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Question
From the top of building AB, a point C is observed on the ground whose angle of depression is 60° and which is at a distance of 40 m from the base of the building. Complete the following activity to find the height of building AB.
From figure, BC = `square`, ∠ACB = `square`
In ΔACB,
tan `square = square/(BC)`
⇒ `square = square/square`
⇒ `square = square`
Hence, the height of the building AB is `square`.
Solution
From figure, BC = 40 m, ∠ACB = 60°
In ΔACB,
tan 60° = `bb(AB)/(BC)`
⇒ `bbsqrt(3) = bb(AB)/bb40`
⇒ `bb(AB) = bb(40sqrt(3))`
Hence, the height of the building AB is `bb(40sqrt(3) m)`.
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