Some Applications of Trigonometry

Given: Length of rope (hypotenuse) = 20 m, angle with ground = 30°.

Let the height of the pole = AB (perpendicular), and the rope AC is the hypotenuse.

Formula used: $\sin\theta = \dfrac{\text{Perpendicular}}{\text{Hypotenuse}}$

Working:
$$\sin 30° = \frac{AB}{AC}$$
$$\frac{1}{2} = \frac{AB}{20}$$
$$AB = \frac{20}{2} = 10 \text{ m}$$

Answer: The height of the pole is 10 m.

Given: The broken part makes an angle of 30° with the ground. The horizontal distance from the foot of the tree to the point where the top touches the ground = 8 m.

Let BC = standing part of the tree (perpendicular), AC = broken part (hypotenuse), AB = 8 m (base).

Step 1: Find BC using $\tan 30°$:
$$\tan 30° = \frac{BC}{AB}$$
$$\frac{1}{\sqrt{3}} = \frac{BC}{8}$$
$$BC = \frac{8}{\sqrt{3}} = \frac{8\sqrt{3}}{3} \text{ m}$$

Step 2: Find AC (broken part) using $\cos 30°$:
$$\cos 30° = \frac{AB}{AC}$$
$$\frac{\sqrt{3}}{2} = \frac{8}{AC}$$
$$AC = \frac{16}{\sqrt{3}} = \frac{16\sqrt{3}}{3} \text{ m}$$

Step 3: Total height of the tree = BC + AC:
$$= \frac{8\sqrt{3}}{3} + \frac{16\sqrt{3}}{3} = \frac{24\sqrt{3}}{3} = 8\sqrt{3} \text{ m}$$

Answer: The height of the tree is $8\sqrt{3}$ m.

Case 1 (Younger children):

Given: Height = 1.5 m, angle of inclination = 30°.

Let the length of the slide = $l_1$ (hypotenuse).

$$\sin 30° = \frac{1.5}{l_1}$$
$$\frac{1}{2} = \frac{1.5}{l_1}$$
$$l_1 = 1.5 \times 2 = 3 \text{ m}$$

Case 2 (Elder children):

Given: Height = 3 m, angle of inclination = 60°.

Let the length of the slide = $l_2$ (hypotenuse).

$$\sin 60° = \frac{3}{l_2}$$
$$\frac{\sqrt{3}}{2} = \frac{3}{l_2}$$
$$l_2 = \frac{3 \times 2}{\sqrt{3}} = \frac{6}{\sqrt{3}} = 2\sqrt{3} \text{ m}$$

Answer: The length of the slide for younger children is 3 m and for elder children is $2\sqrt{3}$ m.

Given: Distance from the point to the foot of the tower = 30 m, angle of elevation = 30°.

Let the height of the tower = $h$ m.

Formula used: $\tan\theta = \dfrac{\text{Height}}{\text{Base}}$

Working:
$$\tan 30° = \frac{h}{30}$$
$$\frac{1}{\sqrt{3}} = \frac{h}{30}$$
$$h = \frac{30}{\sqrt{3}} = \frac{30\sqrt{3}}{3} = 10\sqrt{3} \text{ m}$$

Answer: The height of the tower is $10\sqrt{3}$ m.

Given: Height of kite = 60 m, angle of inclination of string with ground = 60°.

Let the length of the string = $l$ m (hypotenuse).

Working:
$$\sin 60° = \frac{60}{l}$$
$$\frac{\sqrt{3}}{2} = \frac{60}{l}$$
$$l = \frac{60 \times 2}{\sqrt{3}} = \frac{120}{\sqrt{3}} = \frac{120\sqrt{3}}{3} = 40\sqrt{3} \text{ m}$$

Answer: The length of the string is $40\sqrt{3}$ m.

Given: Height of boy = 1.5 m, height of building = 30 m.

Effective height above the boy's eye level = $30 - 1.5 = 28.5$ m.

Let the initial distance of the boy's eyes from the building = $d_1$, and the final distance = $d_2$.

Step 1: When angle of elevation = 30°:
$$\tan 30° = \frac{28.5}{d_1}$$
$$\frac{1}{\sqrt{3}} = \frac{28.5}{d_1}$$
$$d_1 = 28.5\sqrt{3} \text{ m}$$

Step 2: When angle of elevation = 60°:
$$\tan 60° = \frac{28.5}{d_2}$$
$$\sqrt{3} = \frac{28.5}{d_2}$$
$$d_2 = \frac{28.5}{\sqrt{3}} = \frac{28.5\sqrt{3}}{3} = 9.5\sqrt{3} \text{ m}$$

Step 3: Distance walked = $d_1 - d_2$:
$$= 28.5\sqrt{3} - 9.5\sqrt{3} = 19\sqrt{3} \text{ m}$$

Answer: The distance walked by the boy towards the building is $19\sqrt{3}$ m.

Given: Height of building = 20 m. Let height of transmission tower = $h$ m. Let the point on the ground be at distance $d$ from the foot of the building.

Step 1: Angle of elevation of the bottom of the tower (top of building) = 45°:
$$\tan 45° = \frac{20}{d}$$
$$1 = \frac{20}{d}$$
$$d = 20 \text{ m}$$

Step 2: Angle of elevation of the top of the tower = 60°:
$$\tan 60° = \frac{20 + h}{d}$$
$$\sqrt{3} = \frac{20 + h}{20}$$
$$20 + h = 20\sqrt{3}$$
$$h = 20\sqrt{3} - 20 = 20(\sqrt{3} - 1) \text{ m}$$

Answer: The height of the transmission tower is $20(\sqrt{3}-1)$ m.

Given: Height of statue = 1.6 m. Let height of pedestal = $h$ m. Let the distance of the point from the base = $d$ m.

Step 1: Angle of elevation of top of pedestal = 45°:
$$\tan 45° = \frac{h}{d}$$
$$1 = \frac{h}{d}$$
$$d = h \quad \cdots (1)$$

Step 2: Angle of elevation of top of statue = 60°:
$$\tan 60° = \frac{h + 1.6}{d}$$
$$\sqrt{3} = \frac{h + 1.6}{h} \quad [\text{using (1)}]$$
$$\sqrt{3}\, h = h + 1.6$$
$$h(\sqrt{3} - 1) = 1.6$$
$$h = \frac{1.6}{\sqrt{3}-1} = \frac{1.6(\sqrt{3}+1)}{(\sqrt{3}-1)(\sqrt{3}+1)} = \frac{1.6(\sqrt{3}+1)}{2} = 0.8(\sqrt{3}+1) \text{ m}$$

Answer: The height of the pedestal is $0.8(\sqrt{3}+1)$ m.

Given: Height of tower = 50 m. Let height of building = $h$ m. Let the horizontal distance between them = $d$ m.

Step 1: Angle of elevation of top of tower from foot of building = 60°:
$$\tan 60° = \frac{50}{d}$$
$$\sqrt{3} = \frac{50}{d}$$
$$d = \frac{50}{\sqrt{3}} \text{ m}$$

Step 2: Angle of elevation of top of building from foot of tower = 30°:
$$\tan 30° = \frac{h}{d}$$
$$\frac{1}{\sqrt{3}} = \frac{h}{\dfrac{50}{\sqrt{3}}}$$
$$h = \frac{50}{\sqrt{3}} \times \frac{1}{\sqrt{3}} = \frac{50}{3} \text{ m}$$

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

Given: Width of road = 80 m. Let height of each pole = $h$ m. Let the point P be at distance $x$ m from one pole (say pole AB) and $(80 - x)$ m from the other pole (CD).

Step 1: Angle of elevation of pole AB = 60°:
$$\tan 60° = \frac{h}{x}$$
$$\sqrt{3} = \frac{h}{x}$$
$$h = \sqrt{3}\,x \quad \cdots (1)$$

Step 2: Angle of elevation of pole CD = 30°:
$$\tan 30° = \frac{h}{80 - x}$$
$$\frac{1}{\sqrt{3}} = \frac{h}{80 - x}$$
$$h = \frac{80 - x}{\sqrt{3}} \quad \cdots (2)$$

Step 3: Equating (1) and (2):
$$\sqrt{3}\,x = \frac{80 - x}{\sqrt{3}}$$
$$3x = 80 - x$$
$$4x = 80$$
$$x = 20 \text{ m}$$

Step 4: Height of poles:
$$h = \sqrt{3} \times 20 = 20\sqrt{3} \text{ m}$$

Distance from the other pole = $80 - 20 = 60$ m.

Answer: The height of each pole is $20\sqrt{3}$ m. The point is 20 m from one pole and 60 m from the other pole.

Given: Let the height of the tower = $h$ m and the width of the canal = $d$ m.

Let A be the point directly opposite the tower (angle of elevation = 60°), and B be the point 20 m further away (angle of elevation = 30°). So AB = 20 m.

Step 1: From point A:
$$\tan 60° = \frac{h}{d}$$
$$\sqrt{3} = \frac{h}{d}$$
$$h = \sqrt{3}\,d \quad \cdots (1)$$

Step 2: From point B (distance from tower = $d + 20$):
$$\tan 30° = \frac{h}{d + 20}$$
$$\frac{1}{\sqrt{3}} = \frac{h}{d + 20}$$
$$h = \frac{d + 20}{\sqrt{3}} \quad \cdots (2)$$

Step 3: Equating (1) and (2):
$$\sqrt{3}\,d = \frac{d + 20}{\sqrt{3}}$$
$$3d = d + 20$$
$$2d = 20$$
$$d = 10 \text{ m}$$

Step 4: Height of tower:
$$h = \sqrt{3} \times 10 = 10\sqrt{3} \text{ m}$$

Answer: The height of the tower is $10\sqrt{3}$ m and the width of the canal is 10 m.

Given: Height of building = 7 m. Let height of cable tower = $H$ m. Let horizontal distance between them = $d$ m.

Step 1: Angle of depression of the foot of the tower = 45°. The foot of the tower and the base of the building are at the same level.
$$\tan 45° = \frac{7}{d}$$
$$1 = \frac{7}{d}$$
$$d = 7 \text{ m}$$

Step 2: The angle of elevation of the top of the tower from the top of the building = 60°. The height of the tower above the building level = $H - 7$.
$$\tan 60° = \frac{H - 7}{d}$$
$$\sqrt{3} = \frac{H - 7}{7}$$
$$H - 7 = 7\sqrt{3}$$
$$H = 7 + 7\sqrt{3} = 7(1 + \sqrt{3}) \text{ m}$$

Answer: The height of the cable tower is $7(1+\sqrt{3})$ m.

Given: Height of lighthouse = 75 m. Angles of depression of two ships = 45° and 30°.

Let A be the top of the lighthouse, B its base. Let C and D be the positions of the two ships (C is closer, D is farther).

Step 1: For the nearer ship C (angle of depression = 45°):
$$\tan 45° = \frac{75}{BC}$$
$$1 = \frac{75}{BC}$$
$$BC = 75 \text{ m}$$

Step 2: For the farther ship D (angle of depression = 30°):
$$\tan 30° = \frac{75}{BD}$$
$$\frac{1}{\sqrt{3}} = \frac{75}{BD}$$
$$BD = 75\sqrt{3} \text{ m}$$

Step 3: Distance between the two ships:
$$CD = BD - BC = 75\sqrt{3} - 75 = 75(\sqrt{3} - 1) \text{ m}$$

Answer: The distance between the two ships is $75(\sqrt{3}-1)$ m.

Given: Height of girl = 1.2 m. Height of balloon from ground = 88.2 m.

Height of balloon above the girl's eyes = $88.2 - 1.2 = 87$ m.

Let the girl's eye be at point E. Let A and B be the two positions of the balloon.

Step 1: When angle of elevation = 60°, horizontal distance = $d_1$:
$$\tan 60° = \frac{87}{d_1}$$
$$\sqrt{3} = \frac{87}{d_1}$$
$$d_1 = \frac{87}{\sqrt{3}} = \frac{87\sqrt{3}}{3} = 29\sqrt{3} \text{ m}$$

Step 2: When angle of elevation = 30°, horizontal distance = $d_2$:
$$\tan 30° = \frac{87}{d_2}$$
$$\frac{1}{\sqrt{3}} = \frac{87}{d_2}$$
$$d_2 = 87\sqrt{3} \text{ m}$$

Step 3: Distance travelled by balloon:
$$= d_2 - d_1 = 87\sqrt{3} - 29\sqrt{3} = 58\sqrt{3} \text{ m}$$

Answer: The distance travelled by the balloon is $58\sqrt{3}$ m.

Given: Let the height of the tower = $h$ m. The car moves from point A (angle of depression 30°) to point B (angle of depression 60°) in 6 seconds.

Step 1: Let the foot of the tower be O. From the top of the tower:

At position A:
$$\tan 30° = \frac{h}{OA} \implies OA = h\sqrt{3}$$

At position B:
$$\tan 60° = \frac{h}{OB} \implies OB = \frac{h}{\sqrt{3}}$$

Step 2: Distance covered in 6 seconds:
$$AB = OA - OB = h\sqrt{3} - \frac{h}{\sqrt{3}} = h\left(\sqrt{3} - \frac{1}{\sqrt{3}}\right) = h \cdot \frac{3-1}{\sqrt{3}} = \frac{2h}{\sqrt{3}}$$

Step 3: Speed of car:
$$\text{Speed} = \frac{AB}{6} = \frac{2h}{6\sqrt{3}} = \frac{h}{3\sqrt{3}}$$

Step 4: Time to travel from B to O (distance = $OB = \dfrac{h}{\sqrt{3}}$):
$$\text{Time} = \frac{OB}{\text{Speed}} = \frac{\dfrac{h}{\sqrt{3}}}{\dfrac{h}{3\sqrt{3}}} = \frac{h}{\sqrt{3}} \times \frac{3\sqrt{3}}{h} = 3 \text{ seconds}$$

Answer: The car will reach the foot of the tower in 3 seconds from point B.