The diagram shows a parallelogram ABCD on horizontal ground where AC is a path. AB = 15 m and BC = 8 m. The bearing of B from A is 050°.

1. Find $\angle BAC$ in $\triangle ABC$ using the sine rule.

From part (b), $AC \approx 21.1$ m.

In $\triangle ABC$, $\angle ABC = 180^\circ - 50^\circ = 130^\circ$.

Using the sine rule,
$\frac{\sin \angle BAC}{BC}=\frac{\sin 130^\circ}{AC}$
$\frac{\sin \angle BAC}{8}=\frac{\sin 130^\circ}{21.1}$
$\sin \angle BAC=\frac{8\sin130^\circ}{21.1}$
$\angle BAC \approx 16.9^\circ$

2. Use the bearing angle at $A$ to find $\angle DAC$.

Since the bearing of $B$ from $A$ is $050^\circ$, and $AD$ is parallel to the north direction,
$\angle DAB=50^\circ.$

Therefore,
$\angle DAC=\angle DAB-\angle BAC$
$=50^\circ-16.9^\circ$
$=33.1^\circ$

1. Find the reverse bearing from $C$ to $A$.

From part (c), $\angle DAC = 33.1^\circ$. Since $AD$ is in the north direction, the bearing of $C$ from $A$ is $033.1^\circ$.

Therefore the bearing of $A$ from $C$ is
$033.1^\circ + 180^\circ = 213.1^\circ.$

1. Identify the right-angled triangle and choose the trigonometric ratio.

In right-angled triangle $ABT$, $AB=15$ m and $\angle BAT=15^\circ$.

Since
$\tan 15^\circ=\frac{BT}{AB},$
we use the tangent ratio.

2. Calculate the height of the pole.

$\tan 15^\circ=\frac{BT}{15}$
$BT=15\tan 15^\circ$
$BT=4.019\ldots$

So the height of the pole is $4.02$ m, to 3 significant figures.

1. Identify where the minimum angle of elevation occurs.

For any point $P$ on $AC$,
$\tan \theta=\frac{BT}{BP}.$
Since $BT$ is fixed, $\theta$ is smallest when $BP$ is largest.

Along the segment $AC$, the largest distance from $B$ is at $A$, since
$BA=15\text{ m},\qquad BC=8\text{ m}.$
The angle of elevation from $A$ is given as $15^\circ$, so
$\theta_{\min}=15^\circ.$

2. Find the shortest horizontal distance from $B$ to a point on $AC$.

The angle between $AB$ and $AC$ is
$\angle BAC=50^\circ-33.1^\circ=16.9^\circ.$

The shortest distance from $B$ to $AC$ is the perpendicular distance from $B$ to $AC$.

Using right-angled triangle with hypotenuse $AB$,
$\text{shortest distance}=15\sin 16.9^\circ.$

3. Use the pole height from the earlier part to find the maximum angle of elevation.

From part (e),
$BT=4.02\text{ m}.$

Therefore
$\tan \theta_{\max}=\frac{4.02}{15\sin 16.9^\circ}.$

$\theta_{\max}=\tan^{-1}\left(\frac{4.02}{15\sin 16.9^\circ}\right).$

$\theta_{\max}=42.7^\circ\text{ correct to 3 significant figures}.$

4. State the required range of $\theta$.

Hence the range of possible angles of elevation is
$15^\circ \leq \theta \leq 42.7^\circ.$