P, Q, R and S are points on a triangular garden. S R Q is a straight line. It is given that $PQ = 55\ \text{m}$, $RQ = 40\ \text{m}$, $PR = 35\ \text{m}$ and $\angle SPR = 50^\circ$.

1. Find the clockwise bearing of $S$ from $P$ using the given angle from the south line

Since $R$ is due south of $P$, the bearing of $R$ from $P$ is $180^\circ$.

Given $\angle SPR = 50^\circ$, $PS$ is $50^\circ$ west of the south line $PR$.

Therefore, bearing of $S$ from $P$ is

$180^\circ + 50^\circ = 230^\circ.$

1. Identify the side opposite the required angle and apply the cosine rule

In $\triangle PQR$, the angle required is $\angle PRQ$.

The side opposite $\angle PRQ$ is $PQ=55$ m.

Using the cosine rule,
$PQ^2=PR^2+RQ^2-2(PR)(RQ)\cos \angle PRQ.$

2. Substitute the three side lengths and solve for the cosine of the angle

$55^2=35^2+40^2-2(35)(40)\cos \angle PRQ$
$3025=1225+1600-2800\cos \angle PRQ$
$3025=2825-2800\cos \angle PRQ$
$2800\cos \angle PRQ=2825-3025=-200$
$\cos \angle PRQ=-\frac{200}{2800}=-\frac{1}{14}.$

3. Find the angle using inverse cosine and round appropriately

$\angle PRQ=\cos^{-1}\left(-\frac{1}{14}\right)$
$\angle PRQ=94.096\ldots^\circ$

So, to 1 decimal place,
$\angle PRQ=94.1^\circ.$

1. Find the sine of the included angle between $PR$ and $RQ$

For $\angle PRQ$, using the cosine rule,
$\cos \angle PRQ=\frac{35^2+40^2-55^2}{2(35)(40)}=-\frac{1}{14}.$
Hence
$\sin \angle PRQ=\sqrt{1-\left(-\frac{1}{14}\right)^2}=\frac{\sqrt{195}}{14}.$

2. Apply the triangle area formula

$\text{Area of }\triangle PQR=\frac12(35)(40)\sin \angle PRQ$
$=700\left(\frac{\sqrt{195}}{14}\right)$
$=698.2\ldots$
So the area is $698\ \text{m}^2$ to 3 significant figures.

1. Recognise the shortest pathway as the perpendicular height from $R$ to $PQ$

The shortest pathway from $R$ to $PQ$ is perpendicular to $PQ$.

Let this length be $h$ m.

2. Use the area of triangle formula to find the perpendicular height

Using the area from part (a)(iii),
$\text{Area of }\triangle PQR = \frac12 \times PQ \times h$
$698 = \frac12 \times 55 \times h$
$h = \frac{2\times 698}{55}$
$h = 25.3818\ldots$
$h \approx 25.4$

1. Identify the closest viewing point on $PQ$ to maximise the angle of elevation

The greatest angle of elevation occurs when the horizontal distance from the viewer to $R$ is the shortest.

From part (b), the shortest distance from $R$ to $PQ$ is
$25.4\text{ m}.$

2. Use the tangent ratio to find the greatest angle of elevation

Let the greatest angle of elevation be $\theta$.

The tree height is $4\text{ m}$ and the horizontal distance is $25.4\text{ m}$.

$\tan \theta = \frac{4}{25.4}$

$\theta = \tan^{-1}\left(\frac{4}{25.4}\right)$

$\theta = 8.95^\circ$

So, to 1 decimal place,
$\theta = 9.0^\circ.$