Question

A 15 m high eucalyptus tree standing erect on the ground breaks at the height of 5 m from it. The broken part bends such that the top of the tree touches the ground. Find the angle made by the broken part of the tree with the ground.

• A. 30${}^o$
• B. 45${}^o$
• C. 60${}^o$
• D. 90${}^o$

Answer 1

The correct option is A 30${}^o$

Based on the given information, we can draw the figure given above.

Here, $AC$ represents the tree when erect and $BD$ represents the broken part of the tree.

While $BC$ represents the remaining erect tree.

$DC$ is the distance between the top of the tree and the base of the tree.
$\therefore \angle C={90}^o,BC=5m,AC=15m$

$BD=AB=AC-BC$

$=10m$

We know that, $\sin \theta =\frac{\text{Opposite Side}}{\text{Hypotenuse}}$

Hence, $\sin \theta =\frac{BC}{BD}=\frac{5}{10}$

$=\frac{1}{2}$

We know that $\sin {30}^o=\frac{1}{2}$.

Hence, $\sin \theta =\sin {30}^o$

$\therefore \theta ={30}^o$

Hence, the broken part of the tree makes an angle of ${30}^o$ with the ground.