CUET UG 2025 — General Test Geometry

Let the current height of the tower be $h_1$h1​ and the final height be $h_2$h2​. The distance from the base is 75m.

Using the tangent ratio in the first triangle with angle of elevation 30°:

$$\tan(30°) = \frac{h_1}{75}$$tan(30°)=75h1​​

$$\frac{1}{\sqrt{3}} = \frac{h_1}{75}$$3​1​=75h1​​

$$h_1 = \frac{75}{\sqrt{3}}$$h1​=3​75​

Rationalizing:

$$h_1 = \frac{75}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}}$$h1​=3​75​×3​3​​

$$h_1 = \frac{75\sqrt{3}}{3}$$h1​=3753​​

$h_1 = 25\sqrt{3}$h1​=253​ m

---

Using the tangent ratio in the second triangle with angle of elevation 60°:

$$\tan(60°) = \frac{h_2}{75}$$tan(60°)=75h2​​

$$\sqrt{3} = \frac{h_2}{75}$$3​=75h2​​

$h_2 = 75\sqrt{3}$h2​=753​ m

---

The additional height required is the difference between the final height and current height:

Additional height $= h_2 - h_1$=h2​−h1​

$$= 75\sqrt{3} - 25\sqrt{3}$$=753​−253​

$= 50\sqrt{3}$=503​ m

Therefore, the tower must be raised by $50\sqrt{3}$503​ meters.