Question

the angle of elevation of the top of the 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 50 m high , find the height of the building .
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Answer 1

Answer:

16.67 meters.

Step-by-step explanation:

AB ➝ Tower ➝ 50m.

BC ➝ Building ➝ ?m.

Angle of elevation of D from B ➝ ∠DBC = 30°

Angle of elevation of A from C ➝ ∠ACB = 60°

We have to find the height of the building.

In ΔABC,

∠ABC = 90°

Therefore we can say that:

$\Longrightarrow \sf tan60^\circ = \dfrac{Opposite \ Side}{Adjacent \ Side}$

$\Longrightarrow \sf tan60^\circ = \dfrac{AB}{CB}$

$\Longrightarrow \sf \sqrt{3} = \dfrac{50}{CB}$

$\Longrightarrow \sf CB = \dfrac{50}{\sqrt{3}} \ m$

In ΔDCB,

∠DCB = 90°

Therefore we can say that:

$\Longrightarrow \sf tan30^\circ = \dfrac{Opposite \ Side}{Adjacent \ Side}$

$\Longrightarrow \sf \dfrac{1}{\sqrt{3}} = \dfrac{DC}{CB}$

$\Longrightarrow \sf CB = DC \sqrt{3}$

Substitute the value of CB.

$\Longrightarrow \sf \dfrac{50}{\sqrt{3}} = DC\sqrt{3}$

$\Longrightarrow \sf DC = \dfrac{50}{\sqrt{3} \times \sqrt{3}}$

$\Longrightarrow \sf DC = \dfrac{50}{3}$

$\Longrightarrow \sf DC \approx 16.67m.$

Answer 2

Answer:

16.66m

Step by Step explanation:

See image above..