CUET UG 2025 — General Test Geometry

Two towers P and Q have a point O exactly midway between their bases. From point O, the angle of elevation to the top of tower P is 30° and to the top of tower Q is 60°.

Let $d$d = distance from point O to the foot of each tower

Let $h_1$h1​ = height of tower P

Let $h_2$h2​ = height of tower Q

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For tower P, using the tangent ratio:

$$\tan(30°) = \frac{h_1}{d}$$tan(30°)=dh1​​

Since $\tan(30°) = \frac{1}{\sqrt{3}}$tan(30°)=3​1​:

$$\frac{1}{\sqrt{3}} = \frac{h_1}{d}$$3​1​=dh1​​

$$h_1 = \frac{d}{\sqrt{3}}$$h1​=3​d​

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For tower Q, using the tangent ratio:

$$\tan(60°) = \frac{h_2}{d}$$tan(60°)=dh2​​

Since $\tan(60°) = \sqrt{3}$tan(60°)=3​:

$$\sqrt{3} = \frac{h_2}{d}$$3​=dh2​​

$$h_2 = d\sqrt{3}$$h2​=d3​

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The ratio of heights is:

$$h_1 : h_2 = \frac{d}{\sqrt{3}} : d\sqrt{3}$$h1​:h2​=3​d​:d3​

Dividing both sides by $d$d:

$$= \frac{1}{\sqrt{3}} : \sqrt{3}$$=3​1​:3​

Multiplying both sides by $\sqrt{3}$3​:

$$= 1 : 3$$=1:3

Therefore, the ratio of the heights of tower P to tower Q is $1:3$1:3.