Question

The angle of elevation of the top of a tower from the top of a house whose height is 15 m is \(30^\circ\). Also, the angle of elevation of the top of the tower from the ground of the same house is \(60^\circ\). Find the height of the tower and the distance of the tower from the house.

Hint:

When two angles of elevation are given, form two right triangles using \(\tan \theta = \frac{\text{opposite}}{\text{adjacent}}\) and solve simultaneously.

Answer 1

Step 1: Let the height of the tower be \(h\) metres and the distance between the tower and the house be \(x\) metres.
Let the top of the house be \(H\), the bottom be \(G\), and the top of the tower be \(T\). Then, \(HG = 15 \, \text{m}\), \(GT = h\), and the horizontal distance \(GH = x\).
Step 2: From the top of the house, \[ \tan 30^\circ = \frac{(h - 15)}{x} \quad \Rightarrow \quad \frac{1}{\sqrt{3}} = \frac{h - 15}{x} \] \[ \Rightarrow x = \sqrt{3}(h - 15) \]
Step 3: From the ground level of the house, \[ \tan 60^\circ = \frac{h}{x} \quad \Rightarrow \quad \sqrt{3} = \frac{h}{x} \] \[ \Rightarrow x = \frac{h}{\sqrt{3}} \]
Step 4: Equate the two values of \(x\).
\[ \sqrt{3}(h - 15) = \frac{h}{\sqrt{3}} \] \[ 3(h - 15) = h \] \[ 3h - 45 = h \Rightarrow 2h = 45 \Rightarrow h = 22.5 \, \text{m} \]
Step 5: Find the distance \(x\).
\[ x = \frac{h}{\sqrt{3}} = \frac{22.5}{1.732} \approx 13 \, \text{m} \] Step 6: Conclusion.
The height of the tower is \(\boxed{22.5 \, \text{m}}\) and the distance between the tower and the house is \(\boxed{13 \, \text{m (approx.)}}\).