Question

A TV tower stands vertically on the side of a road. From a point on the other side directly opposite to the tower, the angle of elevation of the top of tower is ${60}^o$. From another point $10m$ away from this point, on the line joining this point to the foot of the tower, the angle of elevation of the top of the tower is ${30}^o$.Find the height of the tower and the width of the road.

Answer 1

Let $AB$ be the tower of height $h$.
In this figure; $CD=10m,\angle ACB={60}^{\circ },\angle ADB={30}^{\circ },AB=?$ and $BD=?$

In $△ABC$,

$\tan {60}^{\circ }=\frac{h}{x}$

$\sqrt{3}=\frac{h}{x}$

$h=x\sqrt{3}$ --------- (1)

In $△ABD$,

$\tan {30}^{\circ }=\frac{h}{10+x}$

$\frac{1}{\sqrt{3}}=\frac{h}{10+x}$

$h\sqrt{3}=10+x$ ---------- (2)

From equation (1) and (2);

$x\sqrt{3}=\frac{10+x}{\sqrt{3}}$

$3x=10+x$

$x=5$ which is the width of the road.

From (1),

Height of the tower $h=5\sqrt{3}m$