From a point on the ground 40 m away from the foot of a tower, the angle of elevation of the top of the tower is 30°.

Question:

From a point on the ground 40 m away from the foot of a tower, the angle of elevation of the top of the tower is 30°. The angle of elevation of the top of a water tank (on the top of the tower) is 45°.

Find

(i) the height of the tower,

(ii) the depth of the tank.

Solution:

Let BC be the tower and CD be the water tank.

We have,

$\mathrm{AB}=40 \mathrm{~m}, \angle \mathrm{BAC}=30^{\circ}$ and $\angle \mathrm{BAD}=45^{\circ}$

In $\Delta$ ABD

$\tan 45^{\circ}=\frac{\mathrm{BD}}{\mathrm{AB}}$

$\Rightarrow 1=\frac{\mathrm{BD}}{40}$

$\Rightarrow \mathrm{BD}=40 \mathrm{~m}$

Now, in $\triangle \mathrm{ABC}$,

$\tan 30^{\circ}=\frac{\mathrm{BC}}{\mathrm{AB}}$

$\Rightarrow \frac{1}{\sqrt{3}}=\frac{\mathrm{BC}}{40}$

$\Rightarrow \mathrm{BC}=\frac{40}{\sqrt{3}}$

$\Rightarrow \mathrm{BC}=\frac{40}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}}$

$\Rightarrow \mathrm{BC}=\frac{40 \sqrt{3}}{3} \mathrm{~m}$

(i) The height of the tower, $\mathrm{BC}=\frac{40 \sqrt{3}}{3}=\frac{40 \times 1.73}{3}=23.067 \approx 23.1 \mathrm{~m}$

(ii) The depth of the $\operatorname{tank}, \mathrm{CD}=(\mathrm{BD}-\mathrm{BC})=(40-23.1)=16.9 \mathrm{~m}$