NCERT Solutions for Class 10 Maths Chapter 9 Some Applications of Trigonometry Exercise 9.1#

Let AB be the pole and AC = 20 m be the rope, with ∠ACB = 30°.

In right △ABC, AB is opposite ∠C and AC is the hypotenuse, so:

$$\sin 30° = \frac{AB}{AC}$$

$$\Rightarrow \frac{1}{2} = \frac{AB}{20}$$

$$\Rightarrow AB = 10$$

∴ The height of the pole is 10 m.

Let C be the foot of the tree, B the point where it breaks, and A the point where the top touches the ground. Then BC is the standing part, AB is the broken part, AC = 8 m and ∠BAC = 30°.

In right △BCA, BC is opposite ∠A and AC is adjacent to it, so:

$$\tan 30° = \frac{BC}{AC}$$

$$\Rightarrow \frac{1}{\sqrt{3}} = \frac{BC}{8}$$

$$\Rightarrow BC = \frac{8}{\sqrt{3}}$$

AC is adjacent to ∠A and AB is the hypotenuse, so:

$$\cos 30° = \frac{AC}{AB}$$

$$\Rightarrow \frac{\sqrt{3}}{2} = \frac{8}{AB}$$

$$\Rightarrow AB = \frac{16}{\sqrt{3}}$$

The height of the tree is BC + AB:

$$= \frac{8}{\sqrt{3}} + \frac{16}{\sqrt{3}}$$

$$= \frac{24}{\sqrt{3}}$$

$$= 8\sqrt{3}$$

∴ The height of the tree is $\mathbf{8\sqrt{3}}$ m (about 13.86 m).

In each case, the height is opposite the angle with the ground and the slide is the hypotenuse.

Slide for younger children: Let its length be $l_1$ m. The height is 1.5 m and the angle is 30°, so:

$$\sin 30° = \frac{1.5}{l_1}$$

$$\Rightarrow \frac{1}{2} = \frac{1.5}{l_1}$$

$$\Rightarrow l_1 = 3$$

Slide for elder children: Let its length be $l_2$ m. The height is 3 m and the angle is 60°, so:

$$\sin 60° = \frac{3}{l_2}$$

$$\Rightarrow \frac{\sqrt{3}}{2} = \frac{3}{l_2}$$

$$\Rightarrow l_2 = \frac{6}{\sqrt{3}}$$

$$\Rightarrow l_2 = 2\sqrt{3}$$

∴ The slide for younger children is 3 m long, and the slide for elder children is $\mathbf{2\sqrt{3}}$ m long (about 3.46 m).

Let AB = $h$ m be the tower and P the point on the ground, with BP = 30 m and ∠APB = 30°.

In right △ABP, AB is opposite ∠P and BP is adjacent to it, so:

$$\tan 30° = \frac{AB}{BP}$$

$$\Rightarrow \frac{1}{\sqrt{3}} = \frac{h}{30}$$

$$\Rightarrow h = \frac{30}{\sqrt{3}}$$

$$\Rightarrow h = 10\sqrt{3}$$

∴ The height of the tower is $\mathbf{10\sqrt{3}}$ m (about 17.32 m).

Let the length of the string be $l$ m. The height of the kite, 60 m, is opposite the 60° angle and the string is the hypotenuse, so:

$$\sin 60° = \frac{60}{l}$$

$$\Rightarrow \frac{\sqrt{3}}{2} = \frac{60}{l}$$

$$\Rightarrow l = \frac{120}{\sqrt{3}}$$

$$\Rightarrow l = 40\sqrt{3}$$

∴ The length of the string is $\mathbf{40\sqrt{3}}$ m (about 69.28 m).

Height of the building above the boy’s eyes = 30 − 1.5 = 28.5 m

Let the boy’s first distance from the building be $d_1$ m and his final distance be $d_2$ m.

At first the angle of elevation is 30°, so:

$$\tan 30° = \frac{28.5}{d_1}$$

$$\Rightarrow \frac{1}{\sqrt{3}} = \frac{28.5}{d_1}$$

$$\Rightarrow d_1 = 28.5\sqrt{3}$$

Finally the angle of elevation is 60°, so:

$$\tan 60° = \frac{28.5}{d_2}$$

$$\Rightarrow \sqrt{3} = \frac{28.5}{d_2}$$

$$\Rightarrow d_2 = \frac{28.5}{\sqrt{3}}$$

$$\Rightarrow d_2 = 9.5\sqrt{3}$$

Distance walked = $d_1 - d_2$
$= 28.5\sqrt{3} - 9.5\sqrt{3}$
$= 19\sqrt{3}$

∴ The boy walked $\mathbf{19\sqrt{3}}$ m (about 32.91 m) towards the building.

Let P be the point on the ground, AB = 20 m the building and AT = $h$ m the tower. Let PB = $d$ m.

The angle of elevation of the bottom of the tower, A, is 45°, so in right △ABP:

$$\tan 45° = \frac{AB}{PB}$$

$$\Rightarrow 1 = \frac{20}{d}$$

$$\Rightarrow d = 20$$

The angle of elevation of the top of the tower, T, is 60°, so in right △TBP:

$$\tan 60° = \frac{TB}{PB}$$

$$\Rightarrow \sqrt{3} = \frac{20 + h}{20}$$

$$\Rightarrow 20 + h = 20\sqrt{3}$$

$$\Rightarrow h = 20(\sqrt{3} - 1)$$

∴ The height of the tower is $\mathbf{20(\sqrt{3} - 1)}$ m (about 14.64 m).

Let the height of the pedestal be $h$ m and the distance of the point from its foot be $d$ m.

The angle of elevation of the top of the pedestal is 45°, so:

$$\tan 45° = \frac{h}{d}$$

$$\Rightarrow 1 = \frac{h}{d}$$

$$\Rightarrow d = h$$

The angle of elevation of the top of the statue is 60°, so:

$$\tan 60° = \frac{h + 1.6}{d}$$

$$\Rightarrow \sqrt{3} = \frac{h + 1.6}{h}$$

$$\Rightarrow \sqrt{3}\,h = h + 1.6$$

$$\Rightarrow h(\sqrt{3} - 1) = 1.6$$

$$\Rightarrow h = \frac{1.6}{\sqrt{3} - 1}$$

Multiplying the numerator and denominator by $(\sqrt{3} + 1)$:

$$h = \frac{1.6(\sqrt{3} + 1)}{3 - 1}$$

$$\Rightarrow h = 0.8(\sqrt{3} + 1)$$

∴ The height of the pedestal is $\mathbf{0.8(\sqrt{3} + 1)}$ m (about 2.19 m).

Let T₁T₂ = 50 m be the tower and B₁B₂ = $h$ m the building, with T₁B₁ = $d$ m between their feet.

The angle of elevation of the top of the tower from the foot of the building is 60°, so:

$$\tan 60° = \frac{T_1T_2}{T_1B_1}$$

$$\Rightarrow \sqrt{3} = \frac{50}{d}$$

$$\Rightarrow d = \frac{50}{\sqrt{3}}$$

The angle of elevation of the top of the building from the foot of the tower is 30°, so:

$$\tan 30° = \frac{B_1B_2}{T_1B_1}$$

$$\Rightarrow \frac{1}{\sqrt{3}} = \frac{h}{d}$$

$$\Rightarrow h = \frac{d}{\sqrt{3}}$$

Substituting $d = \dfrac{50}{\sqrt{3}}$:

$$h = \frac{50}{\sqrt{3} \times \sqrt{3}}$$

$$\Rightarrow h = \frac{50}{3}$$

∴ The height of the building is $\mathbf{\dfrac{50}{3}}$ m (about 16.67 m).

Let each pole be $h$ m high, and let the point P be $x$ m from pole A, so it is $(80 - x)$ m from pole B.

The angle of elevation of the top of pole A is 60°, so:

$$\tan 60° = \frac{h}{x}$$

$$\Rightarrow h = \sqrt{3}\,x \quad \cdots (1)$$

The angle of elevation of the top of pole B is 30°, so:

$$\tan 30° = \frac{h}{80 - x}$$

$$\Rightarrow h = \frac{80 - x}{\sqrt{3}} \quad \cdots (2)$$

From equation (1) and equation (2):

$$\sqrt{3}\,x = \frac{80 - x}{\sqrt{3}}$$

$$\Rightarrow 3x = 80 - x$$

$$\Rightarrow 4x = 80$$

$$\Rightarrow x = 20$$

Substituting $x = 20$ in equation (1):

$$h = \sqrt{3}\,x$$

$$h = 20\sqrt{3}$$

Distance from pole B = 80 − 20 = 60 m

∴ Each pole is $\mathbf{20\sqrt{3}}$ m high (about 34.64 m), and the point is 20 m from one pole and 60 m from the other.

Let FT = $h$ m be the tower and FA = $w$ m the width of the canal, with AB = 20 m.

From A, the angle of elevation is 60°, so:

$$\tan 60° = \frac{FT}{FA}$$

$$\Rightarrow h = \sqrt{3}\,w \quad \cdots (1)$$

From B, the angle of elevation is 30°, so:

$$\tan 30° = \frac{FT}{FB}$$

$$\Rightarrow h = \frac{w + 20}{\sqrt{3}} \quad \cdots (2)$$

From equation (1) and equation (2):

$$\sqrt{3}\,w = \frac{w + 20}{\sqrt{3}}$$

$$\Rightarrow 3w = w + 20$$

$$\Rightarrow 2w = 20$$

$$\Rightarrow w = 10$$

Substituting $w = 10$ in equation (1):

$$h = \sqrt{3}\,w$$

$$h = 10\sqrt{3}$$

∴ The height of the tower is $\mathbf{10\sqrt{3}}$ m (about 17.32 m) and the width of the canal is 10 m.

Let AB = 7 m be the building and CD the cable tower. Draw AE ∥ BC, meeting CD at E. Then CE = AB = 7 m and AE = BC. Let DE = $h$ m.

The angle of depression of C is 45°, so ∠ACB = 45° (alternate angles). In right △ABC:

$$\tan 45° = \frac{AB}{BC}$$

$$\Rightarrow 1 = \frac{7}{BC}$$

$$\Rightarrow BC = 7$$

So AE = BC = 7 m.

The angle of elevation of D is 60°, so in right △AED:

$$\tan 60° = \frac{DE}{AE}$$

$$\Rightarrow \sqrt{3} = \frac{h}{7}$$

$$\Rightarrow h = 7\sqrt{3}$$

Height of the tower = CE + DE
$= 7 + 7\sqrt{3}$
$= 7(\sqrt{3} + 1)$

∴ The height of the tower is $\mathbf{7(\sqrt{3} + 1)}$ m (about 19.12 m).

Let LT = 75 m be the lighthouse, S₁ the nearer ship and S₂ the farther ship. The angle of elevation of T from each ship equals the angle of depression (alternate angles).

The angle of depression of S₁ is 45°, so in right △TLS₁:

$$\tan 45° = \frac{LT}{LS_1}$$

$$\Rightarrow 1 = \frac{75}{LS_1}$$

$$\Rightarrow LS_1 = 75$$

The angle of depression of S₂ is 30°, so in right △TLS₂:

$$\tan 30° = \frac{LT}{LS_2}$$

$$\Rightarrow \frac{1}{\sqrt{3}} = \frac{75}{LS_2}$$

$$\Rightarrow LS_2 = 75\sqrt{3}$$

Distance between the ships = $LS_2 - LS_1$
$= 75\sqrt{3} - 75$
$= 75(\sqrt{3} - 1)$

∴ The distance between the two ships is $\mathbf{75(\sqrt{3} - 1)}$ m (about 54.9 m).

Height of the balloon above the girl’s eyes = 88.2 − 1.2 = 87 m

Let the horizontal distances of the balloon from the girl’s eyes be $d_1$ m at first and $d_2$ m later.

At first the angle of elevation is 60°, so:

$$\tan 60° = \frac{87}{d_1}$$

$$\Rightarrow \sqrt{3} = \frac{87}{d_1}$$

$$\Rightarrow d_1 = \frac{87}{\sqrt{3}}$$

$$\Rightarrow d_1 = 29\sqrt{3}$$

Later the angle of elevation is 30°, so:

$$\tan 30° = \frac{87}{d_2}$$

$$\Rightarrow \frac{1}{\sqrt{3}} = \frac{87}{d_2}$$

$$\Rightarrow d_2 = 87\sqrt{3}$$

Distance travelled = $d_2 - d_1$
$= 87\sqrt{3} - 29\sqrt{3}$
$= 58\sqrt{3}$

∴ The balloon travelled $\mathbf{58\sqrt{3}}$ m (about 100.46 m).

Let FT = $h$ m be the tower, A the first position of the car and B its position 6 seconds later. The angle of elevation of T from each position equals the angle of depression (alternate angles).

The angle of depression of A is 30°, so in right △TFA:

$$\tan 30° = \frac{FT}{FA}$$

$$\Rightarrow \frac{1}{\sqrt{3}} = \frac{h}{FA}$$

$$\Rightarrow FA = \sqrt{3}\,h$$

The angle of depression of B is 60°, so in right △TFB:

$$\tan 60° = \frac{FT}{FB}$$

$$\Rightarrow \sqrt{3} = \frac{h}{FB}$$

$$\Rightarrow FB = \frac{h}{\sqrt{3}}$$

Distance covered in 6 seconds = $FA - FB$
$= \sqrt{3}\,h - \dfrac{h}{\sqrt{3}}$
$= \dfrac{3h - h}{\sqrt{3}}$
$= \dfrac{2h}{\sqrt{3}}$

So the car covers $\dfrac{2h}{\sqrt{3}}$ in 6 seconds, and it still has to cover FB = $\dfrac{h}{\sqrt{3}}$, which is half of that distance.

Time to cover FB = $\dfrac{1}{2} \times 6 = 3$ seconds

∴ The car takes 3 seconds to reach the foot of the tower.