1. Tardigrade
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  4. Two trees, A and B are on the same side of a river. From a point C in the river the distance of the trees A and B is 250 m and 300 m, respectively. If the angle C is 45°. Then, the distance between the trees (use √2=1.44 ) is approximately

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Q. Two trees, $A$ and $B$ are on the same side of a river. From a point $C$ in the river the distance of the trees $A$ and $B$ is $250\, m$ and $300\, m$, respectively. If the angle $C$ is $45^{\circ}$. Then, the distance between the trees (use $\sqrt{2}=1.44$ ) is approximately

Trigonometric Functions

Solution:

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Let $A$ and $B$ be the trees and $C$ be a point in the river. Then, $C A=250\, m , C B=300\, m$ and $\angle A C B=45^{\circ}$
Let $A B=x m$
Applying cosine formula on $\triangle A C B$, we get,
$\cos C=\frac{a^2+b^2-c^2}{2 a b}$
$\Rightarrow \cos 45^{\circ}=\frac{(300)^2+(250)^2-x^2}{2 \times 300 \times 250}$
$\Rightarrow \frac{1}{\sqrt{2}}=\frac{90000+62500-x^2}{150000}$
$\Rightarrow \frac{152500-x^2}{150000}=\frac{1}{\sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}}=\frac{\sqrt{2}}{2}$
$\Rightarrow 152500-x^2=75000 \times \sqrt{2}=75000 \times 1.44=108000$
$\Rightarrow x^2=152500-108000=44500$
$\Rightarrow x=\sqrt{44500}=210.95 \,m$
$\approx 210 \,m$ (approx.)