Question

A TV tower stands vertically on a bank of canal. From a point on the other bank directly opposite the tower, the angle of elevation of the top of the tower is ${60}^{\circ }$. From another point 20 m 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 ${30}^{\circ }$. Find the height of the tower and the width of the canal.

Answer 1

In $△ABC$

$\frac{AB}{BC}=tan{60}^{\circ }$.

$\frac{AB}{BC}=\sqrt{3}$

$\therefore BC=\frac{AB}{\sqrt{3}}$

In $△ABD$

$\frac{AB}{BD}=tan{30}^{\circ }$.

$\frac{AB}{BC+CD}=\frac{1}{\sqrt{3}}$

$In\Delta ABD,\frac{AB}{BD}=tan{30}^o\frac{AB}{BC+CD}=\frac{1}{\sqrt{3}}\frac{AB}{\frac{AB}{\sqrt{3}}+20}=\frac{1}{\sqrt{3}}\frac{AB\sqrt{3}}{AB+20\sqrt{3}}=\frac{1}{\sqrt{3}}3AB=AB+20\sqrt{3}2AB=20\sqrt{3}AB=10\sqrt{3}mBC=\frac{AB}{\sqrt{3}}=\frac{10\sqrt{3}}{\sqrt{3}}=10m$

Therefore, the height of the tower is​​​​​​​ $10\sqrt{3}$ m and the width of the canal is 10 m