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.

Answer 1

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

$$\begin{gathered} \mathrm{We}\mathrm{have}, \\ \mathrm{AB}=40\mathrm{m},\angle \mathrm{BAC}=30°\mathrm{and}\angle \mathrm{BAD}=45° \\ \mathrm{In}∆\mathrm{ABD}, \\ \tan 45°=\frac{\mathrm{BD}}{\mathrm{AB}} \\ \Rightarrow 1=\frac{\mathrm{BD}}{40} \\ \Rightarrow \mathrm{BD}=40\mathrm{m} \\ \mathrm{Now},\mathrm{in}∆\mathrm{ABC}, \\ \tan 30°=\frac{\mathrm{BC}}{\mathrm{AB}} \\ \Rightarrow \frac{1}{\sqrt{3}}=\frac{\mathrm{BC}}{40} \\ \Rightarrow \mathrm{BC}=\frac{40}{\sqrt{3}} \\ \Rightarrow \mathrm{BC}=\frac{40}{\sqrt{3}}\times \frac{\sqrt{3}}{\sqrt{3}} \\ \Rightarrow \mathrm{BC}=\frac{40\sqrt{3}}{3}\mathrm{m} \\ \left(\mathrm{i}\right)\mathrm{The}\mathrm{height}\mathrm{of}\mathrm{the}\mathrm{tower},\mathrm{BC}=\frac{40\sqrt{3}}{3}=\frac{40\times 1.73}{3}=23.067\approx 23.1\mathrm{m} \\ \left(\mathrm{ii}\right)\mathrm{The}\mathrm{depth}\mathrm{of}\mathrm{the}\mathrm{tank},\mathrm{CD}=\left(\mathrm{BD}-\mathrm{BC}\right)=\left(40-23.1\right)=16.9\mathrm{m} \end{gathered}$$