Four, the height of a certain U is $\mathrm{CD}$, at point $\mathrm{A}$ due east of the mountain, the angle of elevation to the mountain peak is $60^{\circ}$. From point A, moving 300 meters in a direction 28 degrees west of south, reaching point $\mathrm{B}$, which is exactly southeast of the mountain. Find the height of the mountain $\mathrm{CD}$ (保留三隹有效数字, retain three significant figures).

附: $\sin 62^{\circ}=0.8829, \quad \cos 62^{\circ}=0.4695$, $\left.\sin 73^{\circ}=0.9563, \cos 73^{\circ}=0.2923.\right)$
Four, Solution: As shown in Figure 6, $\because \angle \mathrm{BAE}=28^{\circ}$,
$$
\therefore \angle \mathrm{CAB}=90^{\circ}-28^{\circ}=
$$
$62^{\circ}$.
$$
\begin{array}{l}
\therefore \angle \mathrm{BCA}=45^{\circ}. \\
\therefore \angle \mathrm{ABC}=180^{\circ}-45^{\circ}- \\
62^{\circ}=73^{\circ}.
\end{array}
$$
$$
\begin{array}{l}
A C= 300 \sin 73^{\circ} \\
\operatorname{in} 15^{\circ} \\
=300 \sqrt{2} \times \sin 73^{\circ} \\
=300 \sqrt{2} \times 0.956. \\
\approx 405.7.
\end{array}
$$

In $\triangle \mathrm{ACD}$,
$$
\begin{aligned}
\mathrm{CD} & =\mathrm{AC} \cdot \operatorname{tg} 60^{\circ} \\
& =405.7 \times 1.732 \\
& \approx 702.7 \approx 703.
\end{aligned}
$$

Answer: The height of CD is approximately 703 meters.