Question

From the top of a building 60 m high, the angles of depression of the top and bottom of a tower are observed to be ${30}^{\circ }$ and ${60}^{\circ }$. Find the height of the tower. [4 MARKS]

Answer 1

Let AB be the building and CD be the tower such that $\angle BDE={30}^{\circ },\angle BCA={60}^{\circ }$ and AB = 60 m

Let CA = DE = x metres.



From right $\Delta CAB$, we have

$\frac{CA}{AB}=cot{60}^{\circ }=\frac{1}{\sqrt{3}}$

$\Rightarrow \frac{x}{60}=\frac{1}{\sqrt{3}}\Rightarrow x=60\times \frac{1}{\sqrt{3}}$

$\Rightarrow x=60\times \frac{1}{\sqrt{3}}\times \frac{\sqrt{3}}{\sqrt{3}}=20\sqrt{3}$

$\Rightarrow CA=DE=20\sqrt{3}$

From right $\Delta BED$, we have

$\frac{BE}{DE}=tan{30}^{\circ }=\frac{1}{\sqrt{3}m}\Rightarrow \frac{BE}{20\sqrt{3}}=\frac{1}{\sqrt{3}}$ [using (i)]

$\Rightarrow BE=20\sqrt{3}\times \frac{1}{\sqrt{3}}m=20m$

$\therefore CD=AE=AB-BE=60m-20m=40m$

Hence the height of the tower is 40 m.

Alternative Method,

[$\frac{1}{2}$ MARK]

In $\Delta ACB$

$tan60=\frac{60}{x}$

$\sqrt{3}=\frac{60}{x}$

$x=\frac{60}{\sqrt{3}}\dots \dots \left(1\right)$ [1 MARK]

In $\Delta ADE$

$tan30=\frac{60-h}{x}$

$\frac{1}{\sqrt{3}}=\frac{60-h}{\frac{60}{\sqrt{3}}}$ (from (1)) [1 MARK]

$\frac{1}{\sqrt{3}}=\frac{60\sqrt{3}-\sqrt{3}h}{60}$

60 = 180 - 3h

3h = 120

h = 40m
Hence the height of the tower is 40 m. [$\frac{1}{2}$ MARK]