Question

A tree $AC$ is broken over by over by wind from $B.D$ is the point where the top of the broken tree touches the ground and $BD$ makes an angle of ${45}^{\circ }$ with the ground, if the distance between he base of the tree and the point $D=10$ m. What is the height of the tree ?

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

Answer 1

The correct option is B $10(1+\sqrt{2})$ m

Let the tree $AC$ be broken from point $B$.

Let $BC=x$ metres and $AB=y$ metres.

Then $AB=BD$ ....(broken part)
$\angle BDC={45}^{\circ }$
In $\Delta BDC$, we have
$\tan {45}^{\circ }=\frac{BC}{DC}=\frac{x}{10}$

$\Rightarrow 1=\frac{x}{10}\Rightarrow x=10m$
$\cos {45}^{\circ }=\frac{DC}{BD}=\frac{10}{y}$

$\Rightarrow \frac{1}{\sqrt{2}}=\frac{10}{y}$

$\Rightarrow y=10\sqrt{2}$ m
$\therefore$ Height of the tree $=x+y=(10+10\sqrt{2})$ m $=10(1+\sqrt{2})$ m