Question

The angle of elevation of the top of a building from the foot of the tower is 30° and the angle of
elevation of the top of the tower from the foot of the building is 60°. If the tower is 60 m high
find the height of the building..​

Answer 1

$\blue{\bold{\underline{\underline{Answer:}}}}$

$\green{\therefore{\text{Height\:of\:building=20\:m}}}$

$\orange{\bold{\underline{\underline{Step-by-step\:explanation:}}}}$

• In the given question information given about the angle of elevation of the top of a building from the foot of the tower is 30° and the angle of elevation of the top of the tower from the foot of the building is 60°. If the tower is 60 m high.

• We have to find the height of the building.

$\green{\underline \bold{Given : }} \\ :\implies angle \: of \: elevation \: of \: top \: of \: \\ \: \: \: \: \: \: \: \: \: \: \: building \: from \: foot \: of \:tower \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \theta_{1} = 30 \degree \\ \\ :\implies angle \: of \: elevation \: of \: top \: of \: \\ \: \: \: \: \: \: \: \: \: \: \: tower \: from \: foot \: of \: building \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \theta_{2} = 60 \degree \\ \\ :\implies height \: of \: tower = 60 \: m \\ \\ \red{\underline \bold{To \: Find : }} \\ :\implies height \: of \: building = ?$

• According to given question :

$\bold{For \: angle \: of \: elevation = \theta_{1}} \\ :\implies tan \theta_{1} = \frac{p}{b} \\ \\ :\implies tan \: 30 \degree = \frac{Height \: of \: building}{Distance \: between \: tower \: and \: building} \\ \\ :\implies \frac{1}{\sqrt{3} } = \frac{x}{l} \\ \\: \implies l = \sqrt{3}x - - - - - (1) \\ \\ \bold{For \: angle \: of \: elevation = \theta_{2}} \\ : \implies tan \theta_{2} = \frac{Height \: of \: tower}{Distance \: between \: tower \: and \: building} \\ \\ :\implies \sqrt{3} = \frac{60}{l} \\ \\ \bold{putting \: value \: of \: l \: from \: (1)} \\ :\implies \sqrt{3} = \frac{60}{ \sqrt{3}x } \\ \\ :\implies \sqrt{3} \times \sqrt{3} x = 60 \\ \\ :\implies x = \frac{60}{3} \\ \\ \bold{ :\implies x = 20 \: m} \\ \\ \bold{\therefore Height \: of \: building = 20 \: m}$

Answer 2

Question :--- The angle of elevation of the top of a building from the foot of the tower is 30° and the angle of

elevation of the top of the tower from the foot of the building is 60°. If the tower is 60 m high

find the height of the building..

Solution :---

❁❁ Refer To Image First .. ❁❁

From image we can see that,

→ AB = Tower = 60m.

→ Angle ACB = 60°

→ DC = Building = x m .

→ Angle DBC = 30°..

__________________

In ∆ABC, we have ,

→ Tan60° = Perpendicular / Base

→ Tan60° = AB/BC

→ Tan60° = 60/BC

→ √3 = 60/BC

→ BC = 60/√3

→ BC = (60/√3) * (√3/√3)

→ BC = 60√3/3

→ BC = 20√3 m ------------------- Equation (1)

___________________________

In ∆ DBC, we have,

→ Tan30° = Perpendicular / Base

→ Tan60° = DC/BC

→ Tan30° = x/BC

→ 1/√3 = x/BC

→ BC = x * √3 ----------------- Equation (2) .

__________________________

Putting value of Equation (1) in Equation (2) now, we get,

→ 20√3 = x * √3

(√3 will be cancel From both sides ).

→ x = 20m. = DC

Hence, Height of Building DC is 20m.