W、X、Y 和 Z 是水平地面上的点，PW 是一根垂直的旗杆。$WX = 7\ \text{m}$，$WZ = 3\ \text{m}$，$XY = 10\ \text{m}$，$\angle WYZ = 20^\circ$ 且 $\angle WZY = 28.6^\circ$。

1. Find the third angle in triangle WZY.

In $\triangle WZY$,
$\angle ZWY = 180^\circ - 20^\circ - 28.6^\circ = 131.4^\circ.$

2. Apply the sine rule to find WY.

By the sine rule,
$\frac{WY}{\sin 28.6^\circ}=\frac{WZ}{\sin 20^\circ}.$
Since $WZ=3$,
$WY=\frac{3\sin 28.6^\circ}{\sin 20^\circ}=4.197\ldots$
$WY\approx 4.20\text{ m}.$

1. Find the bearing of Y from Z using the reverse bearing.

The bearing of $Z$ from $Y$ is $308^\circ$.

Therefore the bearing of $Y$ from $Z$ is
$308^\circ - 180^\circ = 128^\circ.$

2. Use the angle at Z to find the bearing of W from Z.

At $Z$,
$\angle WZY = 28.6^\circ.$

From the diagram, $ZW$ is clockwise from $ZY$, so
$\text{bearing of }W\text{ from }Z = 128^\circ + 28.6^\circ = 156.6^\circ.$

To the nearest degree, this is
$157^\circ.$

1. Use Pythagoras' theorem to find the vertical height PW.

Since $PW$ is vertical and $WX$ is on horizontal ground, $\angle PWX = 90^\circ$.

In right-angled triangle $PWX$,
$PX^2 = PW^2 + WX^2$
$8^2 = PW^2 + 7^2$
$PW^2 = 64 - 49 = 15$
$PW = \sqrt{15} = 3.87\text{ m} \text{ correct to 3 s.f.}$

1. Identify the position of T that gives the greatest angle of elevation.

Since $PW$ is fixed, the angle of elevation $\theta$ from $T$ satisfies
$\tan \theta=\frac{PW}{TW}.$
The angle $\theta$ is greatest when $TW$ is as small as possible.

The shortest distance from $W$ to the line segment $XY$ is the perpendicular distance from $W$ to $XY$.

2. Find the shortest possible horizontal distance from W to T on XY.

Using $\angle WXY=20.2^\circ$ from part (b),
$\text{shortest }WT=7\sin 20.2^\circ.$
$WT=2.41\text{ m} \quad \text{(approximately)}$

3. Calculate the greatest angle of elevation.

Using $PW=3.87\text{ m}$ from part (d),
$\tan \theta=\frac{3.87}{2.41}.$
$\theta=\tan^{-1}\left(\frac{3.87}{2.41}\right)$
$\theta=58.0^\circ \quad \text{(to 1 decimal place)}$