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.

Question

A ladder 14 m long just reaches the top of a vertical wall. If the ladder makes an angle of $60^\circ$ with the wall, then the height of the wall is:

Answer 1

Let AB represent the tower and AC represent the canal. C is the point on the opposite side of the canal directly across from the tower.

TV tower stands vertically on a bank of a canal
In right triangle ABC,

$\frac{AB}{ BC} = tan 60^{\degree}$

$\frac{AB}{ BC} = \sqrt3$

$BC = \frac{AB}{\sqrt3}$

In right triangle ABD,

$\frac{AB}{ BD} = tan 30^{\degree}$

$\frac{AB}{BC + CD} = \frac1{\sqrt3}$

$\frac{AB}{\frac{ AB}{\sqrt3} + 20} = \frac{1}{\sqrt3}$

$\frac{ AB \sqrt3}{ AB + 20 \sqrt3} = \frac1{ \sqrt3}$

$3AB = AB + 20 \sqrt3$

$2AB = 20\sqrt3$

$AB = 10 \sqrt3\, m$

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

The height of the tower is $10\sqrt3 \,m$ and the width of the canal is $10\, m$.

Solution and Explanation