OP Malhotra Class 9 Maths Solutions Chapter 19 Trigonometrical Ratios Ex 19(C)

Effective ICSE S Chand Maths Class 9 Solutions Chapter 19 Trigonometrical Ratios Ex 19(C) can help bridge the gap between theory and application.

S Chand Class 9 ICSE Maths Solutions Chapter 19 Trigonometrical Ratios Ex 19(C)

Question 1.
(i) Prove that sin 60° = 2 sin 30° cos 30°.
(ii) Without using trigonometric tables find the value of 3 sin² 45° + 2 cos² 60°.
Solution:
(i) L.H.S. = sin 60° = \(\frac{\sqrt{3}}{2}\)
R.H.S. = 2 sin 30° cos 30°
IMG
= 2 × \(\frac { 1 }{ 2 }\) × \(\frac{\sqrt{3}}{2}\) = \(\frac{\sqrt{3}}{2}\)
Hence sin 60° = 2 sin 30° cos 30°

(ii) 3 sin² 45° + 2 cos² 60°

= \(\frac { 3 }{ 2 }\) + 2 × \(\frac { 1 }{ 4 }\)
= \(\frac { 3 }{ 2 }\) + \(\frac { 1 }{ 2 }\) = \(\frac{3+1}{2}\) = \(\frac { 4 }{ 2 }\) = 2

Question 2.
If 0 ≤ x ≤ 90°, state the numerical value of x for which sin x° = cos x°.
Solution:
sin x° = cos x° ⇒ \(\frac{\sin x}{\cos x}\) = 1
⇒ tan x = 1 = tan 45° {∵ tan 45° = 1}
comparing, we get
x = 45°

Question 3.
Find the value of :
(i) sin² 60° + cos° 45°
(ii) 3 cos° 30° + tan² 60°
(iii) 4 sin² 60° + 3 tan² 30° – 8 sin 45° cos 45°
(iv) 2 sin² 30° – 3 cos² 45° + tan² 60°
(v) \(\frac{\sin 60^{\circ}}{\cos ^2 45^{\circ}}\) – 3 tan 30° + 5 cos 90°
(vi) cos 90° + cos² 45° sin 30° tan 45°
(vii) cos² 45° + sin² 60° + sin² 30°
(viii) sin² 30° + cos² 60°
(ix) \(\frac{\sin ^2 45^{\circ}+\cos ^2 45^{\circ}}{\tan ^2 60^{\circ}}\)
(x) \(\frac{5 \sin ^2 30^{\circ}+\cos ^2 45^{\circ}-4 \tan ^2 30^{\circ}}{2 \sin 30^{\circ} \cos 30^{\circ}+\tan 45^{\circ}}\)
(xi) 2√2 cos 45°. cos 60° + 2√3 sin 30°. tan 60° – cos 0°
(xii) \(\frac { 4 }{ 3 }\) tan² 30° + sin² 60° – 3 cos² 60° + \(\frac { 3 }{ 4 }\) tan² 60° – 2 tan² 45°
(xiii) (cos 0° + sin 45° + sin 30°) × (sin 90° – cos 45° + cos 60°)
Solution:

= \(\frac { 3 }{ 4 }\) + \(\frac { 1 }{ 2 }\)
= \(\frac{3+2}{4}\) = \(\frac{5}{4}\) = \(1 \frac{1}{4}\)

(ii) 3 cos² 30° + tan² 60° = 3 \(\left(\frac{\sqrt{3}}{2}\right)^2\) + \((\sqrt{3})^2\)
= 3 × \(\frac { 3 }{ 4 }\) + 3

= \(\frac { 9 }{ 4 }\) + 3 = \(\frac{9+12}{4}\) = \(\frac{21}{4}\) = \(5 \frac{1}{4}\)

(iii) 4 sin² 60° + 3 tan² 30° – 8 sin 45° cos 45°

(iv) 2 sin² 30° – 3 cos² 45° + tan² 60°

= 2 × \(\frac{1}{4}\) – 3 × \(\frac{1}{2}\) + 3
= \(\frac{1}{2}\) – \(\frac{3}{2}\) + 3
= \(\frac{1-3+6}{2}\) = \(\frac{4}{2}\) = 2

(v) \(\frac{\sin 60^{\circ}}{\cos ^2 45^{\circ}}-3 \tan 30^{\circ}+5 \cos 90^{\circ}\)

(vi) cos 90° + cos² 45° sin 30° tan 45°

(vii) cos² 45° + sin² 60° + sin² 30°

(viii) sin² 30° + cos² 60°

(xii) \(\frac{4}{3}\) tan² 30° + sin² 60° – 3 cos² 60° + \(\frac{3}{4}\) tan² 60° – 2 tan 45°
= \(\frac{4}{3}\) \(\left(\frac{1}{\sqrt{3}}\right)^2\) + \(\left(\frac{\sqrt{3}}{2}\right)^2\) – 3 \(\left(\frac{1}{2}\right)^2\) + \(\frac{3}{4}\) (√3)² – 2(1)²
= \(\frac{4}{3}\) × \(\frac{1}{3}\) + \(\frac{3}{4}\) – 3 × \(\frac{1}{4}\) + \(\frac{3}{4}\) x 3 – 2
= \(\frac{4}{9}\) + \(\frac{3}{4}\) – \(\frac{3}{4}\) + \(\frac{9}{4}\) – \(\frac{2}{1}\)
= \(\frac{16+27-27+81-72}{36}\) = \(\frac{25}{36}\)

(xiii) (cos 0° + sin 45° + sin 30°) × (sin 90° – cos 45° + cos 60°)

Question 4.
ABC is an isosceles right triangle. Assuming AB = BC = x, find the value of each of the following trigonometric ratios.
(i) sin 45°
(ii) cos 45°
(iii) tan 45°
Show your complete working neatly.
Solution:
In △ABC, AB = AC = x
Draw AD ⊥ BC which bisects BC at D

∵ ∠B = 45°, then ∠C = 45°
and ∠A = 90°
∴ ∠BAD = ∠CAD = 45°
( ∵AD is the perpendicular which bisect ∠A)
∴ BD = DC = AD = \(\frac { 1 }{ 2 }\)BC
∵ ∠A = 90°
∴ BC² = AB² + AC²
= x² + x² = 2x²
BC = √2x
∴ BD = DC = AD = \(\frac{\sqrt{2} x}{2}\)
Now (i) sin 45° = \(\frac{\text { AD }}{\text { AB }}\) = \(\frac{\frac{\sqrt{2 x}}{2}}{x}\)
= \(\frac{\sqrt{2}}{2}\) = \(\frac{1}{\sqrt{2}}\)
cos 45° = \(\frac{\text { BD }}{\text { AB }}\) = \(\frac{\frac{\sqrt{2} x}{2}}{x}\) = \(\frac{\sqrt{2}}{2}\) = \(\frac{1}{\sqrt{2}}\)
tan 45° = \(\frac{\text { AD }}{\text { BD }}\) = \(\frac{\frac{\sqrt{2}}{2} x}{\frac{\sqrt{2}}{2} x}\) = 1

Question 5.
Without using tables, find the value of
\(\frac{\sin 30^{\circ}-\sin 90^{\circ}+2 \cos 0^{\circ}}{\tan 30^{\circ} \times \tan 60^{\circ}}\)
Solution:
\(\frac{\sin 30^{\circ}-\sin 90^{\circ}+2 \cos 0^{\circ}}{\tan 30^{\circ} \times \tan 60^{\circ}}\) = \(\frac{\frac{1}{2}-1+2 \times 1}{\frac{1}{\sqrt{3}} \times \sqrt{3}}\)
= \(\frac{\frac{1}{2}-1+2}{1}\) = \(\frac{1}{2}\) – 1 + 2 = \(1 \frac{1}{2}\) = 1.5

Question 6.
Without using tables verify that :
(i) sin 60° = \(\frac{2 \tan 30^{\circ}}{1+\tan ^2 30^{\circ}}\) = \(\frac{\sqrt{3}}{2}\);
(ii) cos 60° = \(\frac{1-\tan ^2 30^{\circ}}{1+\tan ^2 30^{\circ}}\) = \(\frac{1}{2}\);
(iii) cos 60° = cos² 30° – sin² 30°
(iv) cos 60° = 1 – 2 sin² 30° = 2 cos² 30° – 1.
Solution:
(i) sin 60° = \(\frac{2 \tan 30^{\circ}}{1+\tan ^2 30^{\circ}}\) = \(\frac{\sqrt{3}}{2}\)
sin 60° = \(\frac{\sqrt{3}}{2}\)
and \(\frac{2 \tan 30^{\circ}}{1+\tan ^2 30^{\circ}}\) = \(\frac{2 \times \frac{1}{\sqrt{3}}}{1+\left(\frac{1}{\sqrt{3}}\right)^2}\) = \(\frac{\frac{2}{\sqrt{3}}}{1+\frac{1}{3}}\)
= \(\frac{\frac{2}{\sqrt{3}}}{\frac{4}{3}}\) = \(\frac{2}{\sqrt{3}}\) × \(\frac{3}{4}\) = \(\frac{\sqrt{3}}{2}\)
Hence sin 60° = \(\frac{2 \tan 30^{\circ}}{1+\tan ^2 30^{\circ}}\) = \(\frac{\sqrt{3}}{2}\)

(ii) cos 60° = \(\frac{1-\tan ^2 30^{\circ}}{1+\tan ^2 30^{\circ}}\) = \(\frac{1}{2}\)
cos 60° = \(\frac{1}{2}\)
and \(\frac{1-\tan ^2 30^{\circ}}{1+\tan ^2 30^{\circ}}\) = \(\frac{1-\left(\frac{1}{\sqrt{3}}\right)^2}{1+\left(\frac{1}{\sqrt{3}}\right)^2}\) = \(\frac{1-\frac{1}{3}}{1+\frac{1}{3}}\)
= \(\frac{\frac{2}{3}}{\frac{4}{3}}\) = \(\frac{2}{3}\) × \(\frac{3}{4}\) = \(\frac{1}{2}\)
Hence cos 60° = \(\frac{1-\tan ^2 30^{\circ}}{1+\tan ^2 30^{\circ}}\) = \(\frac{1}{2}\)

(iii) cos 60° = cos² 30° – sin² 30°
cos 60° = \(\frac{1}{2}\)
and cos² 30° – sin² 30° = \(\left(\frac{\sqrt{3}}{2}\right)^2\) – \(\left(\frac{1}{2}\right)^2\)
= \(\frac{3}{4}\) – \(\frac{1}{4}\)
= \(\frac{2}{4}\) = \(\frac{1}{2}\)
Hence cos 60° = cos² 30° – sin² 30°

(iv) cos 60° = 1 – 2sin² 30° = 2 cos² 30° – 1
cos 60° = \(\frac{1}{2}\)
1 – 2sin² 30° = 1 – 2 × \(\left(\frac{1}{2}\right)^2\) = 1 – 2 × \(\frac{1}{4}\)
= 1 – \(\frac{1}{2}\) = \(\frac{1}{2}\)
and 2 cos² 30° – 1 = 2 \(\left(\frac{\sqrt{3}}{2}\right)^2\) – 1
= 2 × \(\frac{3}{4}\) – 1
= \(\frac{3}{4}\) – 1 = \(\frac{1}{2}\)
Hence cos 60° = 1 – 2 sin² 30°
= 2 cos² 30° – 1

Question 7.
Prove that:
(i) cos² 30° + sin² 30° + tan² 45° = 2.
(ii) 4 (sin⁴ 30° + cos⁴ 60°) – 3 ( cos² 45° – sin² 90°) = 2.
Solution:
(i) L.H.S. = cos² 30° + sin² 30° + tan² 45°
= \(\left(\frac{\sqrt{3}}{2}\right)^2\) + \(\left(\frac{1}{2}\right)^2\) + (1)²
= \(\frac{3}{4}\) + \(\frac{1}{4}\) + 1 = 1 + 1 = 2 = R.H.S.

(ii) L.H.S. = 4 (sin ⁴ 30° 1+ cos ⁴ 60°) – 3 (cos² 45° – sin² 90°)

Question 8.
A pole 15 m long rests against a vertical wall at an angle of 60° with the ground. Calculate
(i) how high up the wall will the pole reach?
(ii) how far is the foot of the pole from the wall?
Solution:
AC is wall and AB is a pole which rests with wall at 60°
In right △ABC, ∠C = 90°
AB = 15 m
Let height of wall = x m
and away from foot of wall = y m

(i) ∵ sin 60° = \(\frac{\text { AC }}{\text { AB }}\)
⇒ \(\frac{\sqrt{3}}{2}\) = \(\frac{x}{15}\) ⇒ x = \(\frac{15 \sqrt{3}}{2}\)
⇒ x = \(\frac{15(1.732)}{2}\) = \(\frac{25.980}{2}\) = 12.99 m
∴ Height of wall = 12.99 m

(ii) cos 60° = \(\frac{\text { BC }}{\text { AB }}\) ⇒ \(\frac{1}{2}\) = \(\frac{y}{15}\)
⇒ y = \(\frac{15}{2}\) = 7.5 m
∴ Distance from the foot of wall = 7.5 m

Question 9.
In the given triangle, calculate θ if B = 90°, AB = 20 cm and AC = 40 cm.

Solution:
In the △ABC, ∠B = 90° and
∠C = θ, AC = 40 cm, AB = 20 cm
Here sin θ = \(\frac{\text { AB }}{\text { AC }}\) = \(\frac{20}{40}\) = \(\frac{1}{2}\)
= sin 30° {∵ sin 30° = \(\frac{1}{2}\)}
∴ θ = 30°

Question 10.
Without using tables solve each of the following triangle ABC, right-angled at C; given
(i) A = 30° and c = 40;
(ii) B = 60° and c = 15;
(iii) A = 45° and a = 7.
Solution:

(i) In △ABC, ∠c = 90°
∠A = 30 and c = 40
sin θ = \(\frac{\text { BC }}{\text { AB }}\) = \(\frac{a}{c}\)
⇒ sin 30° = \(\frac{a}{c}\) ⇒ \(\frac{1}{2}\) = \(\frac{a}{40}\)
⇒ a = \(\frac{40}{2}\) = 20
But c² = a² + b² (Pythagoras Theorem)
(40)² = (20)² +b² ⇒ 1600 – 400 = 1200
= 400 × 3
∴ b = \(\sqrt{400 \times 3}\) = 20√3 = 20(1.732)
= 34.64 = 34.64
a = 20 and b = 34.64 and
∠B = 90° – 30° = 60°

(ii) ∠B = 60° and c = 15
sin θ = \(\frac{\text { AC }}{\text { AB }}\) ⇒ sin 60° = \(\frac{a}{c}\)
⇒ \(\frac{\sqrt{3}}{2}\) = \(\frac{a}{15}\) ⇒ a = \(\frac{15 \sqrt{3}}{2}\)
⇒ a = \(\frac{15(1.732)}{2}\) = 12.99
We know that by Pythagoras Theorem,
(15)² = \(\left(\frac{15 \sqrt{3}}{2}\right)^2\) + b² ⇒ 225 = \(\frac{675}{4}\) + b²
⇒ b² = 225 – \(\frac{675}{4}\)
= \(\frac{900-675}{4}\) = \(\frac{225}{4}\)
∴ b = \(\sqrt{\frac{225}{4}}\) = \(\frac{15}{2}\) = 7.5
and ∠A = 90° – B = 90° – 60° = 30°

(iii) A = 45° and a = 7
sin θ = \(\frac{\text { BC }}{\text { AC }}\) ⇒ sin 45° = \(\frac{a}{c}\)
⇒ \(\frac{1}{\sqrt{2}}\) = \(\frac{7}{c}\) ⇒ c = \(7 \sqrt{2}\) = 7(1.414)
⇒ c = 9.898 = 9.9
But C² = a² + b²
⇒ (7√2)² = (7)² + b² ⇒ 98 = 49 + b²
⇒ b² = 98 – 49 = 49 = (7)²
∴ b = 7
Hence c = 9.9 and b = 7
and ∠B = 90° – ∠A = 90° – 45° = 45°

Question 11.
If 4 sin² θ – 1 = 0 and angle θ is less than 90°; find the value of θ and hence the value of cos² θ + tan² θ.
Solution:
4 sin² θ – 1 = 0 ⇒ 4 sin ² θ = 1
⇒ sin ² θ = \(\frac{1}{4}\) = \(\left(\frac{1}{2}\right)^2\)

(i) ∴ sin θ = \(\frac{1}{2}\) = sin 30° {∵ sin 30° = \(\frac{1}{2}\)}
∴ θ = 30°
Now cos² θ + tan² θ = cos² 30° + tan² 30°
= \(\left(\frac{\sqrt{3}}{2}\right)^2\) + \(\left(\frac{1}{\sqrt{3}}\right)^2\)
= \(\frac{3}{4}\) + \(\frac{1}{3}\)
= \(\frac{9+4}{12}\) = \(\frac{13}{1}\) = \(1 \frac{1}{12}\)

Question 12.
If θ is an acute angle and sin θ = cos θ, find the value of 2 tan² θ + sin² θ – 1.
Solution:
(i) ∵ sin θ = cos θ ⇒ \(\frac{\sin \theta}{\cos \theta}\) = 1
⇒ tan θ = 1 = tan 45° {∵ tan 45° = 1}
∴ θ = 45°

(ii) Now 2 tan² θ + sin² θ – 1
=2 tan² 45° + sin² 45° – 1

= 2 (1)² + \(\left(\frac{1}{\sqrt{2}}\right)^2\) – 1
= 2 × 1 + \(\frac{1}{2}\) – 1 = 2 + \(\frac{1}{2}\) – 1
= \(1 \frac{1}{2}\) = \(\frac{3}{2}\) = 1.5

Question 13.
A kite, flying at a height of 75 metres from the level ground, is attached to a string inclined at 60° to the horizontal, find the length of the string to the nearest metre.
Solution:
Height of the kite = 75 m
Angle of elevation = 60°

∴ JK is the string tied to the kite
Let JK = x m
Now sin θ = \(\frac{\text { KL }}{\text { JK }}\)
⇒ sin 60° = \(\frac{75}{x}\) ⇒ \(\frac{\sqrt{3}}{2}\) = \(\frac{75}{x}\)
⇒ x = \(\frac{75 \times 2}{\sqrt{3}}\) = \(\frac{150 \times \sqrt{3}}{\sqrt{3} \times \sqrt{3}}\) m
= \(\frac{150 \sqrt{3}}{3}\) = 50√3 = 50(1.732)
= 86.600 m = 87 m

Question 14.
If 2 sin θ – 1 = 0, find :
(i) the value of θ in degrees where θ is an acute angle;
(ii) cos²θ + tan²θ.
Solution:
(i) 2 sin θ – 1 = 0 ⇒ 2 sin θ = 1
⇒ sin θ = \(\frac{1}{2}\) = sin 30° {∵ sin 30° = \(\frac{1}{2}\)}
∴ θ = 30°

(ii) Now, cos²θ + tan²θ = cos² 30° + tan² 30°
= \(\left(\frac{\sqrt{3}}{2}\right)^2\) + \(\left(\frac{1}{\sqrt{3}}\right)^2\) = \(\frac{3}{4}\) + \(\frac{1}{3}\)
= \(\frac{9+4}{12}\) = \(\frac{13}{12}\) = \(1 \frac{1}{12}\)

Question 15.
The altitude AD of a △ABC, in which ∠A is obtuse, is 10 cm. If BD = 10 cm and CD = 10√3 cm, determine ∠A.
Solution:
In △ABC, ∠A is an obtused angle AD ⊥ BC,
BD = 10 cm and CD = 10√3 cm

Let ∠BAD = α and ∠CAD = β
∠A = α + β
Now tan α = \(\frac{\text { BD }}{\text { AD }}\) = \(\frac{10}{10}\) = 1 = tan 45° {∵ tan 45° = 1}
∴ α =45°
and tan β = \(\frac{\text { CD }}{\text { AD }}\) = \(\frac{10 \sqrt{3}}{10}\) = √3 = tan 60° {∵ tan 60° = √3}
∴ β = 60°
Now ∠A = α + β = 45° + 60° = 105°

Question 16.
In a right triangle ABC, right angled at C, if ∠B = 60° and AB = 15 units, find the remaining angles and sides.
Solution:
In △ABC, ∠C = 90°, ∠B = 60°,
AB = 15 units

∴ ∠A = 180° – (90° + 60°)
= 180° – 150° = 30°
Now sin θ = \(\frac{\text { AC }}{\text { AB }}\) ⇒ sin 60° = \(\frac{\text { AC }}{\text { AB }}\)
⇒ \(\frac{\sqrt{3}}{2}\) = \(\frac{\mathrm{AC}}{15}\) ⇒ AC = \(\frac{15 \sqrt{3}}{2}\) = \(\frac{15}{2}\) √3 units
and cos 60° = \(\frac{\text { BC }}{\text { AB }}\) ⇒ \(\frac{1}{2}\) = \(\frac{\mathrm{BC}}{15}\)
⇒ BC = \(\frac{15}{2}\) = 7.5 units
Hence ∠B = 30°, BC = 7.5 units, AC = \(\frac{15}{2}\)√3 units

Question 17.
In a rectangle ABCD, AB = 20 cm, ∠BAC = 60°, calculate side BC and diagonals AC and BD.
Solution:
In reactangle ABCD, AB = 20 cm
∠BAC = 60° and AC and BD are its two diagonals which are equal

tan ∠BAC = \(\frac{\text { BC }}{\text { AB }}\)
⇒ tan 60° = \(\frac{\mathrm{BC}}{20}\) ⇒ √3 = \(\frac{\mathrm{BC}}{20}\)
⇒ BC = 20√3 cm
Similarly cos 60° = \(\frac{\text { AB }}{\text { AC }}\) = \(\frac{20}{A C}\)
⇒ \(\frac{1}{2}\) = \(\frac{20}{A C}\) = AC = 20 × 2 = 40 cm
∵ BD = AC
⇒ BD = 40 cm

Question 18.
A ladder is placed against a wall such that it just reaches the top of the wall. The foot of the ladder is 1.5 m away from the wall and the ladder is inclined at an angle of 60° with the ground. Find the height of the wall. (Given √3 = 1.73)
Solution:
Let AB is wall and AC is ladder which is placed against the wall and makes an angle of 60° with the ground
Foot of the ladder is 1.5 m away from the wall i.e. BC = 1.5 m

Let h be the height of the wall then
tan C = \(\frac{\text { AB }}{\text { AC }}\) ⇒ tan 60° = \(\frac{h}{1.5}\)
⇒ √3 = \(\frac{h}{1.5}\) ⇒ h = 1.5√3
⇒ h = (1.5) (1.73) m = 2.595 m
∴ Height of the wall = 2.595 m = 2.59 m

Question 19.
An electric pole is 10 m high. A steel wire tied to the top of the pole is affixed at a point on the ground to keep the pole upright. If the steel wire makes an angle 45° with the horizontal through the foot of the pole, find the length of the steel wire. (Given √2 = 1.41)

Solution:
BC is the electric pole and AC is the wire tied to it such that
∠A = 45° and BC = 10 m Let AC = x
Now sin A = \(\frac{\text { BC }}{\text { AC }}\)
⇒ sin 45° = \(\frac{10}{x}\)
⇒ \(\frac{1}{\sqrt{2}}\) = \(\frac{10}{x}\)
⇒ x = 10√2 = 10 (1.414) m = 14.14 m
Length of wire = 14.14 = 14.1 m

Question 20.
A vertically straight tree, 15 m high, is broken by the wind in such a way that its top just touches the ground and makes an angle of 60° with the ground. At what height from the ground did the tree break ? (Use √3 = 1.73)

Solution:
Let DB is a tree which is broken from A and makes an angle of 60° with the ground then AD = AC and DB = 15 cm

Let AB = x, then
AC = AD = (15 – x) m
Now, sin θ = \(\frac{\text { Perp. }}{\text { Hyp. }}\) ⇒ sin 60° = \(\frac{\text { AB }}{\text { AC }}\)
⇒ \(\frac{\sqrt{3}}{2}\) = \(\frac{x}{15-x}\) ⇒ 2x = 15√3 = √3 x
⇒ 2x + √3 x = 15√3
⇒ (2 + √3) x = 15√3
∴ x = \(\frac{15 \sqrt{3}}{2+\sqrt{3}}\) = \(\frac{15 \sqrt{3}(2-\sqrt{3})}{(2+\sqrt{3})(2-\sqrt{3})}\)
= \(\frac{30 \sqrt{3}-45}{4-3}\)
= \(\frac{30(1.73)-45}{1}\) = 51.9 – 45 = 6.9 m
∴ Height of tree from which it was broken = 6.9 m

Question 21.
The string of a kite is 150 m long and it makes an angle of 60° with the horizontal. Find the height of the kite from the ground. (Take √3 = 1.73).
Solution:
K is kite and KL is string tied to it which makes an angle of 60° with the ground

∴ ∠L = 60°, LK = 150 m
Let KM = h m
sin θ = \(\frac{\text { Perp. }}{\text { Hyp. }}\)
⇒ sin 60° = \(\frac{\text { KM }}{\text { KL }}\)
⇒ \(\frac{\sqrt{3}}{2}\) = \(\frac{h}{150}\) ⇒ h = \(\frac{150 \sqrt{3}}{2}\) = 75√3 m
∴ Height of teh kite = 75√3 = 75 x 1.73 m = 129.75 = 129.8 m