Question

Two men on either side of a temple 75 m high observed the angle of elevation of the top of the temple to be 30° and 60° respectively. Find the distance between the two men.

Answer 1

Given the height of the temple AB = 75 m.

Now in right-angled ΔABC,

⇒ `"BC"/"AB" = cot 30°`

⇒ `"BC"/"AB" = sqrt3`

⇒ `"BC"/75 = sqrt3`

⇒ BC = 75√3                         ....(i)

Also in right angled ΔABD,

⇒ `"BD"/"AB" = cot 60°`

⇒ `"BD"/75 = 1/sqrt3`

⇒ BD = `75/sqrt3 xx sqrt3/sqrt3`

⇒ BD = 25√3                         ....(ii)

Now the distance between the two men = CD
= BC + BD
= 75√3 + 25√3
= 100√3

Hence, the distance between two men = 100√3 m = 173.2 m