A boy 1.8m tall casts a shadow 2.4m long. Find the distance from the tip of the shadow to the top of his head and the angle of elevation of the sun.

$$
PQ^2 = PR^2 + RQ^2
$$

$$
PQ^2 = (1.8 \text{ m})^2 + (2.4 \text{ m})^2
$$
$$
PQ^2 = 3.24 \text{ m}^2 + 5.76 \text{ m}^2
$$
$$
PQ^2 = 9.00 \text{ m}^2
$$

$$
PQ = \sqrt{9.00 \text{ m}^2}
$$
$$
PQ = 3.0 \text{ m}
$$
The distance PQ is $\boxed{3.0 \text{ m}}$.

$$
\tan \theta = \frac{\text{opposite}}{\text{adjacent}} = \frac{PR}{RQ}
$$

$$
\tan \theta = \frac{1.8 \text{ m}}{2.4 \text{ m}}
$$
$$
\tan \theta = 0.75
$$

$$
\theta = \arctan(0.75)
$$
$$
\theta \approx 36.8698^\circ
$$

$$
\theta \approx 37^\circ
$$
The angle of elevation is $\boxed{37^\circ}$.