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Each side of square subtends an angle of 60^{o} at the top of

Question

Each side of square subtends an angle of $6 0^{o}$ at the top of a tower of $h$ meter 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/are correct?

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

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

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

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

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Scarlett Discussed

Each side of square subtends an angle of

6 0^{o}

at the top of a tower of

h

meter 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/are correct?

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Text solutionVerified

As per question, the tower is standing on the center of square

Then the tower stands the point of intersection of two diagonal of square

Then the base of square of triangle be

\frac{1}{2} a

and the diagonal of square

= \frac{2 a}{2} = \frac{a}{2}

And its perpendicular to be

h

meters

So in this triangle

∴ tan 6 0^{o} = \frac{h}{\frac{a}{2}}

\Rightarrow 3 = \frac{2 h}{a}

\Rightarrow 3 a = 2 h

Squaring both side we get

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

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Question Text	Each side of square subtends an angle of $6 0^{o}$ at the top of a tower of $h$ meter 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/are correct?
Answer Type	Text solution:1
Upvotes	150