Question

From the top of a building 16m high. The angular elevation of the top of a hill is
60 and the angular depression of the foot of the hill is 30'. Find the height of the
hill.​

Answer 1

The height of the hill is 64m.

Explanation:

• In ΔABC,

tan 60=$\dfrac{BC}{AC}$

∵tan 60=$\sqrt{3}$

=>AC=$\dfrac{BC}{\sqrt{3}}$   -equation(1)

• Now, in ACD,

tan 30=$\dfrac{CD}{AC}$

∵tan30=$\dfrac{1}{\sqrt{3} }$

=>$\frac{1}{\sqrt{3} }$=$\frac{16\sqrt{3}}{BC}$  (from equation(1))

(AE=CD=16m)

BC=${16\sqrt{3}} \times{\sqrt{3}}$

BC=48 m

∴Height of hill= BC+CD=48m+16m=64m