A 1.5m tall boy is standing at some distance from a 32m 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 knowns and set up the diagram

Let $PQ$ be the height of the building and $RS$ be the height of the boy. The boy's eye level is $RS = 1.5 \text{ m}$ from the ground. We need to find the height of the building from the boy's eye level, which is $PT$. We subtract the boy's height from the building's height to get $PT = 32 - 1.5 = 30.5 \text{ m}$.

Step 2: Use trigonometry in the first triangle

Consider the right-angled triangle $\triangle PTU$, where $PT$ is the height of the building from the boy's eye level, and $TU$ is the initial distance from the boy to the building. The angle of elevation is $30^{\circ}$. We use the tangent function, $\tan \theta = \frac{\text{opposite}}{\text{adjacent}}$, to find $TU$.

Step 3: Use trigonometry in the second triangle

Next, consider the right-angled triangle $\triangle PTW$, where $PT$ is the same height, and $TW$ is the distance from the boy to the building after he walks closer. The new angle of elevation is $60^{\circ}$. Again, we use the tangent function to find $TW$.

Step 4: Calculate the distance walked

The distance the boy walked towards the building is the difference between his initial distance from the building ($TU$) and his final distance from the building ($TW$). We subtract $TW$ from $TU$ to find $UW$.

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