The angle of elevation of the top of a vertical tower from a point P on the ground is 60° . From another point Q , 10…

Let $AD = h$ be the height of the tower and $AP = x$ be the distance of point $P$ from the foot of the tower.
$Q$ is $10$ m vertically above $P$, so $PQ = 10$ m.
Draw $QE \perp AD$. Then $AE = PQ = 10$ m and $ED = AD - AE = h - 10$.
Also $QE = AP = x$.
In right-angled $\triangle APD$:
$\tan 60^\circ = \frac{AD}{AP} = \frac{h}{x}$
$\sqrt{3} = \frac{h}{x} \Rightarrow h = x\sqrt{3}$ (1)
In right-angled $\triangle QED$:
$\tan 30^\circ = \frac{ED}{QE} = \frac{h-10}{x}$
$\frac{1}{\sqrt{3}} = \frac{h-10}{x} \Rightarrow x = \sqrt{3}(h-10)$ (2)
Substitute $x$ from (1) into (2):
$\frac{h}{\sqrt{3}} = \sqrt{3}(h-10)$
$h = 3(h-10)$
$h = 3h - 30$
$2h = 30 \Rightarrow h = 15$ m.
(a) The height of the tower is $h = 15$ m.
(b) From (1), $x = \frac{h}{\sqrt{3}} = \frac{15}{\sqrt{3}} = \frac{15\sqrt{3}}{3} = 5\sqrt{3}$ m.
The distance of the point $P$ from the foot of the tower is $5\sqrt{3}$ m.
(c) The distance of the point $P$ from the top of the tower is $PD$.
In right-angled $\triangle APD$:
$\cos 60^\circ = \frac{AP}{PD} = \frac{x}{PD}$
$\frac{1}{2} = \frac{5\sqrt{3}}{PD}$
$PD = 2 \times 5\sqrt{3} = 10\sqrt{3}$ m.
The distance of the point $P$ from the top of the tower is $10\sqrt{3}$ m.