Question

The upper part of a tree is broken over by the wind makes an angle of ${30}^{\circ }$ with the ground and the distance from the root to the point where the top of the tree meets the ground is 15 m The height of the broken part is

• A. $15\sin {30}^{\circ }m$
• B. $15\cos {30}^{\circ }m$
• C. $15\tan {30}^{\circ }m$
• D. $15\sec {30}^{\circ }m$

Answer 1

Answer: (D)

$15\sec {30}^{\circ }m$

Let the length of the broken part = $l$
Angle made by the broken part with the ground = ${30}^{\circ }$
the distance on the ground = $15$ m
The ground, the broken part and the unbroken part forms a right angled triangle.
Hence, $\cos 30=\frac{15}{l}$
$l=\frac{15}{\cos {30}^{\circ }}$
$l=15\sec {30}^{\circ }$ m