Exercise 9.1: Some Applications of Trigonometry — NCERT Solutions

Let the height of the pole be $h$ and the shadow length also be $h$.

Let the angle of elevation of the Sun be $\theta$.

Using the right triangle formed by the pole and its shadow: $\tan\theta = \frac{\text{height}}{\text{shadow}} = \frac{h}{h} = 1$

Since $\tan 45^\circ = 1$, $\theta = 45^\circ$

The angle of elevation of the Sun is $45^\circ$.

Let the length of the ladder be $L$ (the hypotenuse). The distance of the foot from the wall is the adjacent side $= 2.5$ m, and the angle with the ground is $60^\circ$.

$\cos 60^\circ = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{2.5}{L}$

Since $\cos 60^\circ = \dfrac{1}{2}$, $\frac{1}{2} = \frac{2.5}{L}$

$L = 2.5 \times 2 = 5 \text{ m}$

The length of the ladder is $5$ m.

Let the height of the tower be $h$. The horizontal distance from the point to the foot of the tower is $30$ m, and the angle of elevation is $30^\circ$.

$\tan 30^\circ = \frac{h}{30}$

Since $\tan 30^\circ = \dfrac{1}{\sqrt{3}}$, $\frac{1}{\sqrt{3}} = \frac{h}{30}$

$h = \frac{30}{\sqrt{3}} = \frac{30}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}} = \frac{30\sqrt{3}}{3} = 10\sqrt{3} \text{ m}$

$h \approx 10 \times 1.732 = 17.32 \text{ m}$

The height of the tower is $10\sqrt{3} \approx 17.32$ m.

The angle of depression of the boat from the top equals the angle of elevation of the top from the boat (alternate angles), so this angle $= 45^\circ$.

Let the horizontal distance of the boat from the foot be $d$. The lighthouse height is the opposite side $= 60$ m.

$\tan 45^\circ = \frac{60}{d}$

Since $\tan 45^\circ = 1$, $1 = \frac{60}{d} \implies d = 60 \text{ m}$

The boat is $60$ m from the foot of the lighthouse.

Let the length of the string be $L$ (the hypotenuse). The height of the kite is the opposite side $= 60$ m, and the string makes $60^\circ$ with the horizontal.

$\sin 60^\circ = \frac{60}{L}$

Since $\sin 60^\circ = \dfrac{\sqrt{3}}{2}$, $\frac{\sqrt{3}}{2} = \frac{60}{L}$

$L = \frac{60 \times 2}{\sqrt{3}} = \frac{120}{\sqrt{3}} = \frac{120\sqrt{3}}{3} = 40\sqrt{3} \text{ m}$

$L \approx 40 \times 1.732 = 69.28 \text{ m}$

The length of the string is $40\sqrt{3} \approx 69.28$ m.

Let the height of the tower be $h$. Let the angle of elevation from the point $4$ m away be $\theta$; then from the point $9$ m away it is $90^\circ - \theta$ (complementary).

From the point $4$ m away: $\tan\theta = \frac{h}{4}$

From the point $9$ m away: $\tan(90^\circ - \theta) = \cot\theta = \frac{h}{9}$

Multiplying the two equations: $\tan\theta \cdot \cot\theta = \frac{h}{4} \cdot \frac{h}{9}$

Since $\tan\theta \cdot \cot\theta = 1$, $1 = \frac{h^2}{36} \implies h^2 = 36 \implies h = 6 \text{ m}$

The height of the tower is $6$ m.

Let the distance between the building and tower be $d$, the height of the tower be $50$ m, and the height of the building be $h$.

Step 1 — Find $d$ using the tower. The angle of elevation of the top of the tower from the foot of the building is $60^\circ$: $\tan 60^\circ = \frac{50}{d} \implies \sqrt{3} = \frac{50}{d} \implies d = \frac{50}{\sqrt{3}}$

Step 2 — Find $h$ using the building. The angle of elevation of the top of the building from the foot of the tower is $30^\circ$: $\tan 30^\circ = \frac{h}{d} \implies \frac{1}{\sqrt{3}} = \frac{h}{d}$ $h = \frac{d}{\sqrt{3}} = \frac{1}{\sqrt{3}} \times \frac{50}{\sqrt{3}} = \frac{50}{3} \approx 16.67 \text{ m}$

The height of the building is $\dfrac{50}{3} \approx 16.67$ m.

Let the tree break at point $B$, at height $AB = x$ above the ground $A$. The broken part $BC$ leans and its top touches the ground at $C$, with $AC = 8$ m and angle $C = 30^\circ$.

Step 1 — Find the standing part $AB = x$ (opposite to $30^\circ$): $\tan 30^\circ = \frac{AB}{AC} = \frac{x}{8}$ $\frac{1}{\sqrt{3}} = \frac{x}{8} \implies x = \frac{8}{\sqrt{3}} = \frac{8\sqrt{3}}{3} \text{ m}$

Step 2 — Find the broken part $BC$ (hypotenuse): $\cos 30^\circ = \frac{AC}{BC} = \frac{8}{BC}$ $\frac{\sqrt{3}}{2} = \frac{8}{BC} \implies BC = \frac{16}{\sqrt{3}} = \frac{16\sqrt{3}}{3} \text{ m}$

Step 3 — Original height $= AB + BC$: $= \frac{8\sqrt{3}}{3} + \frac{16\sqrt{3}}{3} = \frac{24\sqrt{3}}{3} = 8\sqrt{3} \text{ m} \approx 13.86 \text{ m}$

The original height of the tree was $8\sqrt{3} \approx 13.86$ m.

Let the horizontal distance from the point to the building be $d$, the building height be $20$ m, and the tower height be $h$.

Step 1 — Use the $45^\circ$ elevation to the top of the building (bottom of tower): $\tan 45^\circ = \frac{20}{d} \implies 1 = \frac{20}{d} \implies d = 20 \text{ m}$

Step 2 — Use the $60^\circ$ elevation to the top of the tower (total height $= 20 + h$): $\tan 60^\circ = \frac{20 + h}{d} \implies \sqrt{3} = \frac{20 + h}{20}$ $20 + h = 20\sqrt{3}$ $h = 20\sqrt{3} - 20 = 20(\sqrt{3} - 1)$ $h \approx 20(1.732 - 1) = 20 \times 0.732 = 14.64 \text{ m}$

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

Let the constant height be $H = 1500\sqrt{3}$ m. Let $x_1$ and $x_2$ be the horizontal distances from the observer when the angles are $30^\circ$ and $60^\circ$ respectively.

At $30^\circ$: $\tan 30^\circ = \frac{H}{x_1} \implies \frac{1}{\sqrt{3}} = \frac{1500\sqrt{3}}{x_1}$ $x_1 = 1500\sqrt{3} \times \sqrt{3} = 1500 \times 3 = 4500 \text{ m}$

At $60^\circ$: $\tan 60^\circ = \frac{H}{x_2} \implies \sqrt{3} = \frac{1500\sqrt{3}}{x_2}$ $x_2 = \frac{1500\sqrt{3}}{\sqrt{3}} = 1500 \text{ m}$

Distance covered $= x_1 - x_2 = 4500 - 1500 = 3000$ m.

The helicopter covers a horizontal distance of $3000$ m.

Let the height of the multi-storeyed building be $H$ and the horizontal distance between the buildings be $d$. The smaller building is $8$ m tall.

Bottom of small building ($45^\circ$ depression): $\tan 45^\circ = \frac{H}{d} \implies 1 = \frac{H}{d} \implies d = H \quad (1)$

Top of small building ($30^\circ$ depression): the vertical drop from the top of the tall building to the top of the small building is $H - 8$: $\tan 30^\circ = \frac{H - 8}{d} \implies \frac{1}{\sqrt{3}} = \frac{H - 8}{d}$ $d = \sqrt{3}(H - 8) \quad (2)$

Equating (1) and (2): $H = \sqrt{3}(H - 8)$ $H = \sqrt{3}\,H - 8\sqrt{3}$ $8\sqrt{3} = \sqrt{3}\,H - H = H(\sqrt{3} - 1)$ $H = \frac{8\sqrt{3}}{\sqrt{3} - 1} = \frac{8\sqrt{3}(\sqrt{3} + 1)}{(\sqrt{3} - 1)(\sqrt{3} + 1)} = \frac{8(3 + \sqrt{3})}{2} = 4(3 + \sqrt{3})$ $H = 12 + 4\sqrt{3} \approx 12 + 6.93 = 18.93 \text{ m}$

Since $d = H$, the distance between the buildings is also $4(3+\sqrt{3}) \approx 18.93$ m.

Height of the multi-storeyed building $= 4(3+\sqrt3) \approx 18.93$ m; distance $\approx 18.93$ m.