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(D) Let the height of the balloon be $h$ and the distance from the point directly under the balloon to $C$ be $x$. From the geometry of the problem:$x

Vedclass · Accepted Answer

$100\sqrt{3}$ metres

Vedclass · Answer

(D) Let the height of the balloon be $h$ and the distance from the point directly under the balloon to $C$ be $x$. From the geometry of the problem:$x = h \cot(3\alpha) \dots (i)$$x + 100 = h \cot(2\alpha) \dots (ii)$$x + 300 = h \cot(\alpha) \dots (iii)$Subtracting $(i)$ from $(ii)$:$100 = h(\cot(2\alpha) - \cot(3\alpha)) = h \left( \frac{\sin(3\alpha - 2\alpha)}{\sin(2\alpha)\sin(3\alpha)} \right) = h \frac{\sin(\alpha)}{\sin(2\alpha)\sin(3\alpha)} \dots (iv)$Subtracting $(ii)$ from $(iii)$:$200 = h(\cot(\alpha) - \cot(2\alpha)) = h \left( \frac{\sin(2\alpha - \alpha)}{\sin(\alpha)\sin(2\alpha)} \right) = h \frac{1}{\sin(2\alpha)} \dots (v)$Dividing $(iv)$ by $(v)$:$\frac{100}{200} = \frac{\sin(\alpha)}{\sin(3\alpha)} \Rightarrow \frac{\sin(3\alpha)}{\sin(\alpha)} = 2$Using $\sin(3\alpha) = 3\sin(\alpha) - 4\sin^3(\alpha)$:$3 - 4\sin^2(\alpha) = 2$ $\Rightarrow 4\sin^2(\alpha) = 1$ $\Rightarrow \sin(\alpha) = \frac{1}{2}$ $\Rightarrow \alpha = 30^\circ$Substituting $\alpha = 30^\circ$ in $(v)$:$200 = h \frac{1}{\sin(60^\circ)} = h \frac{1}{\sqrt{3}/2} = \frac{2h}{\sqrt{3}}$$h = 100\sqrt{3}$ metres.

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