Question

A tree breaks due to storm and the broken part bends so that the top of the tree first touches the ground, making an angle of 30 with the horizontal. The distance from the foot of the tree to the point where the top touches the ground is 10 m. The height of the tree is

• A. 10(√(3)+1) m
• B. 10√(3) m
• C. 10(√(3)-1) m
• D. \(\frac{10}{√(3)}\) m

Answer 1

The Correct Option is B

In the question it is given that, \(BC = 10\ m\)

and \(∠ACB = 90^0\)

We have to find \(AB\ and\ AC\)

In the right angled triangle =\(ABC\)

\(tan(30^0) = \frac{AB}{BC}\)

\(\frac{1}{\sqrt{3}} = \frac{AB}{10}\)

\(AB = \frac{10}{\sqrt{3}}\)  – (i)

\(cos(30^0) = \frac{BC}{AC}\)

\(\frac{\sqrt{3}}{2} = \frac{10}{AC}\)

\(AC = \frac{20}{\sqrt{3}}\)  – (ii)

The Height of the tree = \(AB + AC\)

= \(\frac{10}{\sqrt{3}}+ \frac{20}{\sqrt{3}}\)

= \(\frac{30}{\sqrt{3}}\) = \(10\sqrt{3}\ m\)

The correct option is (B): 10√(3) m