Trigonometrical Ratios — Exercise 20(D)

A balloon is connected to a meterological station by a cable of length 200 metres, inclined at 60° to the horizontal. Determine the height of the balloon from the ground, assuming that there is no slack in the string. (Take $\sqrt{3}$ = 1.73)

Answer

Let A be the position of balloon and C be the position of meterological station. Then length of cable (AC) = 200 m.

Let height of the ballon from the ground = x

In triangle ABC,

sin 60° = $\dfrac{\text{Perpendicular}}{\text{Hypotenuse}}$

$\dfrac{\sqrt{3}}{2} = \dfrac{x}{200}$

x = $\dfrac{200 \sqrt{3}}{2}$

x = $100\sqrt{3}$

x = 100(1.73)

x = 173 m.

Hence, height of the balloon from the ground = 173 m.

A ladder leaning against a wall, makes an angle of 60° with the horizontal and the foot of tha ladder is 9.5 metres away from the wall. Find the length of the ladder.

Answer

Let AB be the wall and AC be the ladder.

In triangle ABC,

$\Rightarrow \cos 60^\circ = \dfrac{BC}{AC} \\[1em] \Rightarrow \dfrac{1}{2} = \dfrac{9.5}{AC} \\[1em] \Rightarrow AC = 2 \times 9.5 = 19 \text{ m}.$

Hence, length of ladder = 19 m.

A kite is flying with a thread 150 m long. If the thread is assumed stretched straight and makes an angle of 60° with the horizontal, find the height of the kite above the ground. (Take $\sqrt{3}$ = 1.73)

Answer

Let position of kite be A and C be the point on ground.

AC = length of string = 150 m.

AB = Height of the kite above the ground

sin 60° = $\dfrac{AB}{AC}$

$\dfrac{\sqrt{3}}{2} = \dfrac{AB}{150}$

AB = $75\sqrt{3} = 75 \times 1.73$ = 129.75 m

Hence, height of the kite above ground = 129.75 m.

A kite is flying at a height of 120 m from the level ground. It is attached to a string inclined at 60° to the horizontal. Find the length of the string. (Take $\sqrt{3}$ = 1.73)

Answer

Let A be the position of kite and C be the point from where string is attached.

AB = 120 m

In triangle ABC,

$\Rightarrow \sin 60^\circ = \dfrac{AB}{AC} \\[1em] \Rightarrow \dfrac{\sqrt{3}}{2} = \dfrac{120}{AC} \\[1em] \Rightarrow AC = \dfrac{120 \times 2}{\sqrt{3}} \\[1em] \Rightarrow AC = \dfrac{240}{\sqrt{3}} \\[1em] \Rightarrow AC = \dfrac{240 \times \sqrt{3}}{\sqrt{3} \times \sqrt{3}} \\[1em] \Rightarrow AC = \dfrac{240\sqrt{3}}{3} \\[1em] \Rightarrow AC = 80\sqrt{3} = 138.4 \text{ m}.$

Hence, length of the string = 138.4 m.

In a △ABC, right angled at B, if ∠A = 30° and BC = 8 cm, find the remaining angles and sides.

Answer

Given,

∠A = 30°, BC = 8 cm

sin 30° = $\dfrac{\text{Perpendicular}}{\text{Hypotenuse}}$

$\dfrac{1}{2} = \dfrac{BC}{AC}$

$\dfrac{1}{2} = \dfrac{8}{AC}$

AC = 16 cm

Now,

cos 30° = $\dfrac{\text{Base}}{\text{Hypotenuse}}$

$\dfrac{\sqrt{3}}{2} = \dfrac{AB}{AC}$

$\dfrac{\sqrt{3}}{2} = \dfrac{AB}{16}$

AB = $8\sqrt{3}$ cm

By angle sum property of triangle,

∠A + ∠B + ∠C = 180°

30° + 90° + ∠C = 180°

∠C + 120° = 180°

∠C = 180° - 120° = 60°

Hence, ∠C = 60°, AC = 16 cm and AB = $8\sqrt{3}$ cm.

In a rectangle ABCD, AB = 12 cm and ∠BAC = 30°. Calculate the lengths of side BC and diagonal AC.

Answer

In △BAC,

Base = AB = 12 cm and ∠BAC = 30°

Perpendicular = BC

tan 30° = $\dfrac{BC}{AB}$

$\dfrac{1}{\sqrt{3}} = \dfrac{BC}{12}$

BC = $\dfrac{12}{\sqrt{3}} = 4\sqrt{3}$ cm

Now,

cos 30° = $\dfrac{AB}{AC}$

$\dfrac{\sqrt{3}}{2} = \dfrac{12}{AC}$

AC = $\dfrac{24}{\sqrt{3}} = 8\sqrt{3}$ cm.

Hence, BC = $4\sqrt{3}$ cm and AC = $8\sqrt{3}$ cm.