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 the top of the tower from the foot of the hill is 30°. If height of the tower is 50 m, find the height of the hill.

Answer 1

Let AB be height of tower (h)= 50 m.

And,  let the Height of Hill CD = H m.

Distance between The root of the tower and hill = BC 

Now,

In ΔABC

∠C = 30°

    TAN(C) =  AB/BC

⇒ TAN(30) =  50/BC

⇒  1/√3 = 50 /BC

⇒ BC = 50√3 m.

Now,

In ΔBCD,

∠B = 60°

    Tan(B) = CD/BC

⇒ Tan(60) = H/BC

⇒ BC√3 = H

⇒ H = 50√3×√3 = 150 m. (ans.)

Answer 2

$Hey\:!!..$

The answer goes here....

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》To find :

Height of hill

》Given :

The angle of elevation of the top of a hill at the foot of a tower = 60°

the angle of elevation of the top of the tower from the foot of the hill = 30°

Height of tower, AB = 50 m

》Solution :

Let the height of hill be $h$ .

In $\triangle ABC$,

$\angle C=30$°

⟹ tan C = $\frac{AB}{BC}$

⟹ tan 30° = $\frac{50}{BC}$

⟹ $\frac{1}{\sqrt{3}}$ = $\frac{50}{BC}$

⟹ BC = $50\sqrt{3}$

Also,

In $\triangle BCD$,

$\angle B=60$°

⟹ tan B = $\frac{CD}{BC}$

⟹ tan 60° = $\frac{h}{BC}$

⟹ $\sqrt{3}$ = $\frac{h}{BC}$

⟹ h = $BC\times\sqrt{3}$

⟹ h = 150 m

So, the height of hill is 150 m.

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Thanks !!..