The diagram above shows a park $ABCD$ that is on horizontal ground, crossed by a straight path $AC$. Point $B$ is due north of $C$. $AB = 700\ \mathrm{m},\ BC = 550\ \mathrm{m}$ and $CD = 780\ \mathrm{m}$.
Angle $ABC = 115^{\circ}$ and angle $ACD = 29^{\circ}$.

1. Apply the cosine rule to triangle $ABC$ to find $AC^2$.

In $\triangle ABC$,
$AC^2=AB^2+BC^2-2(AB)(BC)\cos\angle ABC$
$AC^2=700^2+550^2-2(700)(550)\cos115^\circ$

2. Evaluate and round the length to two decimal places.

$AC=\sqrt{700^2+550^2-2(700)(550)\cos115^\circ}$
$AC=1057.318\ldots$
$AC=1057.32\ \mathrm{m}\quad \text{(to 2 d.p.)}$

1. Find angle $ACB$ in triangle $ABC$.

Using the sine rule in $\triangle ABC$,

$\frac{\sin \angle ACB}{AB}=\frac{\sin \angle ABC}{AC}$

$\frac{\sin \angle ACB}{700}=\frac{\sin 115^\circ}{1057.32}$

$\sin \angle ACB=\frac{700\sin115^\circ}{1057.32}$

$\angle ACB=36.9^\circ \text{ approximately}$

2. Use the direction of north and the given angle $ACD$ to calculate the bearing of $D$ from $C$.

Since $B$ is due north of $C$, $CB$ is the north line.

The angle from north to $CD$, measured anticlockwise, is

$\angle BCA+\angle ACD=36.9^\circ+29^\circ=65.9^\circ$

Bearings are measured clockwise from north, so

$\text{bearing of }D\text{ from }C=360^\circ-65.9^\circ=294.1^\circ$

To the nearest degree,

$\text{bearing}=294^\circ$

1. Find the area of triangle $ABC$ using the two given sides and the included angle.

$\text{Area of }\triangle ABC=\frac12(700)(550)\sin115^\circ$
$=174464.25\ldots\ \mathrm{m^2}$

2. Find the area of triangle $ACD$ and add the two triangle areas.

Using $AC=1057.32\ \mathrm{m}$ from part (a),
$\text{Area of }\triangle ACD=\frac12(1057.32)(780)\sin29^\circ$
$=199914.26\ldots\ \mathrm{m^2}$

Total area
$=174464.25\ldots+199914.26\ldots$
$=374378.51\ldots$
$\approx 374379\ \mathrm{m^2}$

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

Since the building is vertical and the ground is horizontal, the building height is perpendicular to $CD$.

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

For the angle of elevation from $C$,
$\tan 10.9^\circ=\frac{h}{780}$

2. Calculate the height of the building.

$h=780\tan 10.9^\circ$
$h=150.204\ldots$

So, to 3 significant figures,
$h=150\ \mathrm{m}$