Question

A tree is broken by wind, its upper part touches the ground at a point 10 metre from the foot of the tree and makes an angle of 45° with the ground. Then what is the entire height of the tree?

• A. \(15 \ m\)
• B. \(20 \ m\)
• C. \(10(1+\sqrt 2)\ m\)
• D. \(10(1+\frac {\sqrt 3}{2})\ m\)

Answer 1

The Correct Option is C

Let the height of the tree be $h$. 

The tree breaks at some height $x$ from the ground. 

The broken part touches the ground at a distance of 10 meters from the foot of the tree and makes an angle of $45^{\circ}$ with the ground. 

Let the length of the broken part be $y$. Then $y = h - x$. 

We have a right-angled triangle with base 10 meters, height $x$, and hypotenuse $y$. 

The angle between the base and the hypotenuse is $45^{\circ}$. 

Then, we have: $\cos 45^{\circ} = \frac{\text{base}}{\text{hypotenuse}} = \frac{10}{y}\implies y = \frac{10}{\cos 45^{\circ}} = \frac{10}{\frac{1}{\sqrt{2}}} = 10\sqrt{2}$ $\tan 45^{\circ} = \frac{\text{height}}{\text{base}} = \frac{x}{10}\implies x = 10 \tan 45^{\circ} = 10 \cdot 1 = 10$ 

The entire height of the tree is $h = x + y = 10 + 10\sqrt{2} = 10(1+\sqrt{2})$ meters.