Question

A portion of a 30 m long tree is broken by a tornado and the top strikes the ground making an angle of $30^\circ$ with the ground level. The height of the point where the tree is broken is equal to

• A. $\frac{30}{\sqrt{3}}$ m
• B. 10 m
• C. $30\sqrt{3}$ m
• D. 60 m

Hint:

Use trigonometric ratios to break down height-distance relationships in inclination problems.

Answer 1

The Correct Option is B

Let the unbroken part of the tree be $h$ meters, and let the broken portion that touches the ground be the hypotenuse of a right-angled triangle with length 30 - h meters.
Since the top of the broken part touches the ground at an angle of $30^\circ$, the horizontal projection of the broken part is: $$(30 - h) \cos 30^\circ$$ And the vertical projection of the broken part is: $$(30 - h) \sin 30^\circ$$ Since the total height is given as $h$, the equation for the vertical component is: $$h = (30 - h) \sin 30^\circ$$ Since $\sin 30^\circ = \frac{1}{2}$: $$h = \frac{1}{2} (30 - h)$$ $$2h = 30 - h$$ $$3h = 30$$ $$h = 10 \text{ m}$$ Thus, the height of the break point is 10 m.