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Question
From the top of an 8 m high building, the angle of elevation of the top of a cable tower is 60° and the angle of depression of its foot is 45°. Determine the height of the tower. (Take `sqrt(3)` = 1.732).
Solution
According to given information, we have the following figure
Let BD = h m and AB = x m
Then, CD = h – 8 ...(∵ BC = 8 m)
Also, In ΔEAB, we have
tan45° = `8/x`
⇒ 1 = `8/x`
⇒ x = 8 m
Similarly, In ΔDCE, we have
tan60° = `(CD)/(EC)`
⇒ `sqrt(3) = (h - 8)/x` ...(∵ EC = AB = x m)
⇒ `sqrt(3)x` = h – 8
⇒ `8sqrt(3)` = h – 8 ...(∵ x = 8 m)
⇒ h = `8(sqrt(3) + 1)`
= 8 × 2.732
= 21.856 m
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