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.