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).

Answer 1

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