12. Due to a heavy storm, a part of a banyan tree broke without separating from the main. The top of the tree touched the ground$15m$ from the base making an angle of ${{45}^{\circ }}$with the ground. Calculate the height of the tree before it was broken.

Home » 12. Due to a heavy storm, a part of a banyan tree broke without separating from the main. The top of the tree touched the ground from the base making an angle of with the ground. Calculate the height of the tree before it was broken.

12. Due to a heavy storm, a part of a banyan tree broke without separating from the main. The top of the tree touched the ground $15m$ from the base making an angle of ${{45}^{\circ }}$ with the ground. Calculate the height of the tree before it was broken.

Class 10 | Frank | Heights and Distances | ICSE | Maths

12. Due to a heavy storm, a part of a banyan tree broke without separating from the main. The top of the tree touched the ground $15m$ from the base making an angle of ${{45}^{\circ }}$ with the ground. Calculate the height of the tree before it was broken.

According to the question,

Let us assume that PR was original tree and due to storm it was broken into two parts, the broken part P^’Q is making ${{45}^{\circ }}$ with ground.

In ΔP^’QR,

$\tan {{45}^{\circ }}=\frac{QR}{P'R}$

As we know, $\tan {{45}^{\circ }}=1$

$1=\frac{QR}{15}$

$QR=15$

QR = 15

Then,

$\frac{P'R}{P'Q}=\cos {{45}^{\circ }}$

$\frac{15}{P'Q}=\frac{1}{\sqrt{2}}$

$P'Q=15\sqrt{2}$

Therefore, height of tree $=QR=P'Q$

$=15+15\sqrt{2}$

$=15\left( 1+\sqrt{2} \right)$

$=15\times 2.414$

$=36.21m$

Hence, the height of whole tree is

$36.21m$

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