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(D) Let the original height of the telegraph post be $AB$. Let it bend at point $C$. The top $A$ touches the ground at point $D$.Given: $BD = 10 \sqrt

Vedclass · Accepted Answer

$30$

Vedclass · Answer

(D) Let the original height of the telegraph post be $AB$. Let it bend at point $C$. The top $A$ touches the ground at point $D$.Given: $BD = 10 \sqrt{3} 	ext{ m}$ and $\angle BDC = 30^{\circ}$.In $\Delta BCD$,$	an 30^{\circ} = \frac{BC}{BD}$.$\frac{1}{\sqrt{3}} = \frac{BC}{10 \sqrt{3}} \implies BC = 10 	ext{ m}$.Again,in $\Delta BCD$,$\sin 30^{\circ} = \frac{BC}{CD}$.$\frac{1}{2} = \frac{10}{CD} \implies CD = 20 	ext{ m}$.The total height of the post is $AB = BC + CD$ (since $AC = CD$ as the top touches the ground).$AB = 10 + 20 = 30 	ext{ m}$.

$A$ telegraph post is bent at a point above the ground due to a storm. Its top just touches the ground at a distance of $10 \sqrt{3} \text{ m}$ from its foot and makes an angle of $30^{\circ}$ with the horizontal. Then,the height (in meters) of the telegraph post is:

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