The diagram represents a flat plot of land PQRS.

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

In $\triangle PQR$, $PR=75$, $RQ=90$ and $PQ=40$.

Using the cosine rule,
$40^2=75^2+90^2-2(75)(90)\cos \angle PRQ$
$\cos \angle PRQ=\frac{75^2+90^2-40^2}{2(75)(90)}$
$\angle PRQ=\cos^{-1}\left(\frac{12125}{13500}\right)=26.1^\circ\text{ approximately}$

2. Find and add the areas of $\triangle PQR$ and $\triangle PRS$.

$\text{Area of }\triangle PQR=\frac12(75)(90)\sin 26.1^\circ=1484\text{ m}^2\text{ approximately}$

For $\triangle PRS$, $PR=75$, $RS=60$ and $\angle PRS=70^\circ$.
$\text{Area of }\triangle PRS=\frac12(75)(60)\sin70^\circ=2114\text{ m}^2\text{ approximately}$

Hence,
$\text{Area of }PQRS=1484+2114=3598\text{ m}^2$
$\text{Area of }PQRS\approx 3600\text{ m}^2\quad \text{(3 s.f.)}$

1. Identify the triangle and apply the cosine rule to find $PS$.

In $\triangle PRS$, $PR=75$, $RS=60$ and $\angle PRS=70^\circ$.

Using the cosine rule,
$PS^2=PR^2+RS^2-2(PR)(RS)\cos \angle PRS$
$PS^2=75^2+60^2-2(75)(60)\cos70^\circ$

2. Evaluate and give the length with units.

$PS^2=5625+3600-9000\cos70^\circ$
$PS^2=6146.818\ldots$
$PS=\sqrt{6146.818\ldots}=78.4\ldots$

Therefore, $PS=78.4\text{ m}$

1. Set up the sine rule in triangle $PRS$.

In $\triangle PRS$, $\angle PRS=70^\circ$, $PR=75$ and $PS=78.4$.

Using the sine rule,
$\frac{\sin \angle PSR}{PR}=\frac{\sin 70^\circ}{PS}$
$\frac{\sin \angle PSR}{75}=\frac{\sin 70^\circ}{78.4}$

2. Calculate the angle $PSR$.

$\sin \angle PSR=\frac{75\sin 70^\circ}{78.4}$
$\angle PSR=\sin^{-1}\left(\frac{75\sin 70^\circ}{78.4}\right)$
$\angle PSR=64.0^\circ\text{, to 3 s.f.}$

1. Find the perpendicular distance from S to PR using sine.

Let the shortest distance from $S$ to $PR$ be $d$.

Since the shortest distance is perpendicular to $PR$,
$\sin 70^\circ=\frac{d}{60}$
$d=60\sin70^\circ=56.4\text{ m}$

1. Use the tangent ratio to find the vertical height of the tower.

In right-angled triangle $RST$,

$\tan 10^\circ=\frac{ST}{RS}$

Since $RS=60$,

$ST=60\tan 10^\circ=10.5796\ldots$

$ST\approx 10.6\text{ m}$

1. Use the shortest horizontal distance from $S$ to $PR$ and apply the tangent ratio.

The greatest angle of elevation occurs at the point on $PR$ nearest to $S$.

From part (a)(iv), the shortest distance from $S$ to $PR$ is $56.4\text{ m}$.

From part (b)(i), the height of the tower is $10.6\text{ m}$.

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

$\tan \theta=\frac{10.6}{56.4}$

$\theta=\tan^{-1}\left(\frac{10.6}{56.4}\right)$

$\theta=10.6^\circ\text{, to 1 d.p.}$