The lower window of a house is at \(2\,\text{m}\) above the ground and the upper window is \(4\,\text{m}\) vertically above it. The angles of elevation of a balloon from these windows are \(60^\circ\) and \(30^\circ\), respectively. Find the height of the balloon above the ground.
\(8\,\text{m}.\)
Step 1: Understand the problem
The house has two windows. The lower window is 2 m above the ground. The upper window is 4 m higher than this, so it is at \(2 + 4 = 6\,\text{m}\) above the ground.
From the lower window, the angle of elevation to the balloon is \(60^\circ\). From the upper window, the angle of elevation is \(30^\circ\).
We need to find the total height of the balloon from the ground (let us call this \(H\)).
Step 2: Draw right triangles
Let the horizontal distance between the balloon and the wall of the house be \(x\) metres. This forms two right-angled triangles with vertical sides \((H - 2)\) and \((H - 6)\).
Step 3: Write equation using tan for the lower window
For the lower window (at 2 m):
\(\tan 60^\circ = \dfrac{H - 2}{x}\)
We know \(\tan 60^\circ = \sqrt{3}\). So:
\(\sqrt{3} = \dfrac{H - 2}{x} \;\; \Rightarrow \; H - 2 = \sqrt{3}\,x.\) (Equation 1)
Step 4: Write equation using tan for the upper window
For the upper window (at 6 m):
\(\tan 30^\circ = \dfrac{H - 6}{x}\)
We know \(\tan 30^\circ = \dfrac{1}{\sqrt{3}}\). So:
\(\dfrac{1}{\sqrt{3}} = \dfrac{H - 6}{x} \;\; \Rightarrow \; H - 6 = \dfrac{x}{\sqrt{3}}.\) (Equation 2)
Step 5: Eliminate x
From Equation (1): \(x = \dfrac{H - 2}{\sqrt{3}}\).
Substitute this into Equation (2):
\(H - 6 = \dfrac{1}{\sqrt{3}} \times \dfrac{H - 2}{\sqrt{3}} = \dfrac{H - 2}{3}.\)
Step 6: Solve for H
Multiply both sides by 3:
\(3(H - 6) = H - 2\)
\(3H - 18 = H - 2\)
\(3H - H = -2 + 18\)
\(2H = 16\)
\(H = 8\,\text{m}\)
Final Answer: The height of the balloon above the ground is 8 metres.