Vedanta Opt. Excel in Mathematics Book 8 Final (2078)

Trigonometric Ratios of Some Standard Angles

6.3 Trigonometric Ratios of 0° and 90°

Let ABC be a right angled triangle with ‘ABC = 90°. When ‘ACB becomes smaller,

AB also becomes smaller. When ‘ACB = T is made smaller and smaller then AB is

small and tends to zero. i.e. T = 0°, A

then AB = 0, AC = BC.

Now, sin0° = AB = 0 = 0
cos0° AC AC
tan0° AC AC
= BC = AC = 1 CT B

= AB = 0 = 0
BC BC

Similarly, cosec0° = f, sec0° = 1 and cot0° = f

Note :

Something = f, Nothing = 0 = 0
Zero Something Something

In the above triangle, when ‘ACB becomes greater and greater, BC becomes smaller
and smaller. When ‘ACB = 90° (nearly equal to 90°), AB = AC, BC = 0

Then, sin90° = AB = AB = 1
cos90° AC AB
BC 0 AB AB
= AC = AC = 0, tan90° = BC = 0 = f

Similarly, cosec90° = 1, sec90° = f, cot90° = 0

Way to calculate values of standard angles

Step (i). Write from 0 to 4
Step (ii).
0 123 4
Step (iii).
Divide all of them by 4. 4
0 123 4
4 444 1

1 13 1
=0 4 2 4
Take the square root of all 1

0 1 13
4 24

0 1 13
2 22

vedanta Excel in Additional Mathematics - Book 8 101

Trigonometric Ratios of Some Standard Angles

Now, the values are ready.

If you start from left to right, you get the values of sine.

Angle (T) : 0° 30° 45° 60° 90°

sinT : 0 1 1 3 1
2 2 2

If you start from right to left, you get the values of cos in ascending order of angles.

Angles (T) : 90° 60° 45° 30° 0°

cosT : 0 1 1 3
2 2 21

The values of sin and cos are determined all remaining values can be obtained.

Trigonometric Ratio Value Table

Angle 0° 30° 45° 60° 90°
Ratio 1
0 1 1 3 0
sin 2 2 2 ∞
1
cos 1 3 11 ∞
tan 2 22 0
cosec
sec 0 1 1 3
cot 3

∞2 2 2
3

1 2 22
3

∞ 31 1
3

Worked out Examples

Example 1. Find the value of:
Solution: (a) sin90° + cos90° + tan45°
(b) tan30° . tan60° + sin60° . cos30°
(a) Here, sin90° + cos90° + tan45°

=1+0+1
=2

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Trigonometric Ratios of Some Standard Angles

(b) Here, tan30° . tan60° + sin60° . cos30°

= 1 . 3+ 3 . 3
3 2 2
3
= 1 + 4

= 4 + 3 = 7 = 1 3
4 4 4

Example 2. Find the value of:

(a) 3 tan2 30° + 4 sin2 30° + 2 cos2 45° + tan2 60°

(b) sin Sc + cos Sc sin Sc – cos Sc
3 3 3 3

Solution: (a) Here, 3 . 1 2+4 1 2+ 2 1 2+ ( 3 )2
3 2 2

= 3. 1 +4. 1 +2. 1 +3
3 4 2

=1+1+1+3=6

(b) Here, sin Sc + cos Sc sin Sc – cos Sc
3 3 3 3

Since Sc = 180°

= sin 180° + cos 180° sin 180° – cos 180°
3 3 3 3

= (sin 60° + cos 60°) (sin 60° – cos 60°)

= 3 + 1 3 – 1
2 2 2 2

= 3 – 1 = 3 – 1
4 4 4

= 2 = 1
4 2

Example 3. Prove the following:

sin2 45° + cos2 45° + tan2 45° = 2

Solution: LHS = sin2 45° + cos2 45° + tan2 45°

= 1 2+ 1 2+1
2 2
1 1
= 2 + 2 + 1

= 1 + 1 + 2
2

= 4 = 2 = RHS proved.
2

vedanta Excel in Additional Mathematics - Book 8 103

Trigonometric Ratios of Some Standard Angles

Example 4. If A = 60° and B = 30°, verify that
Solution:
(a) cos (A + B) = cos A . cos B – sinA . sinB.

(b) cot(A + B) = cotA . cotB – 1
cotB + cotA

(a) LHS = cos (A + B)

= cos (60° + 30°)

= cos 90° = 0

RHS = cosA . cosB – sinA . sinB

= cos 60° . cos30° – sin60° . sinB

= 1 × 3 – 3 × 1
2 2 2 2

= 3 – 3 =0
4 4
? LHS = RHS proved.

(b) LHS = cot(A + B)

= cot(60° + 30°)

= cot 90° = 0

RHS = cotA . cotB – 1
cotB + cotA

= cot 60° . cot 30° – 1
cot 60° + cot 30°

1 . 3–1
3
=
1 + 3
3

= 0 =0
3
1 +
3

? LHS = RHS Proved.

Example 5. Find the value of 'x' from the following equations:
Solution:
(a) x + sin 60° + cos 30° = 2 tan 45°

(b) sin2 45° – x tan2 30° = cos2 30°

(a) Here, x + sin 60° + cos 30° = 2 tan 45°

or, x+ 3 + 3 = 2 . 1
2 2

or, x + 23 = 2
2

? x=2– 3

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Trigonometric Ratios of Some Standard Angles

(b) Here, sin2 45° – x tan2 30° = cos2 30°

or, 1 2–x. 1 2= 32
2 3 2

or, 1 – x . 1 = 3
2 3 4

or, – x = 3 – 1
3 4 2

or, – x = 3 – 2
3 4

or, – x = 1
3 4

? x = – 3
4

Example 6. Find the value of T
Solution:
(a) sin T = 1
2
(b) 4 sin2 T – 3 = 0

(a) Here, sin T = 1
2
or, sin T = sin 45°

? T = 45°

(b) Here, 4 sin2 T – 3 = 0

or, 4 sin2 T = 3

or, sin2 T = 32
2

Taking positive sign only,

sin T = 3
2

or, sin T = sin 60°

? T = 60°

Exercise 6.1

1. Find the values of: (b) sin30° + cos60° + tan45°
(a) sin0° + cos0° + tan0°
(d) sin45° + cos45° + 1 tan45°
(c) sin90° + cos90° + tan45° 2
(e) tan60° + cot30° + cos30°
(g) tan60° + sin30° + cos60° (f) tan60° + cot30° – 2 3

vedanta Excel in Additional Mathematics - Book 8 105

Trigonometric Ratios of Some Standard Angles

2. Find the values of:

(a) sin230° + cos260° + tan260° (b) sin245° + sin260° + tan245°

(c) sin230° + sin245° + tan245° (d) 2 sin260° + 3 cos260° + tan245°
3
(e) 5 cosec260° + 3 sec245° + 4 cot245°

(f) tan Sc . sin Sc + cos Sc . cos Sc + cos Sc . sin Sc
3 6 4 4 3 6
Sc Sc Sc Sc
(g) tan2 3 + sin2 3 – cos2 3 + cosec2 4

(h) sin Sc + cos Sc sin Sc – cos Sc
6 6 6 6

3. Prove the following:

(a) cos260° + tan260° = 13
4
7
(b) sin260° + sin245° + cos245° = 4

(c) tan260° + sin260° + cos260° = 4

(d) (sin60° + cos60°) (cos30° – sin30°) = 1
2
S S
(e) 2 . sin 4 + 3 . tan 6 = 2

4. If A = 45° and B = 30°, verify the following:

(a) sin (A – B) = sinA . cosB – cosA . sinB.

(b) cos (A + B) = cosA. cosB – sinA . sinB .

(c) tan (A – B) = 1 tanA – tanB
+ tanA . tanB
cotA . cotB – 1
(d) cot (A + B) = cotB + cotA

5. Find the value of 'x' from the following trigonometrical equations:

(a) 2x + sin90° + tan45° = sec60°

(b) tan60° + sin60° + x = sin30°

(c) x . tan60° . cot260° . sin60° =
2

(d) x sec260° . tan260°.cos260° = 5

(e) x . sin Sc + sin Sc + cos Sc . sin Sc = sin Sc
3 3 4 4 6

(f) sin230° + cos260° + x . tan260° = cot230°

6. Find the value of T (where T is an acute angle).

(a) cosT = (b) 3 tanT – 3 = 0
2 (d) 3 tanT = 3

(c) 2 tan2T = 6

106 vedanta Excel in Additional Mathematics - Book 8

(e) 4 sinT = 3 Trigonometric Ratios of Some Standard Angles

(f) 7 sec2 T + tan2 T = 15

1. (a) 1 (b) 2 (c) 2 (d) 3
2
53
(e) 2 (f) 0 (g) 3 + 1

2. (a) 7 (b) 9 (c) 7 (d) 9
2 4 4 4
50 3+2 3 11 1
(e) 3 (f) 4 (g) 2 (h) – 2

5. (a) 0 (b) 1–3 2 (c) 3 (d) 4
2 15
5
(e) – 1 (f) 6 6.(a) 60° (b) 30°
(c) 60° (e) 60°
(d) 60° (f) 45°

6.4 Trigonometric Ratios of Complementary Angles

Two angles are said to be complementary if their sum is 90°, 65°, and 25° are
complementary angles, 65° + 25° = 90°.

Complementary Angles in Trigonometry

Let ABC be a right-angled triangle where ‘ABC = 90° and ‘ACB = T.

Now, we know,

‘ACB + ‘ABC + ‘BAC = 180° (sum of angles of ')

or, T + 90° + ‘BAC = 180°

or, ‘BAC = 180° – 90° – T = 90° – T

So, ‘ACB and ‘CAB are complementary angles.

Now, taking ‘BAC = 90° – T as the reference angle, we get, A
B
BC = perpendicular (p)

AB = hypotenuse (h)

AB = base (b)

Trigonometric ratios of T are T
C

sin T = AB , cos T = BC , tan T = AB
AC AC BC

cosec T = AC , sec T = AC , cot T = BC
AB BC AB

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Trigonometric Ratios of Some Standard Angles

Taking trigonometric ratios,

sin (90° – T) = p = BC cos (90° – T) = b = AB
h AC h AC

tan (90° – T) = p = BC cot (90° – T) = b = AB
b AB p BC

cosec (90° – T) = h = AC sec (90° – T) = h = AC
p BC b AB

Hence, we have,

sin (90° – T) = cos T cos (90° – T) = sin T
cot (90° – T) = tan T
tan (90° – T) = cot T cosec (90° – T) = sec T

sec (90° – T) = cosec T

Worked out Examples

Example 1. Prove the following :
Solution: (a) tan 30° = cot 60°
(b) sin 35° + cos 45° = cos 55° + sin 45°
(a) Here, tan 30° = cot 60°

LHS = tan 30° = tan (90° – 60°)
= cot 60° = RHS Proved.

(b) Here, sin 35° + cos 45° = cos 55° + sin 45°
LHS = sin 35° + cos 45°
= sin (90° – 55°) + cos (90° – 45°)
= cos 55° + sin 45° = RHS Proved.

Example 2. Prove that :
Solution:
(a) sin T . cos (90° – T) + cos T . sin (90° – T) = 1

(b) sin T + cos (90° – T) = 2
(90° – sin T
cos T)

(c) sin2 (90° – T) + cos2 (90° – T) = cos T + sin T
cos T sin T

(a) Here, sin T . cos (90° – T) + cos T . sin (90° – T) = 1

LHS = sin T . cos (90° – T) + cos T . sin (90° – T)

= sin T . sin T + cos T . cos T

= sin2 T + cos2 T

= 1 = RHS Proved.

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Trigonometric Ratios of Some Standard Angles

(b) Here, sin T + cos (90° – T) = 2
(90° – sin T
cos T) cos T)
sin T (90° –
LHS = cos (90° – T) + sin T

= sin T + sin T = 1 + 1
sin T sin T

= 2 = RHS Proved.

(c) Here, sin2 (90° – T) + cos2 (90° – T) = cos T + sin T
cos T sin T
sin2 (90° – T) cos2 (90° – T)
LHS = cos T + sin T

= cos2 T + sin2 T
cos T sin T

= cos T + sin T = RHS Proved.

Example 3. Prove the following :
Solution:
(a) cos (90° – T) . tan (90° – T) = cos T

(b) cos2 (90° – T) + cos2 T = sin T + cos T
sin T sin (90° –
T)

(a) Here, cos (90° – T) . tan (90° – T) = cos T

LHS = cos (90° – T) . tan (90° – T)

= sin T . cot T = sin T . cos T
sin T

= cos T = RHS Proved.

(b) Here, cos2 (90° – T) + cos2 T = sin T + cos T
sin T sin (90° –
T)
cos2 (90° – T) cos2 T
LHS = sin T + sin (90° –
T)
sin2 T cos2 T
= sin T + cos T

= sin T + cos T = RHS Proved.

Exercise 6.2 109

Short Questions :
1. Prove the following:

(a) sin 70° = cos 20°
(b) cosec 65° = sec 25°
(c) tan 60° + cos 50° = cot 30° + sin 40°
(d) sin 35° + cot 15° = cos 55° + tan 75°

vedanta Excel in Additional Mathematics - Book 8

Trigonometric Ratios of Some Standard Angles

(e) sin 72° – cos 52° = cos 18° – sin 38°

(f) sin 25° . sec 65° = 1

(g) tan 40° . tan 50° = 1

(h) cosec 15° . cos 15° = tan 75°

(i) sin 55° . sec 55° = cot 35°

2. Simplify :

(a) tan (90° – T) . cot (90° – T) – cos (90° – T) . sec (90° – T)

(b) sin T . cos (90° – T) + cos T . sin (90° – T)

(c) sec T . cosec (90° – T) – tan T . cot (90° – T)

(d) tan (90° – T) . cot (90° – T)
cosec T . cos (90° – T)

(e) sin2 T T) + cos2 T T)
cos2 (90° – sin2 (90° –

3. Prove the following :

(a) cos (90° – T) . sec (90° – T) = tan T

(b) cot (90° – T) . sin (90° – T) = sin T

(c) sin (90° – T) . sec (90° – T) = cot T

(d) sin2 T . sin(90° – T) . cot (90° – T) = tan T . sin T
sin2 (90° – tan T
T)

(e) cos T . sin (90° – T) . cot (90° – T) = cot T
sin2 T

2. (a) 0 (b) 1 (c) 1 (d) 1 (e) 2

110 vedanta Excel in Additional Mathematics - Book 8

Solution of Right-angled Triangle and Height and Distance

Solution of Right-angled

Triangle and Height 7
and Distance

7.1 Solution of Right-angled Triangle

It has been stated that a triangle has three sides and three angles. These are
components of a triangle. If three components of a triangle are given, the remaining
three components can be found. The process of finding unknown components of a
triangle is called solution of a triangle.

CASE I

When one angle and two sides of a right angled triangle are given:

the third angle can be calculated by using the relation of sum of angles of triangle.

i.e., A + B + C = 180° A

the third side can be calculated by simply using Pythagoras relation.

h = p2 + b2

p = h2 – b2 CB

b = h2 – p2

In right angled triangle, ABC as shown in the figure.

we write, AB = c, BC = a, and AC = b.

Worked out Examples

Example 1. Solve the right-angled triangle ABC if

(a) ‘B = 90°, ‘C = 30°, AB = 3 , BC = 1

(b) ‘A = 90°, ‘C = 45°, b = 1, a = 2 A
3
Solution: (a) Here, in right-angled 'ABC, ?
‘B = 90°, ‘C = 30° B

AB = 3, BC = 1 C 30° 1 111
‘A = ? , AC = ?

vedanta Excel in Additional Mathematics - Book 8

Solution of Right-angled Triangle and Height and Distance

Now, we know,

‘A + ‘B + ‘C = 180°
or, ‘A + 90° + 30° = 180°
or, ‘A = 180° – 120° = 60°
In a right-angled 'ABC,
112 using Pythagroas theorem, we ge,t

h2 = p2 + b2

or, (AC)2 = (AB)2 + (BC)2

or, (AC)2 = ( 3 )2 + (1)2
or, (AC)2 = 3 + 1

or, (AC)2 = 3 + 1

or, (AC)2 = 4

? AC = 2

Hence, ‘A = 60° and AC = 2 C
(b) Here, in right-angled 'ABC, 45°
‘A = 90°, ‘C = 45°,
b=1
b = AC = 1,
a = BC = 2 A

So, ‘B = ? , a= 2
AB (c) = ? B

We know,

‘A + ‘B +‘C = 180°
or, 90° + ‘B + 45° = 180°
or, ‘B = 180° – 155° = 45°.
Similarly, using Pythagoras theorem, we get,

h2 = p2 + b2

or, (BC)2 = (AC)2 + (BA)2

or, (BC)2 – (AC)2 = (AB)2

or, (AB)2 = ( 2 )2 – (1)2 = 2 – 1 = 1

? AB = 1
Hence, ‘C = 45° and AB (c) = 1

vedanta Excel in Additional Mathematics - Book 8

Solution of Right-angled Triangle and Height and Distance

CASE II

When two angles and a side of a right angled triangle are given:
There are two subcases

(i) When one right angle and one acute angle with one side is given.
To find any one side, we have to use the relation of standard angle with
the given side. Similarly, the third side can be calculated by using Pythagoras
Theorem.

Example 2. Solve the right-angled PQR when Q = 90°, R = 60° and RQ = 3
Solution:
Here, in right-angled 'PQR, P
Q
‘Q = 90°, ‘R = 60°
RQ = 3

We know, In 'PQR, 60°
‘P + ‘Q + ‘R = 180° R
or, ‘P + 90° + 60° = 180°
3

or, ‘P = 180° – 150 = 30°

Now, using ‘R = 60° as the reference angle,

tan R = PQ
RQ
PQ
or, tan 60° = 3
or,
3 = PQ
3
? PQ = 3

Again,

PR2 = PQ2 + RQ2 [ PQR is a right-angled ']

or, PR2 = 32 + ( 3)2

or, PR2 = 9 + 3

or, PR = 12

? PR = 2 3

(ii) When two acute angles and one side is given.
In this case, all three angles and one side of a right angled triangle are given.

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Solution of Right-angled Triangle and Height and Distance

Example 3. Solve ABC where B = 30°, A = 60°, and b = 2 B
30°
Solution: In ' ABC,

‘A = 60°, ‘B = 30° and b = 2.

So, we have,

‘A + ‘B + ‘C = 180° (sum of angles of ') 60° C
A
or, 60° + 30° + ‘C = 180°
b=2

or, ‘C = 180° – 90°

? ‘C = 90°

Now, sinB = AC
AB
2
or, sin 30° = AB

or, 1 = 2
2 AB

or, AB = 4 units

Similarly,

tan 30° = AC
BC
1 2
or, 3 = BC

? BC = 2 3 units

CASE III

When two sides and angle between them is given.
The third side can be simply calculated using Pythagoras theorem.

Angle can be found with the help of sides by using standard angle.

Unknown angle can be found by the help of sides by using trigonometric ratios of
standard angles.

Example 4. (a) In 'ABC if C = 90°, c = 4 3 and b = 2 3 , solve the triangle.
Solution:
(b) Solve 'ABC if ‘C = 90° and a = c = 2 B
114 (a) Here, ‘C = 90° c = 4 3 and b = 2 3

Now, we know using Pythagoras theorem, we get,

(AB)2 = (BC)2 + (AC)2 ( ABC is a c=4 3
right-angled ' )
or, (4 3 )2 = (2 3 )2 + (BC)2

or, 16 × 3 – 4 × 3 = (BC)2 A b=2 3 C

vedanta Excel in Additional Mathematics - Book 8

Solution of Right-angled Triangle and Height and Distance

or, (BC)2 = 48 – 12

or, (BC)2 = 36

? BC = 36 = 6

Now, tanA = BC
AC
6
or, tanA = 23

or, tanA = 3 × 3 [Multiplying by 3 on both
3 3 numerator and denominator.]

or, tanA = 3 3
3

or tanA = 3

or, tanA = tan 60°

? A = 60°

Again, we know,

‘A + ‘B + ‘C = 180°

or, 60° +‘B + 90° = 180°

or, ‘B = 180° – 150°

? ‘B = 30°

(b) In 'ABC,

? ‘C = 45° a = c i.e. AB = BC

So, if two sides are equal, it is an isosceles ' . A
Hence, the angles opposite to them are also equal.

Since 'ABC is an isoceles right angled triangle, c=2

‘CAB = ‘ACB = 45° C a=2 B
Now, using Pythagoras theorem,

AC2 = AB2 + CB2

or, AC2 = 22 + 22 or, AC2 = 4 + 4

or, AC2 = 8 or, AC = 8

? AC = 2 2

Again, tanC = AB = 2 = 1
BC 2

or, tanC = tan45°

or, ‘C = 45°

vedanta Excel in Additional Mathematics - Book 8 115

Solution of Right-angled Triangle and Height and Distance

and ‘A = 180° – (90° + 45°) = 45°
? AC = 2 units, ‘A = 45°, ‘C = 45°

CASE IV

When all three sides are given:
(i) determine the longest side which is hypotenuse.
(ii) the angle opposite to longest side is the right angle.
(iii) the remaining two angles can be calculated using trigonometric ratios.

Example 5. In triangle ABC, if c = 3, a = 2, and b = 1, solve the triangle.
Solution:
In ∆ABC, B

BC = a = 2, CA = b = 1, AB = c = 3 a=2 c= 3
Here, the longest side is BC

Hence, angle opposite to it is right angle. C A
Taking ‘C as a reference angle b=1

tanC = AB = 3 = tan60°
AC 1

? ‘C = 60°

and ‘A + ‘B + ‘C = 180°

or, 90° + ‘B + 60° = 180°

or, ‘B = 180° – 150°

? ‘B = 30°

? ‘C = 60°, ‘B = 30°

Exercise 7.1

1. Solve 'ABC if

(a) ‘A = 90°, ‘B = 60°, BC = 4 and AB = 2 3
AB = AC = 2
(b) ‘A = 90°, ‘C = 45° , b = 3 and a = 1
a = 4 and c = 2
(c) ‘C = 90° , ‘A = 30° ,

(d) ‘A = 90°, ‘B = 60° ,

2. Solve ' ABC if

(a) ‘C = 90°, ‘B = 30° and b = 2 3

(b) ‘B = 90°, ‘C = 45° and AC = 2

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Solution of Right-angled Triangle and Height and Distance

3. Solve ' ABC if

(a) ‘B = 60°, ‘A = 30° and c = 40

(b) ‘A = 60°, ‘B = 30° and BC = 5 3

4. Solve 'ABC if

(a) ‘B = 90°, c = 1 and a = 1

(b) ‘C = 90° , c = 8 3, b = 4 3

(c) ‘C = 90°, a = 10 2 and b = 10 2

(d) ‘A = 90°, b = 6 2, a = 12

5. Solve 'ABC if

(a) a = 4, b = 2 3 and c = 2 (b) a = 3, b = 1 and c = 2

1. (a) ‘C = 30°, AC = 2 (b) ‘B = 45°
(c) ‘B = 60°, c = 2 (d) ‘C = 30°, b = 2 3
(b) a = c = 2, ‘A = 45°
2. (a) ‘A = 60°, a = 6, c = 4 3 (b) ‘C = 90°, AB = 10, AC = 5
3. (a) ‘C = 90°, AC = 20 3, BC = 20 (b) a = 12, ‘B = 30°, ‘A = 60°
4. (a) ‘A = ‘C = 45°, AC = 2 (d) ‘C = ‘B = 45°, c = 6 2
(b) ‘A = 60°, ‘B = 30°, ‘C = 90°
(c) ‘A = ‘B = 45°, c = 20
5. (a) ‘A = 90°, ‘B = 60°, ‘C = 30°

7.2 Height and Distance

Height and distance deals with finding the height or distance of any object without
actually measuring the actual height or actual distance between any two objects.
This is one of the applications of trigonometry.

To find the height of a tall pole or the distance between an object and the eye's of an
observer, there are always three constraints:

(a) The height of the object

(b) The angle made by the observer's eye to the top (angle of elevation) / bottom
of the object (angle of depression). (Top if the object lies above the eyesight
and bottom if the object lies below the eye sight,)

(c) The distance between the foot of the object and the observer.

If any two of these three conditions are given, the third can be obtained easily
using trigonometrical ratio of standard angle.

In figure, a bird is at the top of the tree.

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Solution of Right-angled Triangle and Height and Distance

P

LINE OF SIGHT

O T ANGLE OF ELEVATION

HORIZONTAL LINE T

In the figure,

‘POT = angle of elevation.

OP = line of sight

OT = distance between the man and tree

The angle made by the horizontal line and the line of sight while observing any
object that is above the eye level is called Angle of Elevation.

O HORIZONTAL LINE L
DLINEAONFGSLIGEHOTF DEPTRESSION C
D

The angle made between the horizontal line and the line of sight when the object
lies below the level of the eye, it is called as angle of depression.
In the above figure, a car is below the house.

‘LOC = D = angle of depression.
OL//DC are the horizontal lines.
How to measure the angle of Elevation and angle of Depression ?

We measure the angle of elevation and angle of
depression by using special instruments like
theodolite, sextant, clinometer, hypsometer etc.
But clinometer is usually the most commonly used
instrument which looks as given in the adjoining
figure.

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Solution of Right-angled Triangle and Height and Distance

Note :

To solve the problems related to height and distance :
(i) Construct a right angled triangle.
(ii) Write informations given in problem in the figure (like angles in degree,

height, or distance in metre/cm etc.)
(iii) Use sin, cos, or tan from trigonometric ratios once or more by using the angle

of elevation or angle depression.

Worked out Examples

Example 1. A man observes the top of a pole from 20 meter away from the foot
of the pole and finds the angle of elevation to be 45°. Find the height
of the pole.

Solution: Let A be the position of the man and BC be the height of the pole.

Angle of elevation, ‘BAC = 45°.

Distance between the foot of the pole and the man, (AC) = 20 m.

Height of the pole (BC) = ? B

In right-angled ' ABC,

tan 45° = BC
AC
BC
or, 1 = 20 45°
A 20 m
? BC = 20 m C

? The height of the pole is 20 m.

Example 2. A tree is 25 m tall. A man observed the top of a tree and finds the angle of

elevation to be 30°. Find the distance between the man and the foot
A
of the tree.

Solution: Let C be the position of the man and AB be the tree.

Angle of elevation, ‘ACB = 30° 25m
B
Height of tree, AB = 25 m 30°

Distance between the man and foot of the tree, BC =C ?

We know in right-angled 'ABC,

tan 30° = AB
BC

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Solution of Right-angled Triangle and Height and Distance

or, 1 = 25
3 BC

? BC = 25 m

? The distance between the foot of the tree and the man is 15 3 m.

Example 3. A girl on the top of a building 45 m tall observes a car on the opposite
Solution: side of the road and finds the angle of depression to be 60°. Find
the distance between the foot of the building and the car.

Let PQ be the building and R be the position S 60° P
of the car.

Angle of depression, ‘SPR = ‘PRQ = 60° 60° 45m
R Q
(Alternate angles in parallel lines are equal,
SP//RQ)

Height of the building, PQ = 45 m

Distance between the foot of the building and the car, RQ = ?

We know in right-angled ' PQR,

tan 60° = PQ
RQ

or, 3 = 45
RQ

or, RQ = 45 × 3
3 3

or, RQ = 30 3
3

? RQ = 15 3 m

The distance between the foot of the building and the car is 15 3 m

Example 4 . From the top of a cliff, a ship is observed and found that the angle of
Solution: depression to be 45°. If the ship is 200 m away from the cliff, find the
height of the cliff.

Let AB be the cliff and C be the position of the D A
ship. 45°

Angle of depression from the top of the cliff to
the ship ‘DAC = ‘ACB = 45°

( DA//CB) 45° B
C 200m
The distance between the bottom of the cliff
to the ship, BC = 200m

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The height of the cliff, AB = ?

We know in right-angled ' ABC,

tan 45° = AB
BC
AB
or, 1 = 200

? AB = 200 m

Hence, the height of the cliff is 200 m.

Example 5. A ladder rests on the top of the wall in such a way that it makes an
Solution:
angle of 45° with the floor. If the height of the wall is 12 meters, find

the length of the ladder. L

Let LA be the wall and LD be the ladder.

Angle made by the ladder with the base, 45°240 m ? 12m
‘LDA = 45° D A

Height of the wall, LA = 12 m
Length of the ladder, LD = ?

In right-angled 'LDA,

sin 45° = LA
LD
1 12
or, 2 = LD

or, LD = 12 × 2

? LD = 12 2 m

Hence, the length of the ladder is 12 3 m.

Example 6. A well stretched string of a kite is inclined to the horizon at angle of
30°. If the string laid out is of length 240 m, find the height of the kite.

Solution: Let A be the position of the kite and CA be the length of string laid out.

Angle made by the string with the horizontal level, ‘ACB = 30° A
Height of the kite, AB = ?

Length of the string laid out, CA = 240 m.

In right-angled 'ABC,

sin30° = AB 30° B
AC C
1 AB
or, 2 = 240

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Solution of Right-angled Triangle and Height and Distance

or, 2 × AB = 1 × 240

? AB = 240 = 120 m
2

Hence, the height of the kite is 120 m.

Example 7. A ladder 40 meter long is resting against a wall. If the distance
Solution: between the foot of the ladder from the wall is 20 3 m, find the
angle made by the ladder with the floor.

Let AB be the wall and AC be the ladder.

Length of the ladder, AC = 40 m.

Distance between the foot of the wall and the ladder, BC = 20 3 m.

Angle made by ladder with the floor ‘ACB = ? A

We know in right-angled 'ABC,

cosC = BC 40m
AC

or, cosC = 20 3 C 20 3 m B
40

or, cosC = 3
2

or, cosC = cos30°

? ‘C = 30°

Hence, the angle made by the ladder with the floor is 30°.

Example 8. A lamp - post of height 20 3 m casts a shadow 60 m long due to
Solution: sun’s ray. Find the altitude of the sun.

Let AB be the lamp-post and BC be the length of shadow casted.

Height of the lamp - post, AB = 20 3 m

Length of the shadow, BC = 60 m.

Altitude of sun, ie angle made due to sun rays, ‘ACB = ?

We know in right-angled 'ACB, A

tanC = AB
BC

or, tanC = 20 3 20 3 m
60 C 60 m B

or, tanC = 3 = 3
3 3. 3

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Solution of Right-angled Triangle and Height and Distance

or, tanC = 1
3
or, tanC = tan 30°

? ‘C = 30°

Hence, the altitude of the sun is 30°.

Exercise 7.2

Short Questions :

1. Find the values of x, y, z, T from the given right-angled triangle by using
trigonometric ratios:

(a) A (b) P (c) M

x

x 60 m
T
45° R 30° Q PN
40 m 20 3 m
B 25 m C

(d) P (e) P (f) A

30° 100 m40 m60 m
20 3 m By
RT Q Q TR C
20 m 40 m

(g) M (h) G (i) A

45 m z

N 30° P 45° B 60° C
y 25 m x
(j)
H I

X (k) P 60° S (l) P 45° S

20 3 m z y 100 m
Y QR
Z 60° Qz R
20 3 m

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Solution of Right-angled Triangle and Height and Distance

Long Questions :

2. (a) The angle of elevation of the top of a house from a point on the horizontal
line 60 m away from the foot of the house is 30°. Find the height of the
house.

(b) The angle of elevation of a tower at a distance of 36 meters from the foot of
the tower is 60°. Find the height of the tower.

3. (a) A tower 250 m tall is observed from a point on the horizontal line and
found the angle of elevation to be 30°. Find the distance between the foot
of the tower and the observation point.

(b) The angle of elevation to the top of Dhorohara 80 feet high as observed
from Shahidgate is found to be 30°. Find the distance between Dhorohara and
Shahidgate.

4. (a) From a top of a building of 180 m height the angle of depression of a bus
on the ground is found to be 60°. Find the distance between the foot of the
building and the bus.

(b) From the window of a house of height 60 m, a boy observes a car on the
other side of the road and finds the angle of depression to be 45°. Find the
distance between the foot of the house and the car.

5. (a) A bus is on opposite side of the road of a building. The distance between
the foot of the building and the bus is 20 m. The angle of depression to the
bus as observed from the top of the building is 45°. Find the height of the
building.

(b) From the top of a cliff, the angle of depression to a boat 200 3 m far away
from the cliff is found to be 30°. Find the height of the cliff.

6. (a) A ladder leans against a wall making an angle of 30° at a point 12 m away
from the wall. Find the length of the ladder.

(b) A ladder rests on top of a wall of height 10 m making an angle of 30° with
the ground. Find the length of the ladder.

7. (a) The string of a kite is inclined to the ground at an angle of 30°. If the length
of the string is 300 m, find how high is the kite flying?

(b) The string of a kite is 80 2 m long and it makes an angle of 45° with the
ground. Find the height of the kite from the ground level.

8. (a) A ladder 12 m long rests on a wall 8 m in height. Find the angle made by
the ladder with the ground.

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Solution of Right-angled Triangle and Height and Distance

(b) A wooden plank 15 3 m long rests on a wall in such a way that it makes an
angle with ground at 15 m away from the foot of the wall. Find the angle made
by the plank with the ground.

9. (a) A pillar of height 15 m high casts a shadow of length 15 3 m. Find the
altitude of the sun.

(b) What is the altitude of the sun when a lamp post of height 35 m casts a
shadow of 35 3 m long ?

10. (a) A flagstaff of height 7 m stands on the top of a tower. The angle subtended
by the tower and flagstaff to a point on the ground are 45° and 15°
respectively. Find the height of the tower.

(b) A ladder of 9 m long reaches a point 9 m below the top of a vertical flagstaff.
From the foot of the ladder the angle of elevation of the flagstaff is 60°. Find
the height of the flagstaff.

1. (a) 45 m (b) 46.19 m (c) 60° (d) 60°
(e) 45° (f) 20 3 m (g) 45 3 m (h) 25 m
(i) 50 m (j) 30 m (k) 60 m (l) 100 m
(b) 36 3 m
2. (a) 20 3 m (b) 80 3 m
3. (a) 250 3 m (b) 60 m
4. (a) 60 3 m (b) 200 m
5. (a) 20 m (b) 20 m
6. (a) 8 3 m (b) 80 m
7. (a) 150 m (b) 45°
8. (a) 41.81° (b) 30°
9. (a) 30° (b) 13.5 m
10. (a) 9.56 m

vedanta Excel in Additional Mathematics - Book 8 125

Coordinates Geometry 8

Coordinates Geometry

8.1 Distance formula

The distance between any two points on a plane can be evaluated. Calculation
becomes easy if it moves on the same line. But if it is moving arbitrarily, we need to use
formula.

YY

d = 3 units B(9,7)
A(3, 1)
X' P X X' X d=?

O (3,0)
(0,0)

Y' Y'

Figure (i) Figure (ii)

In figure (i), the distance between O and P is simply 3 units.

In figure (ii), it is not easy to find distance between A and B.

Derivation of distance formula Y

Let XX' be YY' be two mutually perpendicular straight B (x2, y2)
lines intersecting at origin ‘O’. A (x1, y1) and B(x2, y2)
be any two points.

To find distance between A and B, let's draw AP 0 A (x1, y1) C X
perpendicular to OX and BQ perpendicular on OX. X’ PQ
Similarly, AC is drawn perpendicular on BQ.

Such that, Y’

OP = x1 , AP = y1
OQ = x2 , BQ = y2
So, CQ = AP = y1
BC = BQ – CQ = y2 – y1

126 vedanta Excel in Additional Mathematics - Book 8

Coordinates Geometry

and AC = PQ = OQ – OP = x2 – x1
Now, in right-angled ' ABC,

h2 = p2 + b2

or, (AB)2 = (BC)2 + (AC)2

(AB)2 = (y2 – y1)2 + (x2 – x1)2 A(x, y)
? (AB)2 = (x2 – x1)2 + (y2 – y1)2

Hence, AB (d) = (x2 – x1)2+(y2 – y1)2
Note : If one point is at origin O (0, 0) and other point is at A (x, y).

(OA) = (x – o)2+(y – o)2 0 (0, 0)

? OA(d) = x2 + y2

Worked Out Examples

Example 1. Find the distance between the points A(4, 5) and B(7, 9).
Solution:
Let A(4, 5) be (x1, y1)
B(7, 9) be (x2, y2)

Then, distance between AB is given by,

(AB)2 = (x2 – x1)2 + (y2 – y1)2
= (7 – 4)2 + (9 – 5)2

= (3)2 + (4)2

= 9 + 16

= 25

? AB = 25 = 5 units.

Example 2. Find the distance between the points (5, 4) and (– 1, – 4).
Solution: Let A(5, 4) be (x1, y1)

B(– 1, – 4) be (x2, y2)
Then, distance between AB is given by,

(AB)2 = (x2 – x1)2 + (y2 – y1)2
= (– 1 – 5)2 + (– 4 – 4)2
= (– 6)2 + (– 8)2
= 36 + 64 = 100

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Coordinates Geometry

? AB = 100 = 10 units.

Example 3. P(4, –3), Q(6, 3) and R(8, – 3) are three points, Show that PQ = QR.

Solution: Here to show PQ = QR, firstly let's find out PQ and QR
For PQ, distance is given by,

(PQ)2 = (6 – 4)2 + (3 + 3)2
= (2)2 + (6)2

= 4 + 36

= 40
PQ = 40 = 2 10 units.

For QR, distance is given by,
(QR)2 = (8 – 6)2 + (– 3 – 3)2

= (2)2 + (– 6)2

= 4 + 36 = 40
? QR = 40 = 2 10 units.
? PQ = QR = 2 10 units
? PQ = QR proved.

Example 4. If P(x, 3) and Q(7, – 1) and PQ = 5 units find all the possible values
Solution: of x.

Here, P (x , 3) = (x1, y1)
Q(7, – 1) = (x2, y2)

Also, PQ = 5 units.

We know,
(PQ)2 = (x2 – x1)2 + (y2 – y1)2

or, 25 = (7 – x)2 + (– 1 – 3)2
or, 25 = 49 – 14x + x2 + (–4)2
or, 25 = 49 – 14x + x2 + 16
or, x2 – 14x + 49 + 16 – 25 = 0
or, x2 – 14x + 40 = 0
or, x2 – (10 + 4) x + 40 = 0
or, x2 – 10x – 4x + 40 = 0

or, x (x – 10) – 4 (x – 10) = 0

128 vedanta Excel in Additional Mathematics - Book 8

Coordinates Geometry

or, (x – 10) (x – 4) = 0
Either, (x – 10) = 0
? x = 10
Or, x – 4 = 0
? x=4
? The positive values of x are 4 and 10.

Applications of distance formula

Using the distance formula, we can verify various properties of triangles and
quadrilaterals being based on their sides.

To verify the triangle:

Triangles Properties based on sides
(a) Equilateral All sides must be equal.

(b) Isosceles Any two sides must be equal.

(c) Scalene None of the sides equal.

(d) Right-angled triangle h2 = p2 + b2 (longest side is the hypotenuse).

(e) Right-angled isosceles h2 = p2 + b2 and two sides p and b must be equal.

To verify the quadrilaterals:

Quadrilateral Properties based on sides
(a) Parallelogram Opposite sides must be equal.

(b) Rectangle Opposite sides must be equal and diagonals equal.

(c) Rhombus All sides must be equal.

(d) Square All sides must be equal and diagonals equal.

Example 5. Prove that the triangle with the vertices A(–2, 2), B(2, 2) and C(4, 2)
Solution: is a scalene triangle.

To show triangle a scalene triangle,

we need to find all the sides first.

So, for side AB, A(–2, 2) = (x1, y1), B(2, 2) = (x2, y2)
(AB)2 = (x2 – x1)2 + (y2 – y1)2
= (2 + 2)2 + 2 – 2)2
= 42 + 02 = 16

vedanta Excel in Additional Mathematics - Book 8 129

Coordinates Geometry

? AB = 4 units.

Similarly for side BC,

Let B(2, 2) = (x1, y1) and C(4, 2) = (x2, y2)
(BC)2 = (x2 – x1)2 + (y2 – y1)2
= (4 – 2)2 + (2 – 2)2
= 22 + 02 = 4

? BC = 2 units
Again for CA, C(4, 2) = (x1 , y1), A(–2, 2) = (x2, y2)

(CA)2 = (x2 – x1)2 + (y2 – y1)2
= (– 2 – 4)2 + (2 – 2)2
= (– 6)2 + 02

= 36 + 0 = 36

? CA = 36 = 6

Example 6. ? AB z BC z CA i.e. 4 z 2 z 6
Solution:
? ABC is a scalene triangle. proved.

Show that the points (4, 4), (10, 1), and (13, 22) are the vertices of a right-
angled triangle.

Let A(4, 4), B(10, 1), and C(13, 22) be the three points.
For side AB,

Let A(3, 3) be (x1, y1) and B(10, 1) be (x2, y2)
Then, using distance formula,

(AB)2 = (x2 – x1)2 + (y2 – y1)2
= (10 – 4)2 + (1 – 4)2

= (6)2 + (–3)2
= 36 + 9

= 45
? AB = 45 = 3 5 units.
For side BC, B(10, 1) and C(13, 22), using distance formula,

(BC)2 = (13 – 10)2 (22 – 1)2
= (3)2 + (21)2

= 9+ 441

130 vedanta Excel in Additional Mathematics - Book 8

Coordinates Geometry

= 450

? BC = 450 = 15 2 units.

Again, for side CA, C(13, 22) and A(4, 4), using distance formula.

(CA)2 = (4 – 13)2 + (4 – 22)2 B(10, 1)
= (–9)2 + (–18)2

= 81 + 324

= 405

? CA = 405 units.

Now, to prove ABC a right-angled triangle,

h2 = p2 + b2 A(4, 4) C(13, 22)

or, (BC)2 = (CA)2 + (AB)2 [ BC is the longest side]

or, (15 2)2 = ( 405)2 + (3 5)2

or, 450 = 405 + 45

? 450 = 450 (True)

Hence, ABC represents a right-angled triangle.

Example 7. Show that the points P(2, 2), Q(5, 2), R(5, 9), and S(2, 9) are the
vertices of parallelogram.

Solution: Let P(2, 2), Q(5, 2), R(5, 9) and S(2, 9) be any four points.

Now, for side PQ, using distance formula, we get,

(PQ)2 = (5 – 2)2 + (2 – 2)2 P(2, 2) S(9, 9)
= (3)2 + (0)2 = 9

PQ = 9 = 3 units.

Again, for side QR,

(QR)2 = (5 – 5)2 + (9 – 2)2 Q(5, 2) R(5, 9)

= (0)2 + (7)2 = 49

? QR = 49 = 7 units

Similarly, for side RS,

(RS)2 = (2 – 5)2 + (9 – 9)2

= (– 3)2 + (0)2

=9+0=9

? RS = 9 = 3 units.

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Coordinates Geometry

Again, for side PS,
(PS)2 = (2 – 2)2 + (9 – 2)2
= 0 + (7)2 = 49

? PS = 49 = 7 units

Since the opposite sides, PQ = RS = 3 units and QR = PS = 7 units,
PQRS represents a parallelogram.

Example 8. Prove the points P(2, 2), Q(4, 2), R(4, 4) and S(2, 4) represents a
Solution: square.

Let P(2, 2), Q(4, 2), R(4, 4), and S(2, 4) represent a quadrilateral.

We require to show 'PQRS is a square'. P(2, 2) S(2, 4)
So, for side PQ,

(PQ)2 = (4 – 2)2 + (2 – 2)2

= (2)2 + 0 = 4

? PQ = 4 = 2 units Q(4, 2) R(4, 4)
Again, for side QR,

(QR)2 = (4 – 4)2 + (4 – 2)2

= 0 + (2)2

=0+4 =4

? QR = 4 = 2 units

Similarly, for side RS,

(RS)2 = (2 – 4)2 + (4 – 4)2

= (– 2)2 + 0 = 4

? RS = 4 = 2 units

Again, for side PS,

(PS)2 = (2 – 2)2 + (4 – 2)2

= 0 + (2)2 = 4

? PS = 4 = 2 units

For diagonal QS,

(QS)2 = (2 – 4)2 + (4 – 2)2

= (–2)2 + (2)2

=4+4 =8

132 vedanta Excel in Additional Mathematics - Book 8

Coordinates Geometry

? QS = 8 = 2 2 units.
Similarly for the diagonal PR,
(PR)2 = (4 – 2)2 + (4 – 2)2
= (2)2 + (2)2
=4+4 =8
? PR = 8 = 2 2 units.
Since, PQ = QR = RS = PS = 2 units
and the diagonals PR = QS = 2 2 units, PQRS represents a square.

Exercise 8.1

Short Questions :

1. Find the distance between the two points given below:

(a) (3, 4) and (6, 8) (b) (5, –6) and (0, 6)

(c) (p, 0) and (0, q) (d) (–5, 4) and (8, 2)

2. Look at the graph given below and determine the co-ordinate of A and B.

Use it to calculate the distance between them.

(a) Y (b) Y

BA

XO X' X O B X'

A

Y' Y'

Long Questions :

3. (a) O(2, 2), A(5, 4), and B(– 1, 4) are three points. Show that OA = OB.

(b) P(3, –4), Q(5, 2), and R(7, – 4) are three points. Show that PQ = QR.

(c) P(–1, 3), Q(7, –3), and R(4, 1) are three points. Prove that PQ = 2QR.

4. (a) If A is a point on y-axis whose ordinate is 5 and B is a point (–3, 1). Calculate
the distance between A and B.

(b) A is a point on y-axis whose ordinate is 5 and B is a point on x-axis whose
abscissa is –5. Find the distance between A and B.

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Coordinates Geometry

5. (a) The distance between the points P(1, 3) and Q(x, 7) is 5 units. Find the
possible values of x.

(b) The distance between the points A(a, 2) and B(2, –2) is 5 units. Find the
possible values of a.

6. (a) Show that the points A(–5, 2), B(2, 5), and C(5, 2) are the vertices of a scalene
triangle.

(b) Prove that the points P(3, 2), Q (6, 2), and R(6, 6) represents a scalene triangle.

7. (a) Show that the points (5, 5), (5, 0), and (0, 0) are the vertices of an isosceles triangle.

(b) Prove that the points (4, 5), (5, –2), and (1, 1) are the vertices of an isosceles
triangle.

8. Prove that (2, 2), (–2, –2), and (–2 3, 2 3) are the vertices of equilateral triangle.

9. (a) Show that the points (1, 0), (2, 1), and (3, 0) represents a right - angled
triangle.

(b) Prove that the points (0, 0), (0, 3), and (3, 0) are the vertices of right-angled
isosceles triangle.

10. (a) Prove that the points (6, 8), (3, 7), (–2, –2), and (1, –1) are the vertices of
parallelogram.

(b) Prove that the points (4, 8), (0, 2), (3, 0), and (7, 6) are the vertices of parallelogram.

11. (a) Show that the points (0, 2), (1, 1), (4, 4), and (3, 5) are the vertices of a rectangle.

(b) Prove that the points (2, 4), (2, 3), (4, 3), and (4, 4) are the vertices of a rectangle.

12. (a) Prove that the four points (4, 3), (6, 4), (5, 6), and (3, 5) represent a rhombus.

(b) Verify that (7, 7), (5, 0), (–2, –2), and (0, 5) represent a rhombus.

13. (a) Show that the points A(1, 1), B(3, 1), C(3, 3), and D(1, 3) represent a square.

(b) Prove that the points (0, 0), (1, 0), (1, 1), and (0, 1) are the vertices of a
unit square. Also plot the square on graph paper.

1. (a) 5 units (b) 13 units (c) p2 + q2 units (d) 173 units
2. (a) 2 17 units (b) 37 units
5. (a) – 2, 4 (b) – 1, 5 4.(a) 5 units (b) 5 2 units

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Coordinates Geometry

8.2 Section formula

If a piece of bamboo stick of 10 cm long is divided into two equal pieces where does

it break? Obviously the answer is at 5 cm. Similarly, if it has to be divided into two

pieces at the ratio 1:4. We divide it into 2 cm and B
8 cm. 3cm

Internal Division : Let AB be a given line segment 4cm P B
Fig. (i)
and P be a point on it. If P lies within AB, P is said to 3cm
P
divide AB internally in the ratio of AP : PB. It means A
that P is an internal point of AB.

To find the coordinates of a point dividing the line 6cm

segment joining two given points internally in the A Fig. (ii)

given ratio.

Let A(x1, y1) and B(x2, y2) be two given points. Let P(x,y) be any point on the line
joining A and B. Let P divides AB in the ratio of m1 : m2 internally. Then, we write
AP : PB = m1 : m2.

Perpendiculars AL, PN, and BM are drawn from the point A,P, and B respectively on the
X-axis. Y

B(x2, y2)

Again perpendiculars AQ and PR are drawn from

the points A and B to PN and BM respectively. P(x, y) R

Then, AQ = LN = ON – OL = x – x1 A(x1, y1) Q

PR = NM = OM – ON = x2 – x X' O LN MX
QP = NP – NQ = NP – LA = y – y1

BR = MB – MR = MB – NP =y2 – y Y'

Since QP and BR are parallel, triangles PAQ and BPR are similar.

The corresponding sides of similar triangles are proportional.

? AP = AQ = QP
BP PR RB
or, mm21 xx2––xx1 yy2––yy1
Taking the first two ratio, we get,

mm21 xx2––xx1
or, m1x2 – m1x1 = m2x – m2x1

or, m1x2 + m2x = m2x + m1x

or, m1x2 + m2x1 = x(m1 + m2)

? x = m1x2 + m2x1
m1 + m2

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Coordinates Geometry

Again, taking the first and third ratios, we get

mm12 yy2––yy1
or, m1y2 – m1y = m2y – m2y1

or, m1y2 + m2y1 = (m1 + m2) y

? y = m1y2 + m2y1
m1 + m2
m1x2 + m2x1 m1y2 + m2y1
Hence the coordinates of P are m1 + m2 , m1 + m2

Special Cases

(i) If P(x, y) is the mid point of line segment joining A(x1, y1) and B(x2, y2) then
AP = BP and AP : BP = 1:1.

Then, (x, y) = m1x2 + m2x1 , m1y2 + m2y1 = x1 + x2 , y1 + y2
m1 + m2 m1 + m2 2 2
(ii) If P(x, y) is an external point of the line segment
P(x, y)
joining the point A(x1, y1) and B(x2, y2), the
coordinates of C are given by B(x2, y2)

(x, y) = m1x2 – m2x1 , m1y2 – m2y1
m1 – m2 m1 – m2
A(x1, y1)

(iii) If the pont P(x, y) divides the line segment joining the points A(x1, y1) and B(x2,
y2) in the ratio of k : 1 internally, then the coordinates of P are given by

(x, y) = kx2 + x1 , ky2 + y1
k+1 k+1
For the problems in which the ratio is to be calculated by which a point divides

the line segment, the formula is more convenient.

Centroid Formula

Let A(x1, y1), B(x2, y2) and C(x3, y3) be the vertices of triangle A(x1,y1)
∆ABC. Let D,E, and F be the mid-points of the sides BC,

CA, and AB respectively. Then the medians AD, BE, and CF F2 E
are drawn. The medians intersect at G which is called the

centroid of the triangle ABC. Since D is the mid point of BC, 1G
x2 + x3 y2 + y3 D
the coordinates of D are 2 , 2 . B(x2,y2) (x3,y3)

From plane geometry, we know that the centroid of a triangle divides the median in

the ratio 2:1. G divides AD in 2:1 ratio.

2 × x2 + x3 + 1 × x1 2 × y2 + y3 + 1 × y1
2 + 1 2
G (x, y) =G 2 ,
2+1

136 vedanta Excel in Additional Mathematics - Book 8

Coordinates Geometry

=G x1 + x2 + x3 , y1 + y2 + y3
3 3
Definition : The point of intersection of the medians of a triangle is called centroid

of the triangle.

Worked Out Examples

Example 1. Find the coordinates of a point P which divides the line segment
Solution: AB whose coordinates are A(2, 2) and B(4, 5) in the ratio 1 : 2

Example 2. Here, A(2, 2) and B(4, 5) are the end points of line segment AB. P(x, y)
Solution: divides AB in the ratio 1 : 2 i.e. m1 : m2 = 1 : 2

Then, using section formula, B(4, 5)

x = m1x2 + m2x1
m1 + m2

= 1 . 4 + 2 . 2 = 8 P
1 + 2 3

Again, y = m1y2 + m2y1 A(2, 2)
m1 + m2

= 1 . 5 + 2 . 2 = 9 = 3
1 + 2 3

? P(x, y) = 8 , 3
3

Find the coordinate of the point which divides the line joining the
points (5, 5) and (7, 3) externally in the the ratio 3 : 1.

Let A (5, 5) and B(7, 3) be two end points of line segment AB. P(x, y)
divides the line externally in the ratio 3:1 .

i.e. AP : BP = m1 : m2 = 3 : 1 P
B
Then, using section formula,

x = m1x2 – m2x1
Again, y m1 – m2

= 3 . 7 – 1 . 5 = 21 – 5
3 – 1 2
A

= 16 = 8
2

= m1y2 – m2y1
m1 – m2

= 3 . 3 – 1 . 5 = 9 – 5
3 – 1 2

= 4 = 2
2

Hence, the required point is (8, 2).

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Coordinates Geometry

Example 3. In what ratio does the point P(6, 9) divides the line joining the points
Solution: A(1, 3) and B(12, 17) ?

Let the point P(6, 9) divide the line joining the points A(1, 3) and

B(12, 17) in the ratio m1 : m2 B(12, 17)

Then, using section formula,

x-coordinate of P = m1x2 + m2x1 P(6, 9)
m1 + m2

or, 6 = m1 . 12 + m2 . 1 A(1, 3)
m1 + m2
or, 6m1 + 6m2 = 12m1 + m2

or, 5m2 = 6m1

? m1 = 5
m2 6

Hence, the required ratio is 5 : 6.

Example 4. In what ratio does the point P(x, – 7) divide the line joining the points
Solution: A(2, 3) and B(4, 8). Find the value of x.

Let the point P(x, –7) divide the line joining the points A(2, 3) and
B(4, 8) in the ratio m1 : m2

Using section formula,

–y-7co=ormdi1nm.a8t1e++=mmm22m1.y312 + m2y1
+ m2

or,

or, –7m1 – 7m2 = 8m1 + 3m2

or, –7m1 – 8m1 = 7m2 + 3m2

or, –15m1 = 10m2

or, m1 = 10 = 2
m2 –15 –3

? m1 : m2 = 2 : – 3

Hence, the required ratio is 2 : (–3)

Again, to find the value of x, using section formula again,

x = m1x2 + m2x1
m1 + m2

= 2 . 4 + (–3) . 2 = 8 – 6
2 + (–3) 2 – 3

= 2 = –2
–1

The value of x is –2.

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Coordinates Geometry

Example 5. Find the co-ordinates of the mid-point of the line joining the points
Solution: A(7, 8) and B(3, 2).

Let A(7, 8) and B(3, 2) be any two points.

Then, the mid point is determined by

x = x1 + x2 , y = y1 + y2
2 2
7 + 3 8 + 2
or, x = 2 , or, y = 2
or,
or, x = 10 , ? y = 10
2 2

? x=5 y=5

The required mid-point is (5, 5).

Example 6. Find the other end of a line segment whose one end is (8, 4) and the
Solution: mid-point is (4, 3).

Let P(8, 4) and Q(x2, y2) be the two ends of the line whose mid point is
R(4, 3)

Then, using mid point formula,

x = x1 + x2 , y = y1 + y2
2 2
8 + x2 4 + y2
or, 4 = 2 , or, 3 = 2
or,
or, 8 = 8 + x2 6 = 4 + y2

or, 8 – 8 = x2 or, 6 – 4 = y2

? x2 = 0 ?y2 = 2

Hence, the required point is (0, 2).

Example 7. One end of a diameter is (4, –2) and its center has the coordinate
(2, 2). Find the co-ordinate of the
other end of the diameter.

Solution: Here, one end of the diameter is Q(x2, y2) C(2, 2) P(4, –2)
P(4, –2) and the center is C(2, 2).

Let other point be (x2, y2).

Now, using mid-point formula,

x = x1 + x2 , y = y1 + y2
2 2
4 + x2 –2 + y2
or, 2 = 2 , or, 2 = 2

or, 4 = 4 + x2 or, 4 = –2 + y2

or, 4 – 4 = x2 or, 4 + 2 = y2

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Coordinates Geometry

? x2 = 0 ? y2 = 6

? The coordinate of the other end is (0, 6).

Example 8. Find the value of a and b when M(8, 4) is the mid-point of the line
Solution:
joining segment the point of P(6, b) Q(a, –6)
and Q(a, – 6).

Let M(8, 4) be the mid-point of the line M(8, 4)
joining the points P(6, b) and
Q(a, – 6)

Then, using mid point formula, P(6, b)

x = x1 + x2 , y = y1 + y2
2 2
6 + a b + (–6)
or, 8 = 2 , or, 4 = 2

or, 16 = 6 + a or, 8 = b – 6

or, 16 – 6 = a or, 8 + 6 = b

? a = 10 ? b = 14

? The values of a and b are 10 and 14 respectively.

Exercise 8.2

Short Questions :

1. (a) Write the mid-point of line segment joining the points (x1, y1) and (x2, y2).

(b) Write the formulae to find the coordinates of the point P, which divides
join of (x1, y1) and (x2, y2) in ratio of m1 : m2.

(i) internally (ii) externally

2. Find the mid-point of line segments joining the points.

(a) A(7, 8) and B(3, 4) (b) P(–5, –7) and Q(7, –9)

(c) R(8, 8) and (10, 8) (d) T(5, 6) and R(7, 6)

Long Questions :

3. Find the coordinates of a point which divides internally the lines joining.

(a) A(4, 6) and B(4, 3) in the ratio 2 : 1.

(b) P(3, 3) and Q(2, 4) in the ratio 4 : 1.

(c) C(–4, –6) and D(2, 5) in the ratio 2 : 3.

(d) M(6, – 8) and N(0, 2) in the ratio 2 : 1.

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Coordinates Geometry

4. Find the coordinates of a point which divides externally the line joining.

(a) A(–2, 4) and B(4, – 8) in the ratio 3 : 2

(b) P(–3, –4) and Q(–8, 7) in the ratio 5 : 3

5. (a) In what ratio does the point (2, 1) divide the join of the point (1, –3) and
(4, 9) ?

(b) Find the ratio in which the point P(4, 3) divide the line joining the points
A(8, –1) and B(2, 5)?

6. (a) In what ratio does the point (x, 0) divide the line joining the points (4, 12)
and (–3, 5). Hence, find the value of x.

(b) Find the ratio in which the point (3, y) divides the line joining the points
(9, 8) and (– 4, – 6). Hence, find the value of y.

7. (a) The mid-point of a line segment is (2, –2) whose one end is (–4, –6). Find the
other end.

(b) The mid-point of a line segment is (1, 3). If one end has the co-ordinate
(–8, 8) find the co-ordinate of the other end.

8. (a) In the figure alongside AB is the

diameter and C is the center. If the

co-ordinate of A is (–8, –4) and C is C(–6, –4)

(–6, –4), find the co-ordinate of B. A(–8, –4) B(x2, y2)

(b) If the center of the diameter is
(2, 4) whose one end is (1, 2) find the
coordinate of other end.

9. Find the value of a and b if P is the mid point of the line joining AB.

(a) A(a, 4), B(6, b) and P(3, 5) (b) A(b, 2), B(10, a) and P(4, 6)

(c) A (–4, a), B(2, 2) and P b, 5 (d) A(2, 2), B(a, – 2) and P(4, b)
2

2. (a) (5, 6) (b) (1, –8) (c) (9, 8) (d) (6, 6)
3. (a) (4, 4)
(b) 11 , 19 (c) –8 , –8 (d) 2, –4
4. (a) (16, –32) 5 5 5 5 3

6. (a) x = –8 (b) – 31 , 47 5.(a) 1 : 2 (b) 2 : 1
7. (a) (8, 2) 2 2
9. (a) a = 0, b = 6 20
(b) 13

(b) (10, –2) 8.(a) (–4, –4) (b) (3, 6)
(d) a = 6, b = 0
(b) a = 10, b = -2 (c) a = 3, b = -1

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Coordinates Geometry

8.3 Slope

Inclination (T) Y

Let XX' and YY' be the mutually perpendicular lines B
intersecting at O. X

Let AB be a line which makes an angle ‘BAX with T
x-axis in the positive (anti-clock wise) direction. The X' OA
angle made by any line with the x-axis in anti-clock

wise direction is called inclination. It is denoted by T. Y'
The value of T can range from 0° to 180°. The tangent

of the inclination is called slope of the line. It is denoted by m = tan T.

Worked Out Examples

Example 1. Find the inclination in the figure given below:

(a) Y (b) Y
A

B

150° 45° T
T
X' O X X' X
A O

B

Y' Y'

Solution: (a) Here, T is the angle made by line AB in the positive direction.
So, T + 120° = 180° [Linear adjacent angles]
or, T = 180° – 150°
? T = 30°
Hence, inclination is 30°.

(b) Here, T + 45° = 180° (Being linear adjacent angles)
or, T = 180° – 45° = 135°
Hence, the inclination is 135°.

Slope of a line

The measurement of steepness of a line is called slope. The slope or the gradient of
a line is the tangent of the angle of inclination. It is denoted by m and calculated as
m = tanT.

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Coordinates Geometry

Example 2. Calculate the slope of the line whose inclination T = 60°.
Solution:
Here, Y

Inclination (T) = 60° B

? Slope (m) = tan T C 60°

= tan 60° X' O X

=3 A
Y'

Example 3. Find the inclination of the line whose slope is 1.

Solution: Here, slope of the line (m) = 1

We know, slope (m) = tan T

So, tanT = 1

or, tanT = tan 45°

? T = 45°

Hence, the inclination is 45°.

Slope of line when it passes through two points.

Let A(x1, y1) and B(x2, y2) be two points.

Then, AP and BQ are perpendicular drawn on X-axes and AC is drawn perpendicular
Y
on BQ. B(x2, y2)

‘ADP = T is the inclination of the line.

‘BAC = ‘ADP = T (corresponding angles being equal) A(x 1, y 1)T C
Now, in right-angled ' BAC,

tan T = BC X' T X
AC O
y2 – y1 D P Q
= x2 – x1

Also, we know that, Y'

Slope (m) = tan T

? m = y2 – y1
x2 – x1

Example 4. Calculate the slope of a line joining the points (4, 5) and (7, 8).

Solution: Here, let A (–1, 2) and B(3, 7) be any two points.

Slope of line AB = y2 – y1
x2 – x1

= 8 – 54= 3 = 1
7 – 3

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Coordinates Geometry

Example 5. Find the value of x if the slope of the line joining the points (x, 8) and
Solution: (10, 12) is 3.

Here, slope of line joining the points (x, 3) and (10, 12) is given by,

m = y2 – y1 = 12 – 8 = 4 x
x2 – x1 10 – x 10 –

By the question, m = 3

So, 3 = 4 x
10 –

or, 30 – 3x = 4

or, –3x = 4 – 30

or, –3x = – 26

? x = – 26 = 8 2
–3 3

Exercise 8.3

Short Questions :

1. (a) Define inclination of a line.

(b) Define slope of a line.

(c) If a line makes an angle of T with positive X-axis in anti-clockwise direction,
what is its slope?

(d) Find the slope of a line joining the points (7, 8) and (10, 11).

2. Find the inclination T if

(a) Slope (m) = 3 (b) Slope (m) = 1

(c) Slope (m) = 1 (d) Slope (m) = 0
3
3. Find the inclinations in the figure given below:

(a) Y (b) Y

A
A

45° X X' 135° X
X' O B
Y'
O
B

Y'

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(c) Y (d) Coordinates Geometry
Y
X' O X'
150° X A 45° O X
B

AB

Y' Y'

4. Calculate the slope of the line whose inclinations are:

(a) 0° (b) 30° (c) 45° (d) 60°

5. Find the slope of the line passing through the points.

(a) A(2, 2) and B(8, 10) (b) P(–4, 8) and Q(6, 4)
(c) M(2, –2) and N(7, 8) (d) R(6, –7) and S(6, 3)

6. (a) If the slope of the line joining the points (8, 7) and (4, p) is 3, find value of
p.

(b) Find the value of 'p' if the slope of the line joining the points (4, 6) and (8, p)
3
is 2 . Y

7. From the given right-angled isoceles P

triangle PQR whose side QR is parallel

to X-axis, find the slope of the sides of

the triangle. QR

X' O X

Y'

1. (c) tanT (d) 1

2. (a) 60° (b) 45° (c) 30° (d) 0°
(d) 135°
3. (a) 45° (b) 135° (c) 30° (d) 3
(c) 1 (d) f
4. (a) 0 (b) 1 (c) 2
(b) –352 145
5. (a) 4
3
6. (a) – 5 (b) 12

7. slope of PQ = 1, slope of QR = 0, slope of PR = – 1.

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Coordinates Geometry

8.4 Locus

Let us take a graph paper. Draw a circle of radius 5 units with centre at the origin.

Y

X' O X

Y'

Take any point P(x, y) on the circumference of the circle. The point P(x, y) is called
variable point on the circle. Find distance between P and O. We get one equation

x2 + y2 = 25 .................... (i)

Take any five points on the circumference of the circle.

For example : points on the circumference of the circle are (5, 0), (4, 3), (3, 4), (0, 5),
(–4, –3), (–5, 0), etc.

Put the above points in equation (i), check whether

LHS and RHS equal.

The circumference in above circle is called locus of the circle and the distance
between O and P is called radius, the fixed point O is the centre of the circle. The
above equation (i) is called equation of locus (here is circle).

All the points on the locus are satisfied by the equation of the locus.

Definition : The locus of a point is defined as the path traced out by the moving
point under given geometrical conditions. It may also be defined as a set of points
which satisfy the given geometrical conditions.

The following points are to be noted while dealing with a locus:

(i) Every point in the locus must satisfy the given geometrical conditions.

(ii) A point which does not satisfy the given geometrical conditions cannot be on
the locus.

(iii) Every point which lies in the locus satisfies the equation of the locus.

(iv) The points which do not lie on the locus do not satisfy the equation of the
locus.

(v) To find the locus of a moving point plot some points satifying the given
geometrical conditions and then join them.

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Procedure of finding the equations of a locus

To find the equation of a locus with given geometrical conditions, the following are
some general rules:

(i) Draw a figure according to given geometrical condition and take a point P(x, y)
on the locus.

(ii) Write the geometrical conditions according to the points move.

(iii) Express the geometrical conditions in terms coordinates (x, y) by using distance
formula and simplify.

The simplest form of the given condition is the required equation of the locus.

Worked Out Examples

Example 1. Do the points (4, 5), (–5, 4), and (4, 6) lie on the locus x2 + y2 = 41.
Solution:
Given equation of locus is x2 + y2 = 41

Put the point (4, 5), i.e. x = 4 and y = 5 in the equation, we get,

42 + 52 = 41 or, 16 + 25 = 41

or, 41 = 41, which is true

Put the point (–5, 4) in the given equation, we get,

(–5)2 + 42 = 41 or, 25 + 16 = 41

or, 41 = 41, which is true.

Similarly, put the point (4, 6) in the above equation, we get,

42 + 62 = 41 or, 16 + 36 = 41

or, 52 = 41, which is false

? The points (4, 5) and (–5, 4) lie in the locus but (4, 6) does not
lie in x2 + y2 = 41.

Example 2. Find the value of k on the locus x2 + y2 + kx + 4y = 32 if (2, 2) lie in
Solution: the locus.
Given equation of the locus is x2 + y2 + kx + 4y = 32
Put the point (2, 2) i.e. x = 2, y = 2 in the given equation of locus

22 + 22 + k . 2 + 4 . 2 = 32
or, 4 + 4 + 2k + 8 = 32
or, 2k = 32 – 16

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Coordinates Geometry

or, 2k = 16
? k=8

Example 3. Find the equation of the locus of a point which moves at distance of
Solution: 6 units from the point (5, 4).
Let P(x, y) be any point in the locus and given point be A(5, 4).
Then, AP = 6

squaring on both sides, we get,
AP2 = 36 ............... (i)
For AP2, A(5, 4) = (x1, y1) and P(x, y) = (x2, y2)
By using distance formula,
AP2 = (x2 – x1)2 + (y2 – y1)2

= (x – 5)2 + (y – 4)2
= x2 – 2 . x . 5 + 52 + y2 – 2 . y . 4 + 42
= x2 + y2 – 10x – 8y + 41
Now, from (i), we get,
x2 + y2 – 10x – 8y + 41 = 36
or, x2 + y2 – 10x – 8y + 41 – 36 = 0
? x2 + y2 – 10x – 8y + 5 = 0 is the required equation of the locus.

Example 4. Find the equation of the locus of a point which moves so that its
Solution: distance from Y-axis is always 7 units.

Let P(x, y) be any point in the locus Y 7 P(x, y)
and A(0, y) be a point in Y-axis. A(0, y)

Then, AP = 7

squaring on both sides, we get, X' O X
AP2 = 49

or, (x – 0)2 + (y – y)2 = 49 Y'
or, x2 = 49

? x = ± 7 the required equation of the locus.

Example 5. Find the equation of locus a points which move of equal distance
Solution: from points (4, 3) and (–3, 4).

Let the given two points be A(4, 3) and B(–3, 4) as shown in the figure.

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Let P(x, y) be any point in the locus. Y
Then by question, P(x, y)

AP = BP

or, AP2 = BP2 ................... (i)

By using distance formula, B(–3, 4) A(4, 3)

d2 = (x2 – x1)2 + (y2 – y1)2 X' O X
Y'
we get, A(4, 3) = (x1, y1), P(x, y) = (x2, y2)

AP2 = (x – 4)2 + (y – 3)2

= x2 – 2 . x . 4 + 42 + y2 – 2 . y . 3 + 32

= x2 + y2 – 8x – 6y + 25

Similarly, B(x1, y1) = (–3, 4), P(x2, y2) = (x, y)
BP2 = (x + 3)2 + (y – 4)2

= x2 + 2 . x . 3 + 32 + y2 – 2 . y . 4 + 42

= x2 + 6x + y + y2 – 8y + 16

= x2 + y2 + 6x – 8y + 25

putting the values of AP2 and BP2 in equation (i), we get

x2 + y2 – 8x – 6y + 25 = x2 + y2 + 6x – 8y + 25

or, – 8x – 6x = – 8y + 6y

or, – 14x = – 2y

or, 7x = y

? y = 7x

? y = 7x is the required equation of the lacus.

Example 6. Find the equation of the locus of a point which moves so that its
Solution:
distance from the point (2, 3) is always double its distance from the
Y
point (–1, 2).

Let P(x, y) be any point and A(2, 3) P(x, y)

and B(–1, 2) be given point.

Then, for AP, A(2, 3) = (x1, y1) B(–1, 2) O A(2, 3)
and P(x, y) = (x2, y2) X' Y' X

AP2 = (x – 2)2 + (y – 3)2

= x2 – 2 . x . 2 + 22 + y2 – 2 . y . 3 + 32

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Coordinates Geometry

= x2 + y2 – 4x – 6y + 13
Again, for BP, B(–1, 2) = (x1, y1), P(x, y) = (x2, y2)

BP2 = (x + 1)2 + (y – 2)2
= x2 + 2 . x . 1 + 1 + y2 – 2 . y . 2 + 22
= x2 + y2 + 2x – 4y + 5

Now, by question,
AP = 2BP

or, AP2 = 4BP2
or, x2 + y2 – 4x – 6y + 13 = 4(x2 + y2 + 2x – 4y + 5)
or, x2 + y2 – 4x – 6y + 13 = 4x2 + 4y2 + 8x – 16y + 20
? 3x2 + 3y2 + 12x – 10y + 7 = 0 is the required equation of the

locus.

Example 7. If M(2, 0) and N(–5, 3) are two fixed points, find the equation of the
locus of the point P such that MP2 + NP2 = 24.

Solution: Let P(x, y) be any point on the locus.
Given two fixed points are M(2, 0) and N(–5, 3)
For MP2, M(2, 0) = (x1, y1) and P(x, y) = (x2, y2)
MP2 = (x – 2)2 + (y – 0)2
= x2 – 2 . x . 2 + 22 + y2
= x2 + y2 – 4x + 4
Again, for NP2, N(–5, 3) = (x1, y1) and P(x, y) = (x2, y2)
NP2 = (x + 5)2 + (y – 3)2
= x2 + 2 . x . 5 + 52 + y2 – 2 . y . 3 + 32
= x2 + y2 + 10x – 6y + 34
Now, MP2 + NP2 = 24
or, x2 + y2 – 4x + 4 + x2 + y2 + 10x – 6y + 34 = 24
or, 2x2 + 2y2 + 6x – 6y + 14 = 0
? x2 + y2 + 3x – 3y + 7 = 0 is the required equation of the locus.

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