Question

# An aeroplane is flying at a height of $$300\ m$$ above the ground. Flying at this height, the angles of depression from the aeroplane of two points on both banks of a river in opposite directions are $${ 45 }^{ o }$$ and $${ 60 }^{ o }$$ respectively. Find width of the river. $$\left[ \mathrm{Use}\ \sqrt { 3 } =1.732 \right]$$

Solution

## Let $$A$$ be the aeroplane $$BD$$ is the length of the river$$\tan{45}^{o}=\cfrac{AC}{BC}$$$$1=\cfrac{300}{BC}$$$$BC=300\;m$$$$\tan {60}^{o}=\cfrac{AC}{CD}$$$$\sqrt{3}=\cfrac{300}{CD}$$$$CD=\cfrac{300}{\sqrt{3}}\times \cfrac{\sqrt{3}}{\sqrt{3}}$$$$=\cfrac{300\sqrt{3}}{3}=100\sqrt{3}=1.732\times 100=173.2$$width of river $$=BC+CD=300+173.2=473.2\;m$$Mathematics

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