A tree is broken by wind and its upper part touches the ground at a point $10$ metres from the foot of the tree and makes an angle of ${45}^{\circ }$ with the ground. The entire length of the tree is

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

Let the entire length of the tree be h metres and let the vertical part of the tree still standing be y metres.

So, the length of the tree inclined on the ground is $(h-y)m.$

In triangle ABC, we have

$\tan {45}^{\circ }=\frac{AB}{BC}$

$1=\frac{y}{10}$

$\Rightarrow y=10m$

Also, $\cos {45}^{\circ }=\frac{BC}{AC}$

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

$\Rightarrow h-y=10\sqrt{2}$

$\Rightarrow h=10\sqrt{2}+10=10(\sqrt{2}+1)$

Therefore, the length of the tree is $10(\sqrt{2}+1)m$.