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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.
Solution
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
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