A telegraph post is bent at a point above the | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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:

Similar Questions

If $\tan \frac{A}{2} = \frac{3}{2},$ then $\frac{1 + \cos A}{1 - \cos A} = $

The equation $(a + b)^2 = 4ab \sin^2 \theta$ is possible only when

If $\alpha$ is a root of $25\cos^2\theta + 5\cos\theta - 12 = 0$ and $\pi/2 < \alpha < \pi$,then $\sin 2\alpha$ is equal to

$\Delta XYZ$ is right-angled at $Y$. If $\cos X = \frac{3}{5}$,then what is the value of $\operatorname{cosec} Z$?

If $\sec A = a + \left(\frac{1}{4a}\right)$,then $\sec A + \tan A =$

Vedclass Test Series

Exam Paper Generator

Online Exam Module