CUET UG 2025 — General Test Geometry

Let the point on the bridge be $P$P at a height of $9$9 m above the water surface.

Let the banks on opposite sides be $A$A and $B$B.

The angles of depression to banks $A$A and $B$B are $30°$30° and $45°$45° respectively.

By the property of alternate angles, the angles of elevation from the banks to point $P$P are also $30°$30° and $45°$45°.

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Let $x$x be the horizontal distance from the point directly below $P$P to bank $A$A.

Using the tangent ratio:

$$\tan(30°) = \dfrac{9}{x}$$tan(30°)=x9​

$$\dfrac{1}{\sqrt{3}} = \dfrac{9}{x}$$3​1​=x9​

$x = 9\sqrt{3}$x=93​ m

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Let $y$y be the horizontal distance from the point directly below $P$P to bank $B$B.

Using the tangent ratio:

$$\tan(45°) = \dfrac{9}{y}$$tan(45°)=y9​

$$1 = \dfrac{9}{y}$$1=y9​

$y = 9$y=9 m

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The width of the river is the sum of both horizontal distances:

Width $= x + y$=x+y

Width $= 9\sqrt{3} + 9$=93​+9

Width $= 9(\sqrt{3} + 1)$=9(3​+1) m

Therefore, the width of the river is $9(\sqrt{3} + 1)$9(3​+1) m.