Question

A ten meter high tower is standing at the centre of an equilateral triangle and each side of the triangle subtends an angle of ${60}^o$ at the top of the tower. Prove that the length of each side of the triangle is $5\sqrt{6}$ metres.

Answer 1

Let $ABC$ be the equilateral triangle and $PQ$, the tower standing at the centre $P$ of the triangle, as shown in the figure above.

Here, $PQ=10m$.

Since each side of the triangle subtends an angle of ${60}^o$ at Q, the top of the tower, we have:
$\angle BQC=\angle CQA=\angle AQB={60}^o$

Hence, $AQ=QB$

Since, $\angle AQB={60}^o,\Delta AQB$ is equilateral.

Hence, $QA=QB=AB=2a\left(let\right)$

Similarly, $\Delta BQC$ is equilateral.

Hence, $QB=QC=2a$

$\therefore QA=QB=QC=2a$

We know that, a centroid is a point on the median, which is at ${\left(\frac{2}{3}\right)}^{rd}$ distance from the vertex.

Hence, $AP=\frac{2}{3}AD$

We also know that, the median of an equilateral triangle is a perpendicular bisector.

Hence, $\Delta ABD$ is a right-angled triangle.

Also, since $\Delta ABC$ is equilateral, $\angle ABD={60}^o$

We know that, $\sin \theta =\frac{\text{Opposite Side}}{\text{Hypotenuse}}$

Hence, $\sin {60}^o=\frac{AD}{2a}$

$\Rightarrow AD=2a\sin {60}^o$

So, $AP=\frac{2}{3}\times 2a\sin {60}^o$

$=\frac{4a}{3}\times \frac{\sqrt{3}}{2}[\because \sin {60}^o=\frac{\sqrt{3}}{2}]$

$\therefore AP=\frac{2a}{\sqrt{3}}$

$\Delta APQ$ is also a right-angled triangle.
Hence, by Pythagoras theorem, we have,

$A{Q}^2=A{P}^2+P{Q}^2$

$\Rightarrow A{Q}^2-A{P}^2=P{Q}^2$

$\Rightarrow 4a^2-\frac{4a^2}{3}={10}^2$

$\Rightarrow \frac{8a^2}{3}=100$

$\Rightarrow a^2=\frac{75}{2}$

$\Rightarrow a=5\sqrt{\frac{3}{2}}$

Now, the length of the side $AB=2a$

$=2\times 5\sqrt{\frac{3}{2}}$

$=5\sqrt{6}m$

Hence, it is proved that the length of each side of the triangle is $5\sqrt{6}m$.