A telegraph post gets broken at a point against a storm and its top touches the ground at a distance $20$ m from the base of the post making an angle $30^{\circ}$ with the ground. What is the height of the post?

\frac{40}{\sqrt{3}}

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20 \sqrt{3}

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40 \sqrt{3}

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30

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Solution

The correct option is B $20 \sqrt{3}$ m
Height of the telegraph post
$= h + p = 20 tan 30^{\circ} + 20 sec 30^{\circ}$
$= \frac{20}{\sqrt{3}} + 20 \times \frac{2}{\sqrt{3}}$
$= \frac{60}{\sqrt{3}} = 20 \sqrt{3}$

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A telegraph post gets broken at a point against a storm and its top touches the ground at a distance 20 m from the base of the post making an angle 30∘ with the ground. What is the height of the post?

The correct option is B 20√3 mHeight of the telegraph post=h+p=20tan30∘+20sec30∘=20√3+20×2√3=60√3=20√3

A telegraph post gets broken at a point against a storm and its top touches the ground at a distance $20$ m from the base of the post making an angle $30^{\circ}$ with the ground. What is the height of the post?

The correct option is B $20 \sqrt{3}$ m
Height of the telegraph post
$= h + p = 20 tan 30^{\circ} + 20 sec 30^{\circ}$
$= \frac{20}{\sqrt{3}} + 20 \times \frac{2}{\sqrt{3}}$
$= \frac{60}{\sqrt{3}} = 20 \sqrt{3}$