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 = $$\frac{3}{4}$$, what is the height of the tower (√3 = 1.73)?

Solution (By Examveda Team)

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