A tree breaks at a certain height and the upp | vedclass

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

Question

(C) Let the tree be $AB$ and it breaks at point $C$. The upper part $AC$ touches the ground at point $D$.Given: $BD = 10 	ext{ m}$ and $\angle BDC =

Vedclass · Accepted Answer

$10(2+\sqrt{3})$

Vedclass · Answer

(C) Let the tree be $AB$ and it breaks at point $C$. The upper part $AC$ touches the ground at point $D$.Given: $BD = 10 	ext{ m}$ and $\angle BDC = 60^{\circ}$.In $	riangle BCD$,$	an 60^{\circ} = \frac{BC}{BD} \implies \sqrt{3} = \frac{BC}{10} \implies BC = 10\sqrt{3} 	ext{ m}$.Also,$\cos 60^{\circ} = \frac{BD}{CD} \implies \frac{1}{2} = \frac{10}{CD} \implies CD = 20 	ext{ m}$.Since $AC = CD$,the original height of the tree is $AB = BC + AC = BC + CD = 10\sqrt{3} + 20 = 10(2 + \sqrt{3}) 	ext{ m}$.

$A$ tree breaks at a certain height and the upper part touches the ground,making an angle of $60^{\circ}$ with the ground at a distance of $10 \text{ m}$ from its foot. The original height of the tree was (in $\text{m}$):

Similar Questions

If $90^{\circ} < \alpha < 180^{\circ}$,$\sin \alpha = \frac{\sqrt{3}}{2}$ and $180^{\circ} < \beta < 270^{\circ}$,$\sin \beta = -\frac{\sqrt{3}}{2}$,then $\frac{4 \sin \alpha - 3 \tan \beta}{\tan \alpha + \sin \beta} = $

$\sin {163^\circ} \cos {347^\circ} + \sin {73^\circ} \sin {167^\circ} = $

If $\tan \theta = \frac{a}{b},$ then $\frac{{\sin \theta }}{{{{\cos }^8}\theta }} + \frac{{\cos \theta }}{{{{\sin }^8}\theta }} = $

The value of $\frac{\tan x}{\tan 3x}$,whenever defined,never lies between:

$\cot x - \tan x = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module