A helicopter is flying at an altitude of $250 m$ between two banks of a river. From the helicopter it is observed that the angles of a river. From the helicopter its is observed that the angle of depression of two boats on the opposite banks are $45^{o}$ and $60^{o}$ respectively. Find the width of the river. Given your answer correct to nearest metre.

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Solution

Now,
$tan 30^{\circ} = \frac{H C}{A C} = \frac{250}{A C}; tan 45^{\circ} = \frac{H C}{B C} ∴ A C = 250 \sqrt{3} m; H C = B C = 250 m$
$∴$ Width of the river is $= | A B | = 250 \sqrt{3} + 250$
$= 250 (\sqrt{3} + 1) ≅ 683 m$

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Now,tan30∘=HCAC=250AC;tan45∘=HCBC∴AC=250√3m;HC=BC=250m∴ Width of the river is =|AB|=250√3+250=250(√3+1)≅683m

Now,
$tan 30^{\circ} = \frac{H C}{A C} = \frac{250}{A C}; tan 45^{\circ} = \frac{H C}{B C} ∴ A C = 250 \sqrt{3} m; H C = B C = 250 m$
$∴$ Width of the river is $= | A B | = 250 \sqrt{3} + 250$
$= 250 (\sqrt{3} + 1) ≅ 683 m$