ − 3 = 0
2
2(1 − sin θ) + 3 sin θ − 3 = 0
2
2 sin θ − 3 sin θ + 1 = 0
sin (θ) = 1, sin (θ) = 1
2
π

When sin (θ) = 1 ⇒ θ = nπ + (−1) n
2
1
π
When sin (θ) = ⇒ θ = nπ + (−1) n
2 6
(iii) cos θ + cos 3θ = 2 cos 2θ

2
3
cos θ + 4 cos θ − 3 cos θ = 4 cos θ − 2
2
3
4 cos θ − 4 cos θ − 2 cos θ + 2 = 0
3
2
2 cos θ − 2 cos θ − cos θ + 1 = 0
2
2(cos θ)(cos θ − 1) − 1(cos θ − 1) = 0
2
(2 cos θ − 1)(cos θ − 1) = 0
1
cos θ = ±√ or cos θ = 1
2
π
Hence θ = 2nπ ± or θ = 2nπ
4
(iv) sin θ + sin 3θ + sin 5θ = 0

sin θ + sin 5θ + sin 3θ = 0
2 sin 3θ cos 2θ + sin 3θ = 0

sin 3θ(2 cos 2θ + 1) = 0


1
Hence sin 3θ = 0 or cos 2θ = −
2
π π
Solution is θ = nπ or 2θ = 2nπ ± ⇒ θ = nπ ± .
6 3

(v) sin 2θ − cos 2θ − sin θ + cos θ = 0

We have (sin 2θ − sin θ) − (cos 2θ − cos θ) = 0
3θ θ θ 3θ
Hence 2 sin cos + 2 sin sin = 0
2 2 2 2

3θ θ θ
sin cos + sin = 0
2 2 2

θ θ
cos + sin = 0
2 2

θ θ
Divide both sides by cos , cos 6= 0
2 2

60


θ θ
cos + sin
2 2 0
=

θ θ
cos cos
2 2

θ
tan + 1 = 0
2

θ
tan + 1 − 1 = 0 − 1
2

θ
tan = −1
2
θ 3π
= + πn
2 4
3π
θ = + 2nπ
2
π

cos θ = 0 ⇒ θ = (2n + 1)
2
√
(vi) sin θ + cos θ = 2
1
Multiplying both sides by √
2

1 1
√ sin θ + √ cos θ = 1
2 2
π π
cos sin θ + sin cos θ = 1
4 4
π π

sin + θ = sin
4 2
π
θ =
4
π
Hence the solution is θ = nπ + (−1) n
√ 4
(vii) sin θ + 3 cos θ = 1
1
Multiplying both sides by
2
1 1 √ 1
sin θ + 3 cos θ =
2 2 2
π π π
cos sin θ + sin cos θ = sin
3 3 6
π π

sin + θ = sin
3 6
π
θ = −
3
π
Hence the solution is θ = nπ + (−1) n
√ 3
(viii) cot θ + cosecθ = 3
Multiplying both sides by sin θ, We get

61
√
cos θ + 1 = 3 sin θ
√ 1
cos θ − 3 sin θ = −1 Multiplying both sides by
2
1 1 √ 1
cos θ − 3 sin θ = −
2 2 2
π π 2π
cos cos θ − sin sin θ = cos
3 3 3
π 2π

cos θ + = cos
3 3
π
θ =
3
π
Hence the solution is θ = 2nπ ±
√

3
(ix) tan θ + tan θ + π + tan θ + 2π = 3
3 3

π 2π

tan θ + tan θ + + tan θ + = 3
3 3
 π   2π 
tan θ + tan tan θ + tan
 = 3
tan θ +  3 π  +  3 

1 − tan θ tan 1 − tan θ tan 2π
√ 3 √ 3
! !
tan θ + 3 tan θ − 3
tan θ + √ + √ = 3
1 − 3 tan θ 1 + 3 tan θ
√ √ √ √
!
(1 + 3 tan θ)(tan θ + 3) + (1 − 3 tan θ)(tan θ − 3)
tan θ + √ √ = 3
(1 − 3 tan θ)(1 + 3 tan θ)
√ √ √ √
!
2
2
tan θ + 3 + 3 tan θ + 3 + tan θ − 3 − 3 tan θ + 3
tan θ + = 3
2
1 − 3 tan θ

8 tan θ
tan θ + = 3
2
1 − 3 tan θ
3
2
tan θ − 3 tan θ + 8 tan θ = 3 − 9 tan θ
3
2
3 tan θ − 9 tan θ − 9 tan θ + 3 = 0
3
2
tan θ − 3 tan θ − 3 tan θ + 1 = 0
2
3
tan θ + 1 − 3 tan θ − 3 tan θ = 0
2
(tan θ + 1)(tan θ − tan θ + 1) − 3 tan θ(tan θ + 1) = 0
2
(tan θ + 1)(tan θ − 4 tan θ + 1) = 0
√ √
tan θ = −1 tan θ = 2 + 3 tan θ = 2 − 3
π √ √
−1
−1
θ = nπ − θ = nπ − tan (2 + 3) θ = nπ − tan (2 − 3)
4
√
5 + 1
(x) cos 2θ =
4
◦
cos 2θ = cos 36 .

62

π
◦
θ = 18 =
5
π
θ = 2nπ ±
5
2
(xi) 2 cos x − 7 cos x + 3 = 0 (2 cos (x) − 1) (cos (x) − 3) = 0
1
cos(x) = or cos(x) = 3
2
π
Since cos(x) = 3 has no solution,the only solution is x = + 2πn
3
Exercise 3.9:
sin A sin(A − B)
2
2
2
1. In a 4ABC, if = , prove that a , b , c are in Arithmetic Progression.
sin C sin(B − C)
Solution:
sin A sin(B − C) = sin C sin(A − B)

sin(B + C) sin(B − C) = sin(A + B) sin(A − B)
2
2
2
2
sin B − sin C = sin A − sin B
2
2
2
2
2
2
R (b − c ) = R (a − b )
2
2b 2 = a + c 2
2
2
2
Hence a , b , c are in Arithmetic Progression.
√ √
2. The angles of a triangle ABC, are in Arithmetic Progression and if b : c = 3 : 2, find ∠A.
Solution:
Since ∠A, ∠B, ∠C are in A.P, we have 2B = A + C.
π π 2π
But A + B + C = π ⇒ 2B + B = π. That is, B = . Hence A + C = π − =
3 3 3
sin B sin C
From Sine Rule, = .
b c

2π
sin − A
sin B 3
We have =
b c

63


2π c
sin − A = sin B
3 b
√
!
2
π
= √ sin
3 3
√ √
!
2 3
= √
3 2
1
= √
2
π

= sin
4
2π π
− A =
3 4
8π − 5π π
A = =
12 4

sin A
3. In a 4ABC, if cos C = , show that the triangle is isosceles.
2 sin B
Solution:

2 sin B cos C = sin A
sin(B + C) + sin(B − C) = sin A

sin A + sin(B − C) = sin A

sin(B − C) = 0

B − C = 0

B = C
sin B c − a cos B
4. In a 4ABC, prove that = .
sin C b − a cos C
Solution:
c − a cos B 2R sin C − 2R sin A cos B
=
b − a cos C 2R sin B − 2R sin A cos C
2 sin C − sin(A + B) − sin(A − B)
=
2 sin B − sin(A + C) − sin(A − C)
2 sin C − sin C − sin(A − B)
=
2 sin B − sin B − sin(A − C)
sin C − sin(A − B)
=
sin B − sin(A − C)
sin(A + B) − sin(A − B)
=
sin(A + C) − sin(A − C)
2 sin B cos A
=
2 sin C cos A
sin B
=
sin C
5. In a 4ABC, prove that a cos A + b cos B + c cos C = 2a sin B sin C.

64

Solution:
a cos A + b cos B + c cos C = 2R sin A cos A + 2R sin B cos B + 2R sin C cos C Using Sine Rule

= R (sin 2A + sin 2B + sin 2C)

= R (2 sin(A + B) cos(A − B) + sin 2C)

= R (2 sin C cos(A − B) + 2 sin C cos C)

= 2R sin C (cos(A − B) − cos(A + B))
= 2R sin C (2 sin A sin B)

= 2R sin A sin C sin B

= a sin C sin B

B − C
◦
6. In a 4ABC, ∠A = 60 . Prove that b + c = 2a cos .
2
Solution:
b + c (sin B + sin C)
= R
2a 2R sin(A)

B + C B − C
2 sin cos
2 2
=
A A
4 sin cos
2
2
A B − C
cos cos
2 2
=
A A
2 sin cos
2 2
B − C
cos
= 2
2 sin(30) ◦
B − C
b + c = 2a cos
2
7. In a 4ABC, prove the following

A A
(i) a sin + B = (b + c) sin
2 2
Solution:
B + C B − C A B − C
2 sin cos 2 cos cos
b + c k sin B + k sin C sin B + sin C 2 2 2 2
= = = =
a k sin A sin A A A A A
2 sin cos 2 sin cos
2 2 2 2

A B − C B + C A π A
(b + c) sin = a cos = a cos B − = a cos B + − = a sin B +
2 2 2 2 2 2
A
(ii) a(cos B + cos C) = 2(b + c) sin 2
2
Solution:

65

a k sin A
=
b + c k sin B + k sin C
A A
2 sin cos
= 2 2
B + C B − C
2 sin cos
2 2
A A
2 sin cos
= 2 2
A B − C
2 cos cos
2 2
A
sin
= 2
B − C
cos
2
A B + C
sin cos
= 2 2
B + C B − C
cos cos
2 2
A B + C
2 sin cos
= 2 2
B + C B − C
2 cos cos
2 2
A B + C
2 sin cos
= 2 2
cos B + cos C
A
2 sin 2
= 2
cos B + cos C

A
Hence a(cos B + cos C) = 2(b + c) sin 2
2
2
a − c 2 sin(A − C)
(iii) =
b 2 sin(A + C)
Solution:
2
a − c 2 (a + c)(a − c)
=
b sin(A − C) b sin(A − C)
2
2
R 2 sin A − sin C
=
R sin(A + C) sin(A − C)
2
2
sin A − sin C
= R
2
2
sin A − sin C
= R
2
a − c 2 sin(A − C)
Hence = Rb sin(A − C) =
b 2 sin(A + C)
a sin(B − C) b sin(C − A) c sin(A − B)
(iv) = =
2
2
2
b − c 2 c − a 2 a − b 2
Solution:

66

a sin(B − C) R sin A sin(B − C)
=
2
2
2
b − c 2 R 2 sin B − sin C
sin(B + C) sin(B − C)
=
R sin(B + C) sin(B − C)
1
=
R
b sin(C − A)
a sin(B−C)
Because of symmetricity, we can easily show that = =
2
b −c 2 c − a 2
2
c sin(A − B) 1
= .
2
a − b 2 R

a + b A + B A − B
(v) = tan cot .
a − b 2 2
Solution:
a + b R (sin A + sin B)
=
a − b R (sin A − sin B)
A + B A − B
2 sin cos
= 2 2
A − B A + B
2 sin cos
2 2
A + B A − B
   
sin cos
2 2
   
A + B A − B
=    
cos sin
2
2
A + B A − B
= tan cot
2 2
2
2
2
2
2
2
8. In a 4ABC, prove that (a − b + c ) tan B = (a + b − c ) tan C.
Solution:
tan B sin B cos C 2 sin B cos C
= =
tan C cos B sin C 2 cos B sin C
2
2
a + b − c 2
sin B
= 2ab
2
2
a + c − b 2
sin C
2ac
2
2
2
c sin B (a + b − c )
=
2
2
2
b sin C (a + c − b )
2
2
a + b − c 2
=
2
2
a + c − b 2
9. An Engineer has to develop a triangular shaped park with a perimeter 120 m in a village. The park
to be developed must be of maximum area. Find out the dimensions of the park.
Solution:
For a fixed perimeter, the equilateral triangle has the maximum area and the maximum area
s 2
is given by ∆ = √ sq.units. Here s = 120 m. Hence the area of the park is given by
3 3
√
14400 1600 3
∆ = √ = Clearly the sides of the park are equal and given by 40 m.
3 3 .

67



10. A rope of length 12 m is given. Find the largest area of the triangle formed by this rope and find
the dimensions of the triangle so formed.

Solution:

Considering the length of rope as perimeter,the largest area of the triangle formed by this rope is
s 2
an equilateral triangle. The largest area formed is given by ∆ = √ sq.units. Here s = 12 m.
3 3
144 √
Hence the area is given by ∆ = √ = 16 3. Clearly the sides are equal and given by 4 m.
3 3


11. Derive Projection formula from (i) Law of sines, (ii) Law of cosines.

Solution:
Projection Formula gives the relation between angles and sides of a triangle. We can find the length
of a side of the triangle if other two sides and corresponding angles are given using projection
formula. If a, b and c be the length of sides of a triangle and A, B and C are angles opposite to the
sides respectively, then projection formula is given below:
a = b cos C + c cos B

b = c cos A + a cos C

c = a cos B + b cos A

Let us prove the first one using law of cosines.
2
2
2
a = b + c + 2bc cos(A)
2
2
= b + c + 2bc cos(B + C)
2
2
= b + c + 2bc (cos B cos C − sin B sin C) − (b sin C − c sin B) 2
2
2
2
2
2
2
= b + c + 2bc cos B cos C − 2bc sin B sin C − b sin C − c sin B + 2bc sin B sin C
2
2
2
2
2
2
= b + c + 2bc cos B cos C − b sin C − c sin B

2
2

= b 2 1 − sin C + c 2 1 − sin B + 2bc cos C cos B
2
2
2
2
= b cos C + c cos B + 2bc cos C cos B
= (b cos C + c cos B) 2
Taking square root on both sides, the result follows. Let us prove using Law of Sines
a cos B + b cos A = R sin A cos B + R sin B cos A
= R sin(A + B)
= R sin C
= c
Exercise 3.10:

68

1. Determine whether the following measurements produce one triangle, two triangles or no triangle:
◦
∠B = 88 , a = 23, b = 2. Solve if solution exists.
Solution:
◦
Let h = a sin B = 23 sin 88 = 23(0.99939082701) = 22.9859890214..
Since b < h triangle is not possible.
2. If the sides of a 4ABC are a = 4, b = 6 and c = 8, then show that 4 cos B + 3 cos C = 2.

Solution:
2
2
a + c − b 2 16 + 64 − 36
cos B = = = 0.6875
2ac 64
2
2
a + b − c 2 16 + 36 − 64
cos C = = = −0.25
2ab 48
Now,4 cos B + 3 cos C = 4(0.6875) − 3(0.25) = 2
√ √
◦
3. In a 4ABC, if a = 3 − 1, b = 3 + 1 and C = 60 , find the other side and other two angles.
Solution:
√ 2 √ 2 √ √
1
2
2
2
c = a + b − 2ab cos C = 3 − 1 + 3 + 1 − 2 3 − 1 3 + 1
√ √ 2
= 3 + 1 − 2 3 + 3 + 1 + 2 3 − (3 − 1) = 6
. Hence c = 2.
√ √ √
a sin C 3 3 − 1 3 − 1
◦
sin A = = √ = √ = sin 15 .
c 2 6 2 2
◦
◦
B = 180 − (A + C) = 105 .
2
2
b + c − a 2
4. In any 4ABC, prove that the area 4 = .
4 cos A
Solution:
1
Area of the triangle = bc sin A
2
2bc sin A
=
4
2
2
b + c − a 2
= sin A
4 cos A
◦
5. In a 4ABC, if a = 12 cm, b = 8 cm and C = 30 , then show that its area is 24 sq.cm.
Solution:
1 1
◦
Area of the Triangle = ab sin C = (12)(8) sin 30 = 24 sq.cm.
2 2

6. In a 4ABC, if a = 18 cm, b = 24 cm and c = 30 cm, then show that its area is 216 sq.cm.

Solution:
p p
Area of the triangle = s(s − a)(s − b)(s − c) = 36(18)(12)(6) = 216 sq.cm.
7. Two soldiers A and B in two different underground bunkers on a straight road, spot an intruder at
the top of a hill. The angle of elevation of the intruder from A and B to the ground level in the

69

◦
◦
eastern direction are 30 and 45 respectively. If A and B stand 5km apart, find the distance of the
intruder from B.
Solution:
◦
Let CD be the height of the hill. Angle of elevation from A is α = 30 and angle of elevation
◦
from B is β = 45 .Distance between A and B is 5 km.Both elevation are in the same eastern
direction. Let BD be x km. Then AD = 5 + x. (Since elevation from B is more, B is closer to
1 x
the foot of the Hill). Since BD = CD, we have CD = x. Given tan α = √ = . Hence
√ r √ 3 5 + x
5 5 3 + 5 √ 5 3 + 5
x = √ = .Now BC = x = = 2.61km.
3 − 1 2 2
8. A researcher wants to determine the width of a pond from east to west, which cannot be done by
actual measurement. From a point P, he finds the distance to the eastern-most point of the pond
to be 8 km, while the distance to the western most point from P to be 6 km. If the angle between
◦
the two lines of sight is 60 , find the width of the pond.

Solution:


9. Two Navy helicopters A and B are flying over the Bay of Bengal at same altitude from the sea
level to search a missing boat. Pilots of both the helicopters sight the boat at the same time while
they are apart 10 km from each other. If the distance of the boat from A is 6 km and if the line
◦
segment AB subtends 60 at the boat, find the distance of the boat from B.
Solution:



10. A straight tunnel is to be made through a mountain. A surveyor observes the two extremities A and
B of the tunnel to be built from a point P in front of the mountain. If AP = 3km, BP = 5 km
◦
and ∠APB = 120 , then find the length of the tunnel to be built.
Solution:



11. A farmer wants to purchase a triangular shaped land with sides 120feet and 60feet and the
◦
angle included between these two sides is 60 . If the land costs |500 per sq.ft, find the amount he
needed to purchase the land. Also find the perimeter of the land.

Solution:


12. A fighter jet has to hit a small target by flying a horizontal distance. When the target is sighted,
◦
the pilot measures the angle of depression to be 30 . If after 100 km, the target has an angle of
◦
depression of 45 , how far is the target from the fighter jet at that instant?
Solution:


13. A plane is 1 km from one landmark and 2 km from another. From the planes point of view the
◦
land between them subtends an angle of 60 . How far apart are the landmarks?
Solution:

70

14. A man starts his morning walk at a point A reaches two points B and C and finally back to A such
◦
◦
that ∠A = 60 and ∠B = 45 , AC = 4km in the 4ABC. Find the total distance he covered
during his morning walk.
Solution:



15. Two vehicles leave the same place P at the same time moving along two different roads. One
vehicle moves at an average speed of 60km/hr and the other vehicle moves at an average speed
of 80 km/hr. After half an hour the vehicle reach the destinations A and B. If AB subtends 60 ◦
at the initial point P, then find AB.

Solution:



16. Suppose that a satellite in space, an earth station and the centre of earth all lie in the same plane.
Let r be the radius of earth and R be the distance from the centre of earth to the satellite. Let d be
◦
the distance from the earth station to the satellite. Let 30 be the angle of elevation from the earth
station to the satellite. If the line segment connecting earth station and satellite subtends angle α
r
r r
2
at the centre of earth, then prove that d = R 1 + − 2 cos α.
R R
Solution:



Exercise 3.11:
√
1 3 √
−1
1. Find the principal value of (i) sin −1 √ (ii) cos −1 (iii) cosec (−1) (iv) sec −1 − 2
√ 2 2

(v) tan −1 3 .
Solution:
1 π π
(i) Let sin −1 √ = y, where − ≤ y ≤
2 2 2

1 π π
⇒ sin y = √ = sin ⇒ y =
2 4 4
1 π
Thus, the principal value of sin −1 √ =
2 4


1 π π
(i) Let sin −1 √ = y, where − ≤ y ≤
2 2 2
1 π π
⇒ sin y = √ = sin ⇒ y =
2 4 4
1 π
Thus the principal value of sin −1 √ =
√ 2 4
3 π π
(ii) Let cos −1 = y, where − ≤ y ≤
2 2 2
√
3 π π
⇒ cos y = = cos ⇒ y =
2 6 6
2. A man standing directly opposite to one side of a road of width x meter views a circular shaped
traffic green signal of diameter a meter on the other side of the road. The bottom of the green

71

signal is b meter height from the horizontal level of viewer’s eye. If α denotes the angle subtended
by the diameter of the green signal at the viewer’s eye, then prove that


a + b b
α = tan −1 − tan −1 .
x x


Solution:

72

Exercise 3.12:




Choose the correct answer or most suitable answer:
√
1 3
1. − =
cos 80 ◦ sin 80 ◦
√ √
(1) 2 (2) 3 (3) 2 (4) 4
◦
◦
◦
3
2. If cos 28 + sin 28 = k , then cos 17 is equal to
k 3 k 3 k 3 k 3
(1) √ (2) −√ (3) ±√ (4) −√
2 2 x x 2 3
2
2
3. The maximum value of 4 sin x + 3 cos x + sin + cos is
2 2
√ √
(1) 4 + 2 (2) 3 + 2 (3) 9 (4) 4

π 3π 5π 7π

4. 1 + cos 1 + cos 1 + cos 1 + cos =
8 8 8 8
1 1 1 1
(1) (2) (3) √ (4) √
8 2 3 2
3π q √
5. If π < 2θ < , then 2 + 2 + 2 cos 4θ equals to
2
(1) −2 cos θ (2) −2 sin θ (3) 2 cos θ (4) 2 sin θ
◦
tan 140 − tan 130 ◦
◦
6. If tan 40 = λ, then =
◦
1 + tan 140 tan 130 ◦
1 − λ 2 1 + λ 2 1 + λ 2 1 − λ 2
(1) (2) (3) (4)
λ λ 2λ 2λ
◦
◦
◦
◦
7. cos 1 + cos 2 + cos 3 + . . . + cos 179 =
(1) 0 (2) 1 (3) −1 (4) 89
1 k k
8. Let f k (x) = sin x + cos x where x ∈ R and k ≥ 1. Then f 4 (x) − f 6 (x) =
k
1 1 1 1
(1) (2) (3) (4)
4 12 6 3
9. Which of the following is not true?
3 1
(1 ) sin θ = − (2) cos θ = −1 (3) tan θ = 25 (4) sec θ =
4 4
2
2
10. cos 2θ cos 2φ + sin (θ − φ) − sin (θ + φ) is equal to
(1) sin 2(θ + φ) (2) cos 2(θ + φ) (3) sin 2(θ − φ) (4) cos 2(θ − φ)
sin(A − B) sin(B − C) sin(C − A)
11. + + is
cos A cos B cos B cos C cos C cos A

(1) sin A + sin B + (2) 1 (3) 0 (4) cos A + cos B +
sin C cos C
12. If cos pθ + cos qθ = 0 and if p 6= q, then θ is equal to (n is any integer)
π(3n + 1) π(2n + 1) π(n ± 1) π(n + 2)
(1) (2) (3) (4)
p − q p ± q p ± q p + q
sin(α + β)
2
13. If α and β are the roots of tan x + a tan x + b = 0, then is equal to
sin α sin β

73

b a a b
(1) (2) (3) − (4) −
a b b a
2
2
2
14. In a triangle ABC, sin A + sin B + sin C = 2, then the triangle is
(1) equilateral triangle (2) isosceles triangle (3) right triangle (4) scalene triangle.
15. If f(θ) = |sin θ| + |cos θ| , θ ∈ R, then f(θ) is in the interval
√
h i
(1) [0, 2] (2) 1, 2 (3) [1, 2] (4) [0, 1]
cos 6x + 6 cos 4x + 15 cos 2x + 10
16. is equal to
cos 5x + 5 cos 3x + 10 cos x
2
(1) cos x (2) 1+cos x (3) cos 3x (4) cos 2x
17. The triangle of maximum area with constant perimeter 12m
(1) is an equilateral tri- (2) is an isosceles (3) is a triangle with (4) Does not exist.
angle with side 4m triangle with sides sides 3m, 4m, 5m
2m, 5m, 5m
18. A wheel is spinning at 2 radians/second. How many seconds will it take to make 10 complete
revolutions?
(1) 10π seconds (2) 20π seconds (3) 5π seconds (4) 15π seconds
19. If sin α + cos α = b, then sin 2α is equal to
√ √ √
2
2
2
2
(1) b − 1, if b ≤ 2 (2) b − 1, if b > 2 (3) b − 1, if b ≥ 1 (4) b − 1, if b ≥ 2
20. In a 4ABC,
A B C
(i) sin sin sin > 0
2 2 2
(ii) sin A sin B sin C > 0
(1) Both (i) and (ii) are (2) Only (i) is true (3) Only (ii) is true (4) Neither (i) nor (ii)
true is true.




Exercise - 4.1


1. (a) A person went to a restaurant for dinner. In the menu card, the person saw 10 Indian and 7
Chinese food items. In how many ways the person can select either an Indian or a Chinese
food?

Solution:
The person can select an Indian food or a Chinese food in (10+7) ways =17 ways.



(b) There are 3 types of toy car and 2 types of toy train are available in a shop. Find the number
of ways a baby can buy a toy car and a toy train?

Solution:
Number of ways a baby can buy a toy car and a toy train = 3 × 2 = 6 ways.



(c) How many number of two-digit numbers can be formed using 1,2,3,4,5 without repetition of
digits.
Solution:

74

one place is filled in 5 ways and second place is filled in 4 ways. Hence total number of
two-digit numbers can be formed in 20 ways.


(d) Three persons enter in to a conference hall in which there are 10 seats. In how many ways
they can take their places?

Solution:

They can take their places in 10 × 9 × 8 = 720 ways.


(e) In how many ways 5 persons can be seated in a row?

Solution:

5 persons can be seated in a row in 5 × 4 × 3 × 2 × 1 = 5! = 120 ways.


2. (i) A mobile phone has a passcode of 6 distinct digits. What is the maximum number of
attempts one makes to retrieve the passcode?

Solution:

Maximum number of attempts one makes to retrieve the passcode = 10×9×8×7×6×5 =
1, 51, 200 ways.


(ii) Given four flags of different colours, how many different signals can be generated if each
signal requires the use three flags, one below the other?

Solution:

Number of different signals can be generated in 4 × 3 × 2 = 24 ways.


3. Four children are running a race.
(i) In how many ways can the first two places be filled?

Solution:

The first two places can be filled in 4 × 3 = 12 ways.


(ii) In how many different ways could they finish the race?

Solution:

They can finish the race in 4! = 24 ways.



4. Count the number of three-digit numbers which can be formed from the digits 2,4,6,8 if
(i) repetitions of digits is allowed.

Solution:

Number of three-digit numbers can be formed =4 × 4 × 4 = 64 .

75



(ii) repetitions of digits is not allowed.

Number of three-digit numbers can be formed =4! = 24 .



5. How many three-digit numbers are there with 3 in the unit place? (i) with repetition (ii) without
repetition.

Solution:

(i) with repetition

Unit digit is 3. Hundredth place cannot be filled with zero. Hence the total number = 9 × 10 = 90
.
(ii) without repetition.

Hundredth place can be filled in 8 ways using digits (1, 2, 4, 5, 6, 7, 8, 9). Tens place can be filled
in 8 ways. Hence total ways = 8 × 8 = 64 ways.



6. How many numbers are there between 100 and 500 with the digits 0, 1, 2, 3, 4, 5 if
(i) repetition of digits allowed

When the hundredth digit is 1, the remaining two places filled in 36ways

When the hundredth digit is 2, the remaining two places filled in 36ways
When the hundredth digit is 3, the remaining two places filled in 36ways

When the hundredth digit is 4, the remaining two places filled in 36ways

Total number 144ways


(ii) the repetition of digits is not allowed.

When the hundredth digit is 1, the remaining two places filled in 20ways

When the hundredth digit is 2, the remaining two places filled in 20ways

When the hundredth digit is 3, the remaining two places filled in 20ways

When the hundredth digit is 4, the remaining two places filled in 20ways
Total number 80ways



7. How many three-digit odd numbers can be formed by using the digits 0, 1, 2, 3, 4, 5 if

(i) The repetition of digits is not allowed
The unit digit can be filled in 3 ways. The Hundredth digit can be filled in 4 ways. The Tenth digit
can be filled in 4 ways. Hence total numbers = 3 × 4 × 4 = 48.

(ii) The repetition of digits is allowed.

The unit digit can be filled in 3 ways. The Hundredth digit can be filled in 5 ways. The Tenth digit

76

can be filled in 6 ways. Hence total numbers = 3 × 5 × 6 = 90.



8. Count the numbers between 999 and 10000 subject to the condition that there are

(i) no restriction.
Solution:

Total number = 10000 − 999 = 9001.



(ii) no digit is repeated.
unit place filled in 10 ways. Tens place filled in 9 ways. Hundredth place filled in 8 ways.
Thousandth place in 7 ways. Totally 5040 ways.



(iii) at least one of the digits is repeated.

Total number =10000 − 999 = 9001.
None of the digits repeated = 5040.

Atleast one digit is repeated = 9001 − 5040 = 3961



9. How many three-digit numbers, which are divisible by 5, can be formed using the digits 0, 1, 2, 3,
4, 5 if

(i) The repetition of digits are not allowed?

If unit place is 5 , Hundredth place in 4 ways and Tenth place in 4 ways= 4 × 4 = 16 ways.

If unit place is 0 , Remaining two digits are filled in 5 × 4 = 20 ways.

Hence totally 36 ways.
(ii) The repetition of digits are allowed? If unit place is 5. Since hundredth place cannot be 0,
Remaining two digits are filled in 6 × 5 = 30 ways.

If unit place is 0. Since hundredth place cannot be 0, Remaining two digits are filled in 6×5 = 30
ways.

Total number of ways = 60 ways.


10. To travel from a place A to place B, there are two different bus routes B 1 , B 2 , two different train
0
routes T 1 , T 2 and one air route A 1 . From place B to place C there is one bus route say B , two
1
0 0 0
different train routes say T , T and one air route A . Find the number of routes of commuting
1 2 1
from place A to place C via place B without using similar mode of transportation.
Solution:

77

Number of possible mode of transportation are Three.
Air, Bus and Train.

If from A to B, Air route chosen, from B to C, there are
3 ways to travel = 3

If from A to B, Bus route chosen(2 ways), from B to C,
there are 3 ways = 6

If from A to B, Train route chosen ( 2 ways),from B to
C, there are 2 ways= 4
Totally 13 modes.














0 0 0 0 0 0 0 0 0 0 0
They are,{(A 1 B ), (A 1 T ), (A 1 T ), (B 1 A ), (B 1 T ), (B 1 T ), (B 2 A ), (B 2 T ), (B 2 T ), (T 1 A ), (T 1 B ),
2
1
1
2
1
1
1
1
1
2
1
0 0
(T 2 A ), (T 2 B )}.
1 1
11. How many numbers are there between 1 and 1000 (both inclusive) which are divisible neither by
2 nor by 5?
Solution:
Total number which are divisible by 2 = 500.
Total number which are divisible by 5 = 200.
Total number which are divisible by 10 = 100.
Total number which are divisible by 2 or 5 = 500 + 200 − 100 = 600.

Total number which are neither divisible by 2 or 5 = 1000 − 600=400.



12. How many words can be formed using the letters of the word LOTUS if the word

Note: We assume the words formed without repetition of letters.



(i) either starts with L or ends with S?
Solution:

Number of words starts with L = First letter L,Remaining four letters can be arranged in
4! = 24 ways.

Number of words ends with S = Last letter S,Remaining four letters can be arranged in
4! = 24 ways.

Number of words starts with L and ends with S = Remaining three letters can be arranged

78

in 3! = 6 ways.

Number of words either starts with L or ends with S = 24 + 24 − 6 = 42 ways.



(ii) neither starts with L nor ends with S? Total number of words which can be formed =
5! = 120 ways.

Total number of words neither starts with L nor ends with S = 120 − 42 = 78.



13. (i) Count the total number of ways of answering 6 objective type questions, each question
having 4 choices.

Solution:
The total number of ways of answering 6 objective type questions, each question having 4
choices = 6 × 4 = 24.
(ii) In how many ways 10 pigeons can be placed in 3 different pigeon holes ?

Solution:

There are three cases
[i] Two holes empty.

All pigeons can be placed in one hole in one way.


Third hole can be selected in = 3 ways. , ,
[ii] one hole empty.

Two holes can be selected in = 6 ways. 10 pigeons can be placed in these two holes in 5
ways. That is,(9, 1), (8, 2), (7, 3), (6, 4), (5, 5). Totally 6 × 5 = 30 ways.

[iii] no hole is empty.

(8, 1, 1), (7, 2, 1), (6, 3, 1), (6, 2, 2), (5, 4, 1), (5, 3, 2), (4, 4, 2), (4, 3, 3), (3, 3, 4) = 9 ways.

Totally 42 ways.
(iii) Find the number of ways of distributing 12 distinct prizes to 10 students? Number of ways
= (no of places) (no of objects) = 10 12



14. Find the value of
6! = 6 × 5 × 4 × 3 × 2 × 1 = 720

4! + 5! = 24 + 120 = 144

3! − 2! = 6 − 2 = 4

3! × 4! = 6 × 24 = 144
12! 12 × 11 × 10 × 9 × 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1
= = 220
9! × 3! (9 × 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1) (3 × 2 × 1)
(n + 3)! (n + 3)(n + 2)(n + 1)!
2
= = n + 5n + 6
(n + 1)! (n + 1)!

79

n!
15. Evaluate when
r!(n − r)!
6!
(i) n = 6, r = 2 = 15
2!(6 − 2)!
10!
(ii) n = 10, r = 3 = 840
3!(10 − 3)!
n! n(n − 1)
(iii) For any n with r = 2 =
2!(n − 2)! 2

16. Find the value of n if
(i) (n + 1)! = 20(n − 1)!

(n + 1)(n)(n − 1)! = 20(n − 1)!

2
n + n − 20 = 0
n = 4
1 1 n
(ii) + =
8! 9! 10!
9 + 1 n
=
9! 10!
100 n
=
10! 10!
n = 100















Exercise - 4.2





n
1. If (n−1) P 3 : P 4 = 1 : 10, find n.
Solution:
(n−1) P 3 : P 4 = 1 : 10
n
(n − 1)! (n)!
: = 1 : 10
(n − 4)! (n − 4)!
(n − 1)(n − 2)(n − 3)(n − 4)! (n)(n − 1)(n − 2)(n − 3)(n − 4)!
: = 1 : 10
(n − 4)! (n − 4)!
(n − 1)(n − 2)(n − 3) : (n)(n − 1)(n − 2)(n − 3) = 1 : 10

n = 10


10
6
2. If P r−1 = 2 × P r , find r.

80

6
2 × P r = 10 P r−1

6! 10!
2 =
(6 − r)! (10 − r + 1)!
(11 − r)(10 − r)(9 − r)(8 − r)(7 − r)(6 − r)! (10)(9)(8)(7)(6)!
=
(6 − r)! 2(6)!
(11 − r)(10 − r)(9 − r)(8 − r)(7 − r) = (5)(9)(8)(7)

= (5)(3)(3)(2)(4)(7)

= (5)(3)(6)(4)(7)

r = 4
3. (i) Suppose 8 people enter an event in a swim meet. In how many ways could the gold, silver
and bronze prizes be awarded?

Solution:

Number of ways prizes can be awarded = 8 × 7 × 6 = 336.


(ii) Three men have 4 coats, 5 waist coats and 6 caps. In how many ways can they wear them?
We assume the coats,waistcoats, caps are each distinct.
6
5
4
No of ways they can wear them = ( P 3 ) × ( p 3 ) × ( P 3 )
5! 6!
= 4! × ×
2! 3!
(5)(4)(3)(2!) (6)(5)(4)(3!)
= 4!
2! 3!
= 24 ∗ 60 ∗ 120
= 172800



4. Determine the number of permutations of the letters of the word SIMPLE if all are taken at a
time?

Solution:
Number of permutations of the letters taken at a time = 6!= 720 ways.



5. A test consists of 10 multiple choice questions. In how many ways can the test be answered if
(i) Each question has four choices?

Number of ways = 4 10


(ii) The first four questions have three choices and the remaining have five choices?

4
6
Number of ways = 3 × 5 ways.
(iii) Question number n has n+1 choices?
Number of ways =11! ways.

81

6. A student appears in an objective test which contain 5 multiple choice questions. Each question
has four choices out of which one correct answer.
(i) What is the maximum number of different answers can the students give? Maximum number
5
of different answers can be given = P 4 = 120
(ii) How will the answer change if each question may have more than one correct answers?
Maximum number of different answers = 120.

Exactly only one correct answer = 5.

More than one correct answer = 120 − 5 = 115.
7. How many strings can be formed from the letters of the word ARTICLE, so that vowels occupy
the even places?

Solution:
There are 3 vowels and 4 consonants. The vowels occupy in the even places in 3! = 6 ways.
Remaining 4 places filled by 4 consonants in 4! = 24 ways. Totally 6 ∗ 24 = 144 ways.



8. 8 women and 6 men are standing in a line.
(i) How many arrangements are possible if any individual can stand in any position?

Totally 14 individuals are standing.Hence the number of arrangements possible = 14!.
(ii) In how many arrangements will all 6 men be standing next to one another?
Consider all 6 men to be one object. totally 8 women + 1 object =9 objects. Hence the total
number of arrangements = 9! ways.



(iii) In how many arrangements will no two men be standing next to one another?

Total number of arrangements of any two men standing next to one another
8 women, 4 men, 1 object = 13! ways.
9. Find the distinct permutations of the letters of the word MISSISSIPPI?

There are 11 letters in the word ”MISSISSIPPI”. ”I” and ”S” repeated 4 times. ”P” repeated 2
11! 11 × 10 × 9 × 8 × 7 × 6 × 5 × 4!
times. Hence the number of distinct permutations is =
4!4!2! 4!4!2!
11 × 10 × 9 × 8 × 7 × 6 × 5
= = 11 × 10 × 9 × 7 × 5 = 34650.
48
2 3 4
10. How many ways can the product a b c be expressed without exponents?
Number of distinct ways =
9! 9 × 8 × 7 × 6 × 5
= = 1260.
2!3!4! 12


11. In how many ways 4 mathematics books, 3 physics books, 2 chemistry books and 1 biology book
can be arranged on a shelf so that all books of the same subjects are together.

Considering all books of the same subject as single object, the number of ways of arrangement =
4! = 24 ways.
12. In how many ways can the letters of the word SUCCESS be arranged so that all Ss are together?

82

5!
Number of ways = = 60
2!


13. A coin is tossed 8 times,
(i) How many different sequences of heads and tails are possible?
Number of different sequences possible = 2 8
(ii) How many different sequences containing six heads and two tails are possible?

8
Total sequences = C 2 = 28


14. How many strings are there using the letters of the word INTERMEDIATE, if
(i) The vowels and consonants are alternative

There are 6 vowels and 6 consonants. 6 vowels can be placed with 6 gaps in 6! ways. In the
6 gaps 6 consonants can be placed in 6! ways. The two types vowels and consonants can be
arranged in 2! ways. Hence totally 2! × 6! × 6! ways.


(ii) All the vowels are together

The vowels can be together considered as one object. Including 6 consonants there are 7
objects which can be arranged in 7! ways.


(iii) Vowels are never together

6 vowels can be arranged among themselves in 6! ways. Filling from left as consonant,
vowel, consonant,...vowel and filling from right end as vowel,.... consonant, vowel conso-
nant, there are two ways. Hence total possible ways = 2(6!)
(iv) No two vowels are together.

This means between any two vowels there should be one or more consonants.

Exactly one consonant between any two vowel.
6! 7!
There are 7 places and 6 consonants. Total possible ways = × =
2! 2!3!
1 × 2 × 3 × 4 × 5 × 6 × 1 × 2 × 3 × 4 × 5 × 6 × 7
= 21600.
24


15. Each of the digits 1, 1, 2, 3, 3 and 4 is written on a separate card. The six cards are then laid out
in a row to form a 6-digit number.
(i) How many distinct 6-digit numbers are there?

6!
Total number of distinct numbers = = 180.
2!2!

(ii) How many of these 6-digit numbers are even?

5!
Last digit can be arranged by 2, 4 in 2 ways. Remaining places can be arranged in
2!2!
ways. Totally 2 × 30 = 60.

83



(iii) How many of these 6-digit numbers are divisible by 4?
4!
Last two places will be 12, 24, 32 = 3 ways. Remaining four places can be filled in +
2!
4! 4!
+ ways.
2!2! 2!
16. If the letters of the word GARDEN are permuted in all possible ways and the strings thus formed
are arranged in the dictionary order, then find the ranks of the words (i) GARDEN (ii) DANGER.



Solution:

The word ”GARDEN” has 6 letters in which no letters are repeating.

The first word is ”ADEGNR”.
Number of words starting with A = 5! = 120

Number of words starting with D = 5! = 120

Number of words starting with E = 5! = 120

Number of words starting with GAD = 3! = 6
Number of words starting with GAE = 3! = 6

Number of words starting with GAN = 3! = 6

Number of words starting with GARDEN = 1! = 1


The rank of the word is 120 + 120 + 120 + 6 + 6 + 6 + 1 = 379.


17. Find the number of strings that can be made using all letters of the word THING. If these words
th
are written as in a dictionary, what will be the 85 string?
Solution:

Number of strings using all letters = 5! = 120.
When the letters are arranged in dictionary order, the first string will be ”GHINT”.

Number of words starting with G = 4! = 24

Number of words starting with H = 4! = 24

Number of words starting with I = 4! = 24
Number of words starting with NG = 3! = 6

Number of words starting with NH = 3! = 6


The string begining with ”NI” will be 84 th string. The word is ”NIGHT”. The 85 th string is
”NIGTH”.

84

18. If the letters of the word FUNNY are permuted in all possible ways and the strings thus formed
are arranged in the dictionary order, find the rank of the word FUNNY.

Solution:

The first string is ”FNNUY”.
Number of words starting with FN = 3! = 6

Number of words starting with FUNNY = 1 = 1

Hence the rank is 7.



19. Find the sum of all 4-digit numbers that can be formed using digits 1, 2, 3, 4, and 5 repetitions not
allowed?

Solution:
Sum of all the numbers which can be formed by using the n digits without repetition is:
(n − 1)! × (sum of the digits) × (111..n times).

4! × 15 × (11111) = 3999960.



20. Find the sum of all 4-digit numbers that can be formed using digits 0, 2, 5, 7, 8 without repetition?
Solution:

If you fix 8 as the last digit, you see that there are 4 × 3 × 2 ways to complete the number. Thus,
8 appears 24 times as the last digit. By the same logic, if we enumerate all possible numbers
using these 4 digits, each number appears 24 times in each of the 3 positions. That is, the digit 8
contributes (24×8+240×8+2400×8). In total, we have (0+2+5+7+8)(24+240+2400) =
58608 as our total sum.

But Wait!, Since 4-digit numbers cannot start with 0, then we have overcounted. Now we have to
subtract the amount by which we overcounted, which is found by answering: ”What is the sum of
all 3-digit numbers formed by using digits 2, 5, 7, and 8?” Now if 8 appears as the last digit, then
there are 6 ways to complete the number, so 8 contributes (6 × 8 + 60 × 8 + 600 × 8). In total, we
have (2 + 5 + 7 + 8)(6 + 60 + 600) = 14652.Subtracting this from the above gives us 43956.


Exercise - 4.3


n
n
1. If C 12 = C 9 find 21 C n .
Solution:
n = C n−12 .
n
C 12
n − 12 = 9

n = 21.

21
C 21 = 1.
15
2. If C 2r−1 = 15 C 2r+4 , find r.
Solution:

85

15 = 15
C 2r−1 C 15−2r+1
15 − 2r + 1 = 2r + 4

r = 3.


n
n
3. If P r = 720, and C r = 120, find n, r.
Solution:
n!
n P r = = 720
(n − r)!
n! 720
n C r = =
r!(n − r)! r!
720
= 120
r!
r! = 6

r = 3


15
15
15
4. Prove that C 3 + 2 × C 4 + C 5 = 17 C 5
Solution:
15 15 15 15 15 15 15
C 3 + 2 × C 4 + C 5 = ( C 3 + C 4 ) + ( C 4 + C 5 )
16
= 16 C 4 + C 5
= 17 C 5


35
5. Prove that C 5 + P 4 (39−r) C 4 = 40 C 5 .
r=0
35 P 4 (39−r) 35 39 38 37 36 35
C 5 + C 4 =
r=0 C 5 + C 4 + C 4 + C 4 + C 4 + C 4
38
35
35
37
36
= 39 C 4 + C 4 + C 4 + C 4 + C 4 + C 5
37
36
38
36
= 39 C 4 + C 4 + C 4 + C 4 + C 5
37
37
38
= 39 C 4 + C 4 + C 4 + C 5
38
38
= 39 C 4 + C 4 + C 5
39
= 39 C 4 + C 5
= 40 C 5
6. If (n+1) C 8 : (n−3) P 4 = 57 : 16, find n.

86

16 (n+1) C 8 = 57 (n−3) P 4


(n + 1)! (n − 3)!
16 = 57
8!(n − 7)! (n − 7)!
16(n + 1)! = 57(8!)(n − 3)!

16(n + 1)(n)(n − 1)(n − 2) = (3)(19)(8!)
= (3)(19)(8)(7)(6)(5)(4)(3)(2)(1)

= (7)(3)(5)(4)(19)(8)(2)(6)(3)

= (21)(20)(19)(16)(18)

(n + 1)(n)(n − 1)(n − 2) = (21)(20)(19)(18)

n = 20

n
2 × 1 × 3 × · · · (2n − 1)
7. Prove that 2n C n =
n!
Solution:

(2n)!
2n C n =
(n!)(n!)
(2n)! = (2n)(2n − 1)(2n − 2)(2n − 3)(2n − 4)(2n − 5) . . . (5)(4)(3)(2)(1)

= [(2n)(2n − 2)(2n − 4) . . . (4)(2)] [(2n − 1)(2n − 3)(2n − 5) . . . (5)(3)(1)]

= [2(n)2(n − 1)2(n − 2) . . . (2)(2)(2)(1)] [(2n − 1)(2n − 3)(2n − 5) . . . (5)(3)(1)]

n
= 2 (n!) [(2n − 1)(2n − 3)(2n − 5) . . . (5)(3)(1)]
n
2 [(2n − 1)(2n − 3)(2n − 5) . . . (5)(3)(1)]
2n C n =
(n!)
n
8. Prove that if 1 ≤ r ≤ n then n × (n−1) C r−1 = (n − r + 1) C r−1 .
Solution:
n
n
We know that by the property, C r = × (n−1) C r−1 .
r
n × (n−1) C r−1 = (r) × C r
n
n!
= (r) ×
r!(n − r)!
n!
=
(r − 1)!(n − r)!
(n − r + 1)n!
=
(r − 1)!(n − r + 1)(n − r)!
n!
= (n − r + 1) ×
(r − 1)!(n − r + 1)!
n
= (n − r + 1) C r−1

Method 2

87

(n − 1)!
n × (n−1) C r−1 = n ×
(r − 1)!(n − 1 − r + 1)!
n!
=
(r − 1)!(n − r)!
n!
= (n − r + 1) ×
(r − 1)!(n − r + 1)(n − r)!
n!
= (n − r + 1) ×
(r − 1)!(n − r + 1)!
n
= (n − r + 1) C r−1


9. (i) A Kabaddi coach has 14 players ready to play. How many different teams of 7 players could
the coach put on the court?

Solution:
1
Number of teams to select 7 players from 14 players = 4C 7 = 3432.


(ii) There are 15 persons in a party and if, each 2 of them shakes hands with each other, how
many handshakes happen in the party?

Solution:
15 people shake hands with 14 other people. Two people are required for a handshake. If A
and B are two people A shakes hands with B, B will not shake hands with A again. So This
(15!) 15 × 14
1
can be done in 5C 2 = = 105. Hence there are 105 handshakes in all
(2!)(13!) 1 × 2
(iii) How many chords can be drawn through 20 points on a circle?
Solution:

As all 20 points are on a circle and a chord is formed by joining any 2 of them, there can be
2
as many as 0C 2 chords i.e. 190 chords.


(iv) In a parking lot hundred one year old cars are parked . Out of them five are to be chosen at
random and tested to see how well their pollution devices are working. How many different
set of five cars can be chosen?

Solution:

Number of set of five cars that can be chosen =
100 × 99 × 98 × 97 × 96
1 00C 5 = = 75287520.
1 × 2 × 3 × 4 × 5


(v) How many ways can a team of 3 boys,2 girls and 1 transgender be selected from 5 boys, 4
girls and 2 transgenders?

Solution:

5
3 boys selected from 5 boys in C 3 ways = 10.
4
2 girls selected from 4 girls in C 2 ways = 6.
2
1 transgender selected from 2 in C 1 ways = 2.

88

Totally 10 × 6 × 2 = 120 ways.



10. Find the total number of subsets of a set with
(i) 4 elements (ii) 5 elements (iii) n elements


Solution:

(i) 4 elements
4
The number of subsets with 0 is C 0 = 1.
4
The number of subsets with 1 is C 1 = 4.
4
The number of subsets with 2 is C 2 = 6.
4
The number of subsets with 3 is C 3 = 4.
4
The number of subsets with 4 is C 4 = 1.
Totally 16 subsets.
n
n
n
n
Note: To find number of subsets, note that C 0 + C 1 + · · · + C n = 2 . Hence the solution for
4
the above question is 2 = 16.
5
(ii) 5 elements 2 = 32.
n
(iii) n elements = 2 .
11. A trust has 25 members.

(i) How many ways 3 officers can be selected?
Solution:

25
3 officers can be selected in C 3 ways = 2300.


(ii) In how many ways can a President, Vice President and a Secretary be selected?
Solution:

President can be selected in 25 C 1 = 25 ways. Vice President can be selected in 24 C 1 ways.
23
Secretary can be selected in C 1 ways. Hence totally 25 × 24 × 23 = 13800 ways.


12. How many ways a committee of six persons from 10 persons can be chosen along with a chair
person and a secretary?

Solution:

Six persons can be selected from 10 persons in 10 C 6 ways. Along with this committee, from
8
remaining 8 persons two persons viz., chair person and secretary can be selected in P 2 ways.
10 × 9 × 8 × 7 8!
8
10
Hence totally C 6 × P 2 = = 210 × 56 = 11760
1 × 2 × 3 × 4 6!
13. How many different selections of 5 books can be made from 12 different books if,
(i) Two particular books are always selected?

89

Solution:

Two particular books can be considered as one object. Remaining 3 books can be selected
10
from remaining 10 different books in C 3 = 120 ways.
(ii) Two particular books are never selected?

Solution:

10
All the 5 books are selected from remaining 10 books in C 5 = 252 ways.


14. There are 5 teachers and 20 students. Out of them a committee of 2 teachers and 3 students is to
be formed. Find the number of ways in which this can be done. Further find in how many of these
committees

(i) a particular teacher is included?

Solution:
If a particular teacher is included, the selection should be made for the remaining 1 teacher
4
20
and 3 students only from 4 teachers and 20 students in C 1 × C 3 = 4×1140 = 4560 ways.

(ii) a particular student is excluded?

Solution:
If a particular student is excluded, then 2 teachers and 3 students can be selected from 5
5
19
teachers and 19 students in C 2 × C 3 = 9690 ways.

15. In an examination a student has to answer 5 questions, out of 9 questions in which 2 are
compulsory. In how many ways a student can answer the questions?

Solution:
7
Remaining 3 questions can be answered out of 7 questions in C 3 = 35 ways.


16. Determine the number of 5 card combinations out of a deck of 52 cards if there is exactly three
aces in each combination.

Solution:
4
Three aces can be selected from 4 aces in C 3 = 4 ways. Remaining 2 cards are selected from 48
48
cards in C 2 = 1128 ways. Hence total number of combination is 4 × 1128 = 4512 ways.


17. Find the number of ways of forming a committee of 5 members out of 7 Indians and 5 Americans,
so that always Indians will be the majority in the committee.

Solution:

The possible formation is listed as below.

90


Indian American Combination Value

5
5 0 7 C 5 × C 0 21
5
4 1 7 C 4 × C 1 175
5
3 2 7 C 3 × C 2 350
Total 546

Hence total number of forming the committee = 546.



18. A committee of 7 peoples has to be formed from 8 men and 4 women. In how many ways can this
be done when the committee consists of
(i) exactly 3 women?

Solution:
4
Three women can be selected from 4 women in C 3 = 4 ways. Remaining 4 people can be
8
selected from 8 men in C 4 = 70 ways.Totally 4 × 70 = 280 ways.


(ii) at least 3 women?


Men Women Combination Value

4
4 3 8 C 4 × C 3 280
4
3 4 8 C 3 × C 4 56

Total 336


(iii) at most 3 women?


Men Women Combination Value

4
7 0 8 C 7 × C 0 8
4
6 1 8 C 6 × C 1 112
4
5 2 8 C 5 × C 2 336
4
4 3 8 C 4 × C 3 280
Total 736



19. 7 relatives of a man comprises 4 ladies and 3 gentlemen, his wife also has 7 relatives; 3 of them
are ladies and 4 gentlemen. In how many ways can they invite a dinner party of 3 ladies and 3
gentlemen so that there are 3 of men’s relative and 3 of the wife’ s relatives?

Solution:

91

Men’s relative Women’relative Combination Value

Men Women Men Women

3
4
4
3
8 0 0 3 ( C 3 × C 0 ) × ( C 0 × C 3 ) 16
3
4
4
3
2 1 1 2 ( C 2 × C 1 ) × ( C 1 × C 2 ) 324
4
3
4
3
1 2 2 1 ( C 1 × C 2 ) × ( C 2 × C 1 ) 144
3
4
3
4
0 3 3 0 ( C 0 × C 3 ) × ( C 3 × C 0 ) 1
20. A box contains two white balls, three black balls and four red balls. In how many ways can three
balls be drawn from the box, if at least one black ball is to be included in the draw?
Solution:
Black White or Red Combination Value
6
1 2 3 C 1 × C 2 45
6
2 1 3 C 2 × C 1 18
6
3 0 3 C 3 × C 0 1
Total number of ways 64



21. Find the number of strings of 4 letters that can be formed with the letters of the word
EXAMINATION.

Solution:

In the word Examination,E-1,X-1,A-2,M-1,I-2,N-2,T-1,O-1 there are 8 distinct words.

Case I: All words are distinct. abcd
8
we can take 4 words from 8 distinct in C 4 and we can arrange the 4 words in 4! ways.


8
Total number of ways= C 4 × 4! ways=1680
Case II:1 word is alike and 2 is distinct. aabc
7
3
There are 3 group of alike words .We can choose in C 1 ways and the distinct words in C 2 .We
4!
can arrange the words in 4! ways but two words are same so ways.
2!
4!
3
7
Total number of ways= C 2 × C 1 × = 756
2!
Case III: Two are alike. aabb


4!
3
Total number of ways: C 2 × × 2! = 18
2!

92

Total of ways in which 4 letter words can be made out of the word EXAMINATION:1680+756+
18 = 2454


22. How many triangles can be formed by joining 15 points on the plane, in which no line joining any
three points?

Solution:

No line joining any three points means none of any three points are collinear. Hence number of
15
triangles which can be formed = C 3 = 455.


23. How many triangles can be formed by 15 points, in which 7 of them lie on one line and the
remaining 8 on another parallel line?
Solution:

A triangle needs 3 distinct/non-collinear points to form the three vertices.
Line 1 has 7 points and Line 2 has 8. We cannot select all 3 points from the same line to form a
triangle.
we have 2 options. Select 2 points from Line1 and 1 from Line2 (OR) Select 1 point from Line1
and 2 from Line2.
8
7
7
8
in other words, C 2 × C 1 + C 1 × C 2 = 364
Method 2
There are 7 + 8 = 15 points in all. ”combination” of any 3 points taken at a time will give
us a triangle so 15 C 3 = 455. BUT if all the 3 points we choose lie on the same line then they
will be co-linear and hence wont form a triangle. SO we subtract all such combination’s i.e
7
8
( C 3 + C 3 ) = 91. Therefore 455 − 91 = 364.
Method 3
Each point out of 7 on the 1st line can be joined with 8 × 7 pairs on the 2nd, and any out of 8 the
2nd we join with 7 × 6 pairs on the 1st. Just don’t forget to divide the sum by 2, otherwise each
(7 × 8 × 7) + (8 × 7 × 6)
triangle will counted twice. = 364.
2
all the triangles which can be formed have equal height



24. There are 11 points in a plane. No three of these lies in the same straight line except 4 points,
which are collinear. Find,
(i) the number of straight lines that can be obtained from the pairs of these points?

Solution:
Total points = 11

Collinear points = 4

11
No of lines formed from 11 points = C 2
(Since we require two points to form one line.)

4
Four points are collinear = C 2

93

4
11
Required no of straight line = C 2 − C 2 + 1 = 55 − 6 + 1 = 50

(ii) the number of triangles that can be formed for which the points are their vertices?
Solution:

Method 1

Total points = 11

Collinear points = 4
11
No of triangles formed from 11 points = C 3

(Since we require three points to form one triangle.)
4
But Four points are collinear. No of triangles with this four points = C 3 .( which have zero
area)

11
Required no of triangles = C 3 − 4 = 165 − 4 = 161
Method 2


Out of 11 points, 4 are collinear, 7 are non collinear.


7
(i) triangles formed from 7 non collinear points C 3 = 35.


(ii) triangles formed from 1 point of non collinear and 2 points of collinear.
4
7 C 1 × C 2 = 42.


(iii) triangles formed from 2 points of non collinear and 1 point of collinear points.
7 4
C 2 × C 1 = 84.


Let’s add all triangles = 35 + 42 + 84 = 161.



25. A polygon has 90 diagonals. Find the number of its sides?

Solution:
n
Two vertices required to form a side or diagonal. Hence the number of diagonals or sides = C 2 .
n(n − 3)
n
Subtracting n sides, the number of diagonals = C 2 − n = .
2
n(n − 3)
= 90 ⇒ n = 15.
2









Exercise - 4.4

94



1. By the principle of mathematical induction, prove that, for n ≥ 1
2
n(n + 1)
3
3
3
3
1 + 2 + 3 + · · · + n =
2
Solution:
3
3
3
Let P(n) = 1 + 2 + 3 + · · · + n 3
CASE N = 1
2
1(1 + 1)
3
1 =
2
= 1
LHS = RHS

Hence P(1) is true.

CASE N = k

Assume P(k) is true. That is,
2
k(k + 1)
3
3
3
3
P(k) = 1 + 2 + 3 + · · · + (k) =
2
CASE N = k+1
3
3
3
3
3
3
3
3
1 + 2 + 3 + · · · + (k + 1) = 1 + 2 + 3 + · · · + (k) + (k + 1) 3
3
3
3
3
= [1 + 2 + 3 + · · · + (k) ] + (k + 1) 3
= P(k) + (k + 1) 3
2
k(k + 1)
= + (k + 1) 3
2
2
k (k + 1) 2
= + (k + 1) 3
4
2
2
(k + 1) [k + 4(k + 1)]
=
4
2
(k + 1) (k + 2) 2
=
4
= P(k + 1)
This implies, P(k + 1) is true whenever P(k) is true. Hence, by the principle of mathematical
induction, P(n) is true for all natural numbers.
2. By the principle of mathematical induction, prove that, for n ≥ 1
n(2n − 1)(2n + 1)
2
2
2
2
1 + 3 + 5 + · · · + (2n − 1) = .
3
Solution:
2
2
2
Let P(n) = 1 + 3 + 5 + · · · + (2n − 1) 2
CASE N = 1

95

1(1)(3)
2
1 =
3
= 1

= 1
LHS = RHS


Hence P(1) is true.

CASE N = k

Assume P(k) is true. That is,
k(2k − 1)(2k + 1)
2
2
2
2
P(k) = 1 + 3 + 5 + · · · + (2k − 1) =
3
CASE N = k+1
2
2
2
2
2
2
2
2
1 + 3 + 5 + · · · + (2k + 1) = 1 + 3 + 5 + · · · + (2k − 1) + (2k + 1) 2
2
2
2
2
= [1 + 3 + 5 + · · · + (2k − 1) ] + (2k + 1) 2
= P(k) + (2k + 1) 2
k(2k − 1)(2k + 1)
= + (2k + 1) 2
3
(2k + 1) [k(2k − 1) + 3(2k + 1)]
=
3 This implies,
2
(2k + 1) [2k − k + 6k + 3]
=
3
(2k + 1)(k + 1)(2k + 3)
=
3
(k + 1) [2(k + 1) − 1] [2(k + 1) + 1]
=
3
= P(k + 1)
P(k + 1) is true whenever P(k) is true. Hence, by the principle of mathematical induction, P(n)
is true for all natural numbers.
2
3. Prove that the sum of the first ’n’ non-zero even numbers is n + n.
2
Let P(n) = n + n. CASE N = 1
2
2 = 1 + 1
= 2

LHS = RHS


Hence P(1) is true.

CASE N = k
Assume P(k) is true. That is,

2
P(k) = 2 + 4 + 6 + · · · + 2k = k + k

96

CASE N = k+1
2 + 4 + 6 + · · · + 2(k + 1) = 2 + 4 + 6 + · · · + 2k + 2(k + 1)

= P(k) + 2(k + 1)

2
= k + k + 2k + 2
This implies, P(k + 1) is true
2
= k + 2k + 1 + k + 1
2
= (k + 1) + (k + 1)
= P(k + 1)

whenever P(k) is true. Hence, by the principle of mathematical induction, P(n) is true for all
natural numbers.


4. By the principle of mathematical induction, prove that, for n ≥ 1
n(n + 1)(n + 2)
1.2 + 2.3 + 3.4 + · · · + n.(n + 1) = .
3
Let P(n) = 1.2 + 2.3 + 3.4 + · · · + n.(n + 1)

CASE N = 1
1(2)(3)
1.2 =
3
= 2

LHS = RHS

Hence P(1) is true.

CASE N = k

Assume P(k) is true. That is,

k(k + 1)(k + 2)
P(k) = 1.2 + 2.3 + 3.4 + · · · + k.(k + 1) =
3
CASE N = k+1
1.2 + 2.3 + 3.4 + · · · + (k + 1).(k + 2) = 1.2 + 2.3 + 3.4 + · · · + k.(k + 1) + (k + 1).(k + 2)

= [1.2 + 2.3 + 3.4 + · · · + k.(k + 1)] + (k + 1).(k + 2)

= P(k) + (k + 1).(k + 2)
k(k + 1)(k + 2)
= + (k + 1).(k + 2)
3
(k + 1)(k + 2) [k + 3]
=
3
= P(k + 1)

This implies, P(k + 1) is true whenever P(k) is true. Hence, by the principle of mathematical
induction, P(n) is true for all natural numbers.



5. Using the Mathematical induction, show that for any natural number n ≥ 2,

97


1 1 1 1 n + 1
1 − 1 − 1 − · · · 1 − = .
2 2 3 2 4 2 n 2 2n
n + 1
Let P(n) = CASE N = 2
2n

1 2 + 1
1 − =
2 2 4
3
=
4
LHS = RHS


Hence P(2) is true.

CASE N = k

Assume P(k) is true. That is,

1 1 1 1 k + 1
P(k) = 1 − 1 − 1 − · · · 1 − =
2 2 3 2 4 2 (k) 2 2k
CASE N = k+1
1 1 1 1 1 1 1

1 − 1 − 1 − · · · 1 − = 1 − 1 − 1 − · · ·
2 2 3 2 4 2 (k + 1) 2 2 2 3 2 4 2
1 1

1 − 1 −
(k) 2 (k + 1) 2

1
= P(k) 1 −
(k + 1) 2

k + 1 1
= 1 −
2k (k + 1) 2
2
k + 1 (k + 1) − 1
=
2k (k + 1) 2
2
k + 1 k + 2k
=
2k (k + 1) 2

(k + 1) k(k + 1)
=
2k (k + 1) 2
1
=
2
(k + 2)
=
2(k + 2)
(k + 1 + 1)
=
2(k + 1)
= P(k + 1)


This implies, P(k + 1) is true whenever P(k) is true. Hence, by the principle of mathematical
induction, P(n) is true for all natural numbers n ≥ 2,.


6. Using the Mathematical induction, show that for any natural number n ≥ 2,

1 1 1 1 n − 1
+ + + · · · + = .
1 + 2 1 + 2 + 3 1 + 2 + 3 + 4 1 + 2 + 3 + · · · + n n + 1

98

Solution:
1 1 1 1
Let P(n) = + + + · · · +
1 + 2 1 + 2 + 3 1 + 2 + 3 + 4 1 + 2 + 3 + · · · + n
CASE N = 2
1 2 − 1
=
1 + 2 2 + 1
1
=
3
LHS = RHS


Hence P(2) is true.
CASE N = k

Assume P(k) is true. That is,

k − 1
P(k) = 1 + 1 + 1 + · · · + 1 =
1+2 1+2+3 1+2+3+4 1+2+3+···+k k + 1
CASE N = k+1
1 +
1+2
1 + 1 + · · · + 1 = 1 + 1 + 1 + · · · + 1 + 1
1+2+3 1+2+3+4 1+2+3+···+(k+1) 1+2 1+2+3 1+2+3+4 1+2+3+···+k 1+2+3+···+(k+1)
= P(k) + 1
1+2+3+···+(k+1)
k − 1 1
= +
k + 1 1 + 2 + 3 + · · · + (k + 1)
k − 1 2
= +
k + 1 (k + 1)(k + 2)
(k + 2)(k − 1) + 2
=
(k + 1)(k + 2)
2
k + k
=
(k + 1)(k + 2)
k(k + 1)
=
(k + 1)(k + 2)
k
=
k + 2
(k + 1) − 1
=
(k + 1) + 1
= P(k + 1)

This implies, P(k + 1) is true whenever P(k) is true. Hence, by the principle of mathematical
induction, P(n) is true for all natural numbers n ≥ 2,.



7. Using the Mathematical induction, show that for any natural number n,
1 1 1 1 n(n + 3)
+ + + · · · + = .
1.2.3 2.3.4 3.4.5 n.(n + 1).(n + 2) 4(n + 1)(n + 2)
Solution:

99

1 1 1 1
Let P(n) = + + + · · · + .
1.2.3 2.3.4 3.4.5 n.(n + 1).(n + 2)
CASE N = 1
1 1(1 + 3)
=
1.2.3 4(1 + 1)(1 + 2)
1
=
6
LHS = RHS


Hence P(1) is true.
CASE N = k

Assume P(k) is true. That is,

1 1 1 1 k(k + 3)
P(k) = + + + · · · + =
1.2.3 2.3.4 3.4.5 k(k + 1)(k + 2) 4(k + 1)(k + 2)
CASE N = k+1
1 1 1 1 1 1 1
+ + + · · · + = + + + · · · +
1.2.3 2.3.4 3.4.5 (k + 1)(k + 2).(k + 3) 1.2.3 2.3.4 3.4.5
1 1
+
k(k + 1)(k + 2) (k + 1)(k + 2).(k + 3)
1
= P(k) +
(k + 1)(k + 2).(k + 3)
k(k + 3) 1
= +
4(k + 1)(k + 2) (k + 1)(k + 2).(k + 3)

1 k(k + 3) 1
= +
(k + 1)(k + 2) 4 k + 3
2
1 k(k + 3) + 4
=
(k + 1)(k + 2) 4(k + 3)
3 2
1 k + 6k + 9k + 4
=
(k + 1)(k + 2) 4(k + 3)
2
1 (k + 1) (k + 4)
=
(k + 1)(k + 2) 4(k + 3)
(k + 1)(k + 4)
=
4(k + 2)(k + 3)
= P(k + 1)
This implies, P(k + 1) is true whenever P(k) is true. Hence, by the principle of mathematical
induction, P(n) is true for all natural numbers.



8. Using the Mathematical induction, show that for any natural number n,
1 1 1 1 n
+ + + · · · + = .
2.5 5.8 8.11 (3n − 1)(3n + 2) 6n + 4

Solution:
1 1 1 1
Let P(n) = + + + · · · + .
2.5 5.8 8.11 (3n − 1)(3n + 2)
CASE N = 1

100

1 1
=
2.5 6 + 4
LHS = RHS

Hence P(1) is true.

CASE N = k

Assume P(k) is true. That is,

1 1 1 1 k
P(k) = + + + · · · + =
2.5 5.8 8.11 (3k − 1)(3k + 2) 6k + 4
CASE N = k+1
1 1 1 1 1 1 1
+ + + · · · + = + +
2.5 5.8 8.11 (3k + 2)(3k + 5) 2.5 5.8 8.11
1 1
+ · · · + +
(3k − 1)(3k + 2) (3k + 2)(3k + 5)
1
= P(k) +
(3k + 2)(3k + 5)
k 1
= +
6k + 4 (3k + 2)(3k + 5)
k(3k + 2)(3k + 5) + 6k + 4
=
(6k + 4)(3k + 2)(3k + 5)
3
2
9k + 21k + 16k + 4
=
(6k + 4)(3k + 2)(3k + 5)
(k + 1)(3k + 2) 2
=
(6k + 4)(3k + 2)(3k + 5)
(k + 1)(3k + 2)
=
2(3k + 2)(3k + 5)
k + 1
=
6k + 10
k + 1
=
6(k + 1) + 4
= P(k + 1)


This implies, P(k + 1) is true whenever P(k) is true. Hence, by the principle of mathematical
induction, P(n) is true for all natural numbers.



9. Prove by Mathematical Induction that 1! + (2 × 2!) + (3 × 3!) + ... + (n × n!) = (n + 1)! − 1.

Solution:
Let P(n) = 1! + (2 × 2!) + (3 × 3!) + ... + (n × n!)

CASE N = 1

1! = (1 + 1)! − 1
= 1

LHS = RHS