该图显示了水平地面上的四个点 P、Q、R 和 S。
Q 位于 P 的正东方向。
角 SPQ = 40°，角 PSR = 15°。
PQ = 30 m，PR = 35 m，RS = 28 m。

1. Find angle $SPR$ in triangle $PRS$ using the sine rule.

In $\triangle PRS$, $PR=35$ is opposite $\angle PSR=15^\circ$, and $RS=28$ is opposite $\angle SPR$.

$\frac{\sin \angle SPR}{28}=\frac{\sin 15^\circ}{35}$

$\sin \angle SPR=\frac{28\sin15^\circ}{35}$

$\angle SPR=11.95^\circ \text{ approximately}$

2. Find the angle of $PR$ above the east direction $PQ$.

Since $\angle SPQ=40^\circ$ and $\angle SPR\approx 11.95^\circ$,

$\angle QPR=40^\circ-11.95^\circ=28.05^\circ$

So $PR$ is $28.05^\circ$ north of east.

3. Convert the direction from east into a three-figure bearing from north.

Bearings are measured clockwise from north. Since east has bearing $090^\circ$,

$\text{bearing of }R\text{ from }P=90^\circ-28.05^\circ=61.95^\circ$

To the nearest degree, this is $062^\circ$.

1. Find the included angle $QPR$ in triangle $PQR$.

From part (a)(i), the bearing of $R$ from $P$ is $062^\circ$.

Since $PQ$ is due east of $P$, the direction of $PQ$ has bearing $090^\circ$.

So
$\angle QPR = 90^\circ - 62^\circ = 28^\circ.$

2. Apply the cosine rule to find $QR$.

In triangle $PQR$,
$QR^2 = PQ^2 + PR^2 - 2(PQ)(PR)\cos \angle QPR.$

$QR^2 = 30^2 + 35^2 - 2(30)(35)\cos 28^\circ$

$QR = 16.4\ldots$

Using the unrounded angle from part (a)(i) gives
$QR \approx 16.5.$

1. Find the included angle between $PR$ and $RS$.

From the earlier working in part (a)(i),
$\angle SPR \approx 11.95^\circ$.

In triangle $PRS$,
$\angle PRS=180^\circ-15^\circ-11.95^\circ=153.05^\circ.$

2. Calculate the area of triangle $PRS$ using two sides and the included angle.

$\text{Area of }\triangle PRS=\frac12(35)(28)\sin153.05^\circ$
$=222.1\ldots$
$\approx 222\text{ m}^2$

1. Identify the right triangle and choose the tangent ratio.

Let the height of the mast be $h$ m.

In the vertical right triangle formed by $P$, $R$ and the top of the mast,

$\tan 18^\circ = \frac{h}{35}$

2. Calculate the height of the mast.

$h = 35\tan 18^\circ$

$h = 11.371\ldots$

$h \approx 11.4\text{ m}$

1. Identify the point on $PQ$ produced giving the largest angle of depression.

The angle of depression is largest when the horizontal distance from $R$ to the point on $PQ$ produced is smallest.

From part (a)(ii), the shortest horizontal distance from $R$ to $PQ$ produced is
$QR = 16.5\text{ m}.$

2. Use the tangent ratio to find the angle of depression.

Let the largest angle of depression be $\theta$.

The height of the mast from part (b)(i) is $11.4\text{ m}$.

$\tan \theta = \frac{11.4}{16.5}$

$\theta = \tan^{-1}\left(\frac{11.4}{16.5}\right)$

$\theta = 34.6^\circ$

3. Round the angle appropriately.

$\theta \approx 35^\circ$

So the largest angle of depression is $35^\circ$.