A 1.5m tall boy is standing at some distance from a 30m 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.

Step 1: Identify given values and set up the diagram

First, we identify the given heights. The building is 30 meters tall, and the boy is 1.5 meters tall. Since the angle of elevation is measured from the boy's eyes, we need to find the effective height of the building from his eye level. This is calculated by subtracting the boy's height from the building's height.

Step 2: Use trigonometry for the initial position

Let $x_1$ be the initial distance of the boy from the building. We can use the tangent function, which relates the angle of elevation to the opposite side (effective height $H$) and the adjacent side ($x_1$). We know that $\tan(30^\circ) = 1/\sqrt{3}$. Substituting the values, we find the initial distance $x_1$.

Step 3: Use trigonometry for the final position

Let $x_2$ be the final distance of the boy from the building after walking towards it. Similarly, we use the tangent function with the new angle of elevation, $60^\circ$. We know that $\tan(60^\circ) = \sqrt{3}$. Substituting the values, we find the final distance $x_2$.

Step 4: Calculate the distance walked

The distance the boy walked towards the building is the difference between his initial distance ($x_1$) and his final distance ($x_2$). We substitute the calculated values for $x_1$ and $x_2$ and simplify the expression to find the distance walked, $d$. We rationalize the denominator by multiplying the numerator and denominator by $\sqrt{3}$.

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