A pole 6m high is fixed on the top of a tower. The angle of elevation of the top of the pole observed from a point P on the ground is $\displaystyle 60^{\mathrm{o}}$ and the angle of depression of the point P from the top of the tower is $\displaystyle 45^{\circ}$. Find the height of the tower and the distance of point P from the foot of the tower. (Use $\displaystyle \sqrt{3}=1.73$ )

Let BC be the pole and AB be the tower of height 'h' m. \[\begin{aligned} & \tan 45^{\circ}= \\ & \Rightarrow \mathrm{h}=\mathrm{x} \end{aligned} \] \[\tan 60^{\circ}=\sqrt{3}=\frac{\mathrm{h}+6}{\mathrm{x}} \] Solving (i) & (ii) to get \[h=3(\sqrt{3}+1)=8.19 \] and \(\displaystyle \mathrm{x}=8.19\) Therefore, the height of tower is $\displaystyle 8.19$ m and the distance of point P from the foot of the tower is $\displaystyle 8.19$ m