Question

The angle of elevation of the top of a hill from 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 the tower is 50 m high,
then find the height of the hill.

Answer 1

Here is your answer mate ↙️↙️☺️☺️✌️✌️✅✅

Ans ➡️ Let AB is the Tower of height = 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.

Hope this helps you mate ☺️☺️✌️✌️✅✅✅⭐⭐⭐

Answer 2

Answer:

The angle of elevation of the top of a building from the foot of the tower is 30°.

The angle of elevation of the top of tower from the foot of the building is 60°.

Height of the tower is 50 m.

Let the height of the building be h.

Step-by-step explanation:

$\sf In \: \Delta BDC$

$:\implies \sf tan \: \theta = \dfrac{Perpendicular}{Base}$

$:\implies \sf tan \: 60^{\circ} = \dfrac{CD}{BD}$

$:\implies \sf \sqrt{3} = \dfrac{50}{BD}$

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

___________________

$\sf In \: \Delta ABD,$

$\dashrightarrow\:\: \sf tan \: 30 ^{\circ} = \dfrac{AB}{BD}$

$\dashrightarrow\:\: \sf \dfrac{1}{ \sqrt{3} } = \dfrac{AB}{BD}$

$\dashrightarrow\:\: \sf \dfrac{1}{ \sqrt{3} } = \dfrac{h}{ \dfrac{50}{ \sqrt{3} } }$

$\dashrightarrow\:\: \sf h = \dfrac{ 50 }{ \sqrt{3} \times \sqrt{3} }$

$\dashrightarrow\:\: \underline{ \boxed{\sf Height= \dfrac{50}{3} \: m }}$

$\therefore\:\underline{\textsf{Height of the building is \textbf{50/3 meter}}}.$