Question 5 of 12intermediate🔧 ApplyNumerical4 marks

A 1.5 m tall boy is standing at some distance from a 30 m tall building. The angle of elevation from his eyes to the top of the building increases from 30° to 60° as he walks towards the building. Find the distance he walked towards the building.

Correct Answer

The distance walked towards the building is 193\sqrt{3} m.

Exercise: EXERCISE 9.1 | Q: 6 | (Page Page 9)
For More Understanding

Explanation

This is a classic heights and distances problem from NCERT Class 10 Mathematics. The key insight is that the effective height from the boy's eyes to the top of the building remains constant (30 - 1.5 = 28.5 m). We need to find the difference between the initial and final distances from the building using tangent ratios at two different angles of elevation.

Solution Steps

  1. Step 1: Find the effective height from the boy's eyes to the top of the building. Effective height = 30 - 1.5 = 28.5 m

  2. Step 2: Let the initial distance of the boy from the building be x m. At angle of elevation 30°: tan 30° = 28.5/x 1/3\sqrt{3} = 28.5/x Therefore, x = 28.53\sqrt{3} m

  3. Step 3: Let the final distance of the boy from the building be y m. At angle of elevation 60°: tan 60° = 28.5/y 3\sqrt{3} = 28.5/y Therefore, y = 28.5/3\sqrt{3} m

  4. Step 4: Find the distance walked towards the building. Distance walked = Initial distance - Final distance = 28.53\sqrt{3} - 28.5/3\sqrt{3} = 28.5(3\sqrt{3} - 1/3\sqrt{3}) = 28.5 × (3-1)/3\sqrt{3} = 28.5×228.5 \times 2/3\sqrt{3} = 57/3\sqrt{3} = 193\sqrt{3} m

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