Question

Two trees A and B are on the same side of a river. From a point C in the river, the distance of trees A and B are 250 m and 300 m respectively. If $\angle C={45}^{\circ },$ find the distance between the trees (Use $\sqrt{2}$ = 144.)

Answer 1

Let A and B be the trees and C be a point in the river.

Then, CA = 250 m, CB = 300 m and $\angle ACB={45}^{\circ }$.

Let AB = x metres.

Applying costine formula on $\Delta ACB,$ we get

$CosC\frac{a^2+b^2-c^2}{2ab}$

$\Rightarrow cos{45}^{\circ }=\frac{(300{)}^2+(250{)}^2-x^2}{2\times 300\times 250}$

$\Rightarrow \frac{1}{\sqrt{2}}=\frac{90000+62500-x^2}{150000}$

$\Rightarrow \frac{152500-x^2}{150000}=\frac{1}{\sqrt{2}}\times \frac{\sqrt{2}}{\sqrt{2}}\times \frac{\sqrt{2}}{2}$

$\Rightarrow 152500-x^2=75000\times \sqrt{2}=75000\times 1.44=108000$

$\Rightarrow x^2=152500-108000=44500$

$\Rightarrow x=\sqrt{44500}=210.95m.$