From the top of a 12 m high building, The angle of elevation of the top of a tower is 60° and the angle of depression of the foot of the tower is θ. Such that tan θ = 3/4 , what is the height of the tower (√3 = 1.73)?

Answer: Option A

Solution

Answer: Option A

Solution:

Let AB be the building and EC be the tower.
$$\begin{array}{rl}& \text{In }\Delta ADE\\ & \tan 60^{\circ }=\frac{ED}{AD}\\ & \sqrt{3}=\frac{ED}{16}\\ & ED=16\sqrt{3}=16\times 1.73=27.68\text{m}\end{array}$$
Hence the height of the tower
EC = CD + ED
= 12 + 27.68
= 39.68 m