Height and Distance Problem Solution

This problem involves trigonometry to find the distance walked by a boy towards a building.

Effective Height Calculation

• Building Height = \(12 \, \text{m}\)
• Boy's Height = \(1.8 \, \text{m}\)
• The relevant height for calculations is the difference: Height \((h) = 12 \, \text{m} - 1.8 \, \text{m} = 10.2 \, \text{m}\).

Trigonometric Setup

Let the initial distance from the boy to the building be \(x\) meters and the final distance be \(y\) meters. The distance walked is \(d = x - y\). We use the tangent function (\(\tan\)) which relates the angle of elevation to the opposite side (height) and adjacent side (distance).

• Initial Position: The angle of elevation is \(45^\circ\).
  \(\tan(45^\circ) = \frac{\text{Height}}{\text{Initial Distance}} = \frac{h}{x}\)
• Final Position: The angle of elevation is \(60^\circ\).
  \(\tan(60^\circ) = \frac{\text{Height}}{\text{Final Distance}} = \frac{h}{y}\)

Calculating Distances

• From the initial position:
  \(1 = \frac{10.2}{x}\)
  \(x = 10.2 \, \text{m}\)
• From the final position:
  \(\sqrt{3} = \frac{10.2}{y}\)
  \(y = \frac{10.2}{\sqrt{3}} \, \text{m}\)

Distance Walked Calculation

The distance the boy walked is the difference between the initial and final distances (\(d = x - y\)).

• \(d = 10.2 - \frac{10.2}{\sqrt{3}}\)
  Factor out $10.2$:
  \(d = 10.2 \left( 1 - \frac{1}{\sqrt{3}} \right)\)
  Combine terms inside the parenthesis:
  \(d = 10.2 \left( \frac{\sqrt{3} - 1}{\sqrt{3}} \right)\)
  Express $10.2$ as a fraction: \(10.2 = \frac{102}{10} = \frac{51}{5}\).
  Substitute the fraction back into the equation:
  \(d = \frac{51}{5} \left( \frac{\sqrt{3} - 1}{\sqrt{3}} \right)\)
  \(d = \frac{51(\sqrt{3} - 1)}{5\sqrt{3}} \, \text{m}\)

This result matches Option B.