$A$ tree stands vertically on the bank of a river. The angle of elevation of the top of the tree from a point exactly opposite to the tree on the other bank of the river is $60^{\circ}$. The angle of elevation of the top of the tree from a point $40 \ m$ away from that point is $30^{\circ}$. Find the width of the river and the height of the tree.

(N/A) Let $\overline{PM}$ be the tree and $\overline{OM}$ be the width of the river. $O$ and $A$ are the points of observation.
Given: $OA = 40 \ m$,$m\angle A = 30^{\circ}$,$m\angle O = 60^{\circ}$,and $m\angle M = 90^{\circ}$.
Let $OM = x \ m$ and $PM = h \ m$.
Then,$MA = OM + OA = (x + 40) \ m$.
In $\Delta PMO$,$m\angle M = 90^{\circ}$.
$\tan O = \frac{PM}{OM} \implies \tan 60^{\circ} = \frac{h}{x} \implies \sqrt{3} = \frac{h}{x} \implies h = \sqrt{3}x \quad ...(1)$
In $\Delta PMA$,$m\angle M = 90^{\circ}$.
$\tan A = \frac{PM}{AM} \implies \tan 30^{\circ} = \frac{h}{x + 40} \implies \frac{1}{\sqrt{3}} = \frac{h}{x + 40} \implies h = \frac{x + 40}{\sqrt{3}} \quad ...(2)$
From $(1)$ and $(2)$:
$\sqrt{3}x = \frac{x + 40}{\sqrt{3}} \implies 3x = x + 40 \implies 2x = 40 \implies x = 20 \ m$.
Now,$h = \sqrt{3} \times 20 = 20\sqrt{3} \approx 20 \times 1.732 = 34.64 \ m$.
Thus,the width of the river is $20 \ m$ and the height of the tree is $34.64 \ m$.