Introduction to Trigonometry and Its Applications Question - CBSE

Question

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

Accepted Answer

Height $h = 30 - 1.5 = 28.5 = \dfrac{57}{2} ext{ m}$. [0.5 Mark]
In $\Delta 1$: $x_1 = 28.5 \cot 30^\circ = 28.5 \sqrt{3}$.
In $\Delta 2$: $x_2 = 28.5 \cot 60^\circ = \dfrac{28.5}{\sqrt{3}}$. [1.0 Mark]
Distance walked $= x_1 - x_2 = 28.5\left(\sqrt{3} - \dfrac{1}{\sqrt{3}} ight) = 28.5 imes \dfrac{2}{\sqrt{3}} = \dfrac{57}{\sqrt{3}} = 19\sqrt{3} ext{ m}$. [0.5 Mark]

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🎯 Official CBSE Marking Scheme:
Eye-level height adjustment $h = 28.5 ext{ m}$: 0.5 Mark
Distance expressions using $\cot 30^\circ$ and $\cot 60^\circ$: 1.0 Mark
Calculating distance walked $= 19\sqrt{3} ext{ m}$: 0.5 Mark

Question:

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