Question 46

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)

Solution

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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