Question

A square tower stands upon a horizontal plane. From a point in this plane, from which three of its upper corners are visible, their angular elevations are respectively are ${45}^{\circ },{60}^{\circ }$, and ${45}^{\circ }$. Show that the height of the tower is to the breadth of one of its sides as $\sqrt{6}(\sqrt{5}+1)$ to $4$.

Answer 1

One face of the tower with two upper corners $D$ and $E$ is shown in the figure.

$D$ makes ${60}^{\circ }$ with a point $A$ in the plane.

Hence, in $\Delta ACD$

$tan{60}^{\circ }=\frac{CD}{AC}$

$\Rightarrow \sqrt{3}=\frac{H}{AC}$

$\Rightarrow AC=\frac{H}{\sqrt{3}}$

The corner $E$ makes an angle of $45^\circ$ with point $A$ in the plane.

Hence, in $\Delta ABE$

$tan{45}^{\circ }=\frac{EB}{BA}$

$\Rightarrow 1=\frac{H}{AB}$

$\Rightarrow AB=H$

In $\Delta ABC$, $\angle ACB={135}^{\circ }$ (using symmetry, since the angle is same ${45}^{\circ }$ with two upper corners)

Triangle $ABC$ is redrawn below:

Using sine rule in the above triangle,

$\frac{\sin {135}^{\circ }}{H}=\frac{\sin \theta }{H/\sqrt{3}}$

$\Rightarrow \sin \theta =\frac{1}{\sqrt{6}}$

$\therefore cos\theta =\frac{\sqrt{5}}{\sqrt{6}}$

From the figure, $\angle A={180}^{\circ }-({135}^{\circ }+\theta )$

$\Rightarrow \angle A={45}^{\circ }-\theta$

$\sin A=\sin ({45}^{\circ }-\theta )=\frac{1}{\sqrt{2}}\times \frac{\sqrt{5}}{\sqrt{6}}-\frac{1}{\sqrt{2}}\times \frac{1}{\sqrt{6}}=\frac{\sqrt{5}-1}{2\sqrt{3}}$

Again using sine rule we get,

$\frac{\sin A}{a}=\frac{\sin {135}^{\circ }}{H}$

$\Rightarrow \frac{H}{a}=\frac{\sin {135}^{\circ }}{\sin A}$

$\Rightarrow \frac{H}{a}=\frac{1\left(2\sqrt{3}\right)}{\sqrt{2}(\sqrt{5}-1)}$

$=\frac{\sqrt{6}}{\sqrt{5}-1}$

$=\frac{\sqrt{6}(\sqrt{5}+1)}{(\sqrt{5}-1)(\sqrt{5}+1)}$

$\Rightarrow \frac{H}{a}=\frac{\sqrt{6}(\sqrt{5}+1)}{4}$

Hence the height of the tower is to the breadth of one of its sides as $\sqrt{6}(\sqrt{5}+1)$ to $4$.