CUET UG 2025 — General Test Geometry

Consider a tower with two observation points along the same straight line from its base:

• Point A: 5 meters from the base
• Point B: 20 meters from the base

The angles of elevation from these points are complementary (they add up to 90°).

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Let the height of the tower be $h$h meters.

Let the angle of elevation from Point A be $\alpha$α.

Since the angles are complementary, the angle of elevation from Point B is $(90° - \alpha)$(90°−α).

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From Point A (5m away):

$$\tan(\alpha) = \frac{h}{5}$$tan(α)=5h​

From Point B (20m away):

$$\tan(90° - \alpha) = \frac{h}{20}$$tan(90°−α)=20h​

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Using the complementary angle property: $\tan(90° - \alpha) = \cot(\alpha) = \frac{1}{\tan(\alpha)}$tan(90°−α)=cot(α)=tan(α)1​

Therefore:

$$\frac{1}{\tan(\alpha)} = \frac{h}{20}$$tan(α)1​=20h​

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Since $\tan(\alpha) = \frac{h}{5}$tan(α)=5h​:

$$\frac{1}{\frac{h}{5}} = \frac{h}{20}$$5h​1​=20h​

$$\frac{5}{h} = \frac{h}{20}$$h5​=20h​

$$5 \times 20 = h^2$$5×20=h2

$$100 = h^2$$100=h2

$$h = 10$$h=10

Therefore, the height of the tower is 10 meters.