CUET UG 2025 — General Test Geometry

Let $h$h = height of the tower

Let $x$x = length of shadow when Sun's altitude = 60°

Let $x + 40$x+40 = length of shadow when Sun's altitude = 30°

The shadow is 40 m longer when the Sun is at 30° compared to when it's at 60°.

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When the Sun is at 60° altitude:

$$\tan(60°) = \frac{h}{x}$$tan(60°)=xh​

$$\sqrt{3} = \frac{h}{x}$$3​=xh​

$x = \frac{h}{\sqrt{3}}$x=3​h​ ... (Equation 1)

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When the Sun is at 30° altitude:

$$\tan(30°) = \frac{h}{x + 40}$$tan(30°)=x+40h​

$$\frac{1}{\sqrt{3}} = \frac{h}{x + 40}$$3​1​=x+40h​

$x + 40 = h\sqrt{3}$x+40=h3​ ... (Equation 2)

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Substituting Equation 1 into Equation 2:

$$\frac{h}{\sqrt{3}} + 40 = h\sqrt{3}$$3​h​+40=h3​

$$40 = h\sqrt{3} - \frac{h}{\sqrt{3}}$$40=h3​−3​h​

$$40 = h\left(\sqrt{3} - \frac{1}{\sqrt{3}}\right)$$40=h(3​−3​1​)

$$40 = h\left(\frac{3}{\sqrt{3}} - \frac{1}{\sqrt{3}}\right)$$40=h(3​3​−3​1​)

$$40 = h\left(\frac{2}{\sqrt{3}}\right)$$40=h(3​2​)

$$40 = \frac{2h}{\sqrt{3}}$$40=3​2h​

$$40\sqrt{3} = 2h$$403​=2h

$h = 20\sqrt{3}$h=203​ m

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Therefore, the height of the tower is $20\sqrt{3}$203​ m.