Question

A TV tower stands vertically on a bank of a canal. From a point on the other bank directly opposite the tower, the angle of elevation of the top of the tower is 60°. From another point 20 m away from this point on the line joing this point to the foot of the tower, the angle of elevation of the top of the tower is 30° (see Fig. 9.12). Find the height of the tower and the width of the canal.

Answer 1

let AB be the tower and AC be the canal. C is the point on the other side of the canal directly opposite the tower.

In ∆ABC,

\(\frac{AB}{ BC} = tan 60^{\degree}\)

\(\frac{AB}{ BC} = \sqrt3\)

\(BC = \frac{AB}{\sqrt3}\)

In ∆ABD, 

\(\frac{AB}{ BD} = tan 30^{\degree}\)

\(\frac{AB}{BC + CD} = \frac1{\sqrt3}\)

 \(\frac{AB}{\frac{ AB}{\sqrt3} + 20} = \frac{1}{\sqrt3}\)

\(\frac{ AB \sqrt3}{ AB + 20 \sqrt3} = \frac1{ \sqrt3}\)

 \(3AB = AB + 20 \sqrt3\)

\(2AB = 20\sqrt3\)

\(AB = 10 \sqrt3\, m\)

\(BC = \frac{AB}{ \sqrt3} = (\frac{10  \sqrt3}{\sqrt3})m = 10\,m\)

Therefore, the height of the tower is \(10\sqrt3 \,m\) and the width of the canal is \(10\, m\).