From the top of 75 m high tower, the angle of depression of two points P and Q on opposite side of the base of the tower on level ground is $$\theta $$ and $$\phi $$, such that $$\tan \theta = \frac{3}{4}$$   and $$\tan \phi = \frac{5}{8}.$$   What is the distance between the points P and Q?

$$\eqalign{ & {\text{Let }}AB{\text{ be the tower}}{\text{.}} \cr & {\text{Given that;}} \cr & \tan \theta = \frac{3}{4},\,\tan \phi = \frac{5}{8} \cr & {\text{In }}\Delta ABP, \cr & \tan \theta = \frac{{AB}}{{PB}} \cr & \frac{3}{4} = \frac{{75}}{{PB}} \cr & PB = 100{\text{ m}} \cr & {\text{In }}\Delta ABQ, \cr & \tan \phi = \frac{{AB}}{{BQ}} \cr & \frac{5}{8} = \frac{{75}}{{BQ}} \cr & BQ = 120{\text{ m}} \cr & {\text{Hence, the distance between }}P{\text{ and }}Q \cr & = 100 + 120 \cr & = 220 \cr} $$