The diagram shows the positions of four checkpoints W, X, Y and Z in a jungle reserve area. The checkpoints are connected by the irregular tracks shown. X is due north of Y and is on a bearing of $077^\circ$ from W. Angle $XWY = 53^\circ$ and $WY$ is 60 metres.

1. Find the angle at X in triangle WXY using the bearing information.

Since $X$ is due north of $Y$, the direction from $X$ to $Y$ is due south, bearing $180^\circ$.

The bearing of $X$ from $W$ is $077^\circ$, so the bearing of $W$ from $X$ is
$077^\circ+180^\circ=257^\circ.$

Hence
$\angle WXY=257^\circ-180^\circ=77^\circ.$

2. Use the sine rule to calculate XY.

In $\triangle WXY$,
$\angle XWY=53^\circ,$
$WY=60$ m, and $WY$ is opposite $\angle WXY=77^\circ$.

By the sine rule,
$\frac{XY}{\sin 53^\circ}=\frac{60}{\sin 77^\circ}.$

So
$XY=\frac{60\sin 53^\circ}{\sin 77^\circ}=49.17\ldots$

$XY\approx 49.2\text{ m}.$

1. Find the reverse bearing from $Y$ to $Z$.

The bearing of $Y$ from $Z$ is $152^\circ$.

The bearing of $Z$ from $Y$ is the reverse bearing:

$152^\circ + 180^\circ = 332^\circ$

1. Find the smallest direction from $Y$ that must be included in the camera view.

From $W$, the bearing of $X$ is $077^\circ$. Since $\angle XWY=53^\circ$, the bearing of $Y$ from $W$ is

$077^\circ+53^\circ=130^\circ.$

Hence the reverse bearing, from $Y$ to $W$, is

$130^\circ+180^\circ=310^\circ.$

From part (ii), the bearing of $Z$ from $Y$ is $332^\circ$.

Since $310^\circ<332^\circ$ and the difference is

$332^\circ-310^\circ=22^\circ<50^\circ,$

the camera can cover both directions if the lower boundary line $CP$ is pointed at $310^\circ$. This is the minimum possible bearing.