From a point on a bridge across the river, the angles of depression of the banks on opposite sides of the river are 30° and 45° respectively. If the bridge is at a height of 2.5m from the banks, then the width of the river is (take √3 = 1.732)
AC is the height of the bridge = 2.5 m
Width of river = BD = ?
In $$\triangle$$ ACD,
=> $$tan(\angle ACD)=\frac{AC}{CD}$$
=> $$tan(45^\circ)=1=\frac{2.5}{CD}$$
=> $$CD=2.5$$ m
Similarly, in $$\triangle$$ ABC,
=> $$tan(30^\circ)=\frac{2.5}{BC}$$
=> $$\frac{1}{\sqrt{3}}=\frac{2.5}{BC}$$
=> $$BC = 2.5 \times 1.732 = 4.33$$ m
$$\therefore$$ BD = BC + CD
= $$2.5 + 4.33=6.83$$ m
=> Ans - (B)
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