A $1.5 \text{ m}$ tall boy is standing at some distance from a $30 \text{ m}$ tall building. The angle of elevation from his

Step 1: Understand the Geometry and Set up Variables

First, let's visualize the situation. The boy is $1.5 \text{ m}$ tall, and the building is $30 \text{ m}$ tall. Since the angle of elevation is measured from the boy's eyes, we need to consider the effective height of the building above his eye level. This is the total height of the building minus the boy's height.

Step 2: Calculate Effective Height

The effective height of the building from the boy's eye level is $28.5 \text{ m}$. Let $x$ be the initial distance of the boy from the building and $y$ be the distance he walked towards the building. So, the final distance from the building will be $x-y$.

Step 3: Apply Tangent for Initial Position

When the angle of elevation is $30^\circ$, the boy is at an initial distance $x$ from the building. We can use the tangent function, which relates the opposite side (effective height) to the adjacent side (distance from the building). We know that $\tan(30^\circ) = \frac{1}{\sqrt{3}}$.

Step 4: Apply Tangent for Final Position

As the boy walks $y$ meters towards the building, his new distance from the building is $x-y$. The angle of elevation increases to $60^\circ$. We apply the tangent function again. We know that $\tan(60^\circ) = \sqrt{3}$.

Step 5: Calculate the Distance Walked

To find the distance the boy walked, $y$, we subtract the final distance from the initial distance. We substitute the values we found for $x$ and $x-y$ and simplify. We can also approximate the value using $\sqrt{3} \approx 1.732$.

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