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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 $$=BC+CD=300+173.2=473.2\;m$$

1348326_1231434_ans_f65601ef0f08494c8235b8b1ee1245bd.PNG

Mathematics

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