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Each side of square subtends an angle of $60^{o}$ at the top of a tower of $h$ m height standing in the centre of the square. If $a$ is the length of each side of the square then which of the following is correct?

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2 a^{2} = h^{2}

2 h^{2} = a^{2}

3 a^{2} = 2 h^{2}

2 h^{2} = 3 a^{2}

answer is B.

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Detailed Solution

It is given that the tower is standing in the centre of the square.
Then the tower stands at the point of intersection of two diagonals of square.
M is the midpoint of the side BC, so, CM

= \frac{a}{2}

Each side of a square ABCD subtends an angle of 60 at the top of a tower of. height h standing at the centre of a square. If a be the length

Since, PC = PB and BC subtends an angle of

60

° at P, the ΔPBC becomes an equilateral triangle.
And PM is the angle bisector of the angle P.
So, ∠CPM = 30°.
In

CPM

\tan 3 0^{\circ} = \frac{CM}{PM}

⇒

PM = \frac{\frac{a}{2}}{\frac{1}{\sqrt{3}}}

⇒

PM = \frac{\sqrt{3}}{2} a

m
In triangle POM,
By Pythagoras theorem,

P M^{2} = P O^{2} + O M^{2}

⇒

P O^{2} = P M^{2} - O M^{2}

⇒

h^{2} = {(\frac{\sqrt{3}}{2} a)}^{2} - {(\frac{a}{2})}^{2}

⇒

h^{2} = \frac{3}{4} a^{2} - \frac{1}{4} a^{2}

⇒

h^{2} = {\frac{1}{2} a}^{2}

Therefore,

2 h^{2} = a^{2}

.
Hence, the correct option is 2.

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