Question

The angle of elevation of the top of a hill at the foot of a tower is 60° and the angle of elevation of top of the tower from the foot of the hill is 30°.If the tower is 50 m high, what is the height of the hill?​

Answer 1

Answer:

Let AB be the tower and CD be the hill. Then,

angle ACB=30°, angle CAD=60° and AB=50m.

Let CD= x m

In right triangle BAC, we have,

cos=30° = AC/AB

root 3= AC/50

AC= 50 root 3 m

In right triangle ACD, we have,

tan 60° = CD/AC

root 3 = x / 50 root 3

x = 50 × 3 = 150 m.

Therefore, the height of the hill is 150 m.

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Answer 2

Answer:

Let AB be the tower and CD be the hill. Then, ∠ACB=30°

∠CAD=60°

AB=50 m

Let CD=x m

In right △BAC,

we have,

$\sf \: cot \: 30° = \frac{AC}{AB}$

$\sf \implies \: \sqrt{3} = \frac{AC}{50}$

$\sf \implies \: AC = 50 \sqrt{3} \: m$

In right △ACD, we have,

$\sf \: tan \: 60°= \frac{CD}{AC}$

$\sf \implies \: \sqrt{3} = \frac{x}{50 \sqrt{3} }$

$\sf \implies \: x=50×3 \: m$

$\sf \implies \: x=150 \: m$

Therefore, the height of the hill is 150 m.