The angle of elevation of the top of a hill at the foot of a tower is 60∘ and the angle of depression from the top of the tower to the foot of the hill is 30∘. If the tower is 50 m high, find the height of the hill.
Let AB be the hill and CD be the tower.
Angle of elevation of the hill at the foot of the tower is 60∘.i.e.,∠ADB=60∘ and the angle of depression of the foot of hill from the top of the tower is 30∘, i.e., ∠CBD=30∘.
Now in right angled ΔCBD:
tan 30∘=CDBD⇒BD=CDtan 30∘⇒BD=50[1√3]⇒BD=50√3m
In right ΔABD:
tan 60∘=ABCD
⇒AD=BD×tan 60∘=50√3×√3m=50×3 m
= 150 m
Hence, the height of the hill is 150 m.