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A lamp-post stands at the center of a circular park. Let $P$ and $Q$ be two points on the boundary such that $PQ$ subtends and angle $90 °$ at the foot of the lamp-post and the angle of elevation of the top of the lamp post from $P$ is $30 °$ . If $PQ = 30$ m, then find the height of the lamp post.

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5 \sqrt{6}

4 \sqrt{6}

6 \sqrt{5}

8 \sqrt{3}

answer is A.

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Detailed Solution

It is given that the lamp-post stands at the center of a circular park and

P

and

Q

are two points on the boundary such that

PQ

subtends an angle

90 °

at the foot of the lamp-post. The angle of elevation of the top of the lamp post from

P

30 °

and PQ=30m.
According to the given data the figure has been drawn below.
Question Image

Let the center of park be

O

which is at the bottom of the lamp post.
Let

R

be the top point of the lamp post.
Here

OP

and

OQ

are radius of park,

⟹ OP = OQ

And

PQ = 30 m

Since the

△ POQ

is right angled triangle, therefore using pythagoras theorem,

⟹ P Q^{2} = O P^{2} + O Q^{2}

⟹ 900 = 2 O P^{2}

(∵

OP = OQ

)

⟹ O P^{2} = 450

⟹ OP = \sqrt{450}

∴

OP = 15 \sqrt{2} m

Now,

△ POR

will also form a right triangle.

tan θ = P e r p e n d i c u l a r b a s e

\tan ∠ OPR = \frac{OR}{OP}

\Rightarrow \tan 30 ° = \frac{OR}{15 \sqrt{2}}

\Rightarrow OR = \frac{1}{\sqrt{3}} \times 15 \sqrt{2}

∵ tan 30 ° = 13

\Rightarrow OR

= \frac{15 \sqrt{6}}{3}

\Rightarrow OR = 5 \sqrt{6} m

So the height of lamp post is

5 \sqrt{6} m

.
So, the correct option is 1.

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