The diagram shows five towns A, B, C, D and E. Towns B, C, D lie on a straight line. Town E is due east of Town D. It is given that $\angle ABC = 32^\circ$, AC = 8 km and AD = 10 km.

1. Identify the equal angle and choose the correct right-angled triangle

Since $B, C, D$ lie on a straight line, $DB$ is along the same line as $DC$.

So
$\angle ADB = \angle ADC.$

2. Apply a trigonometric ratio in right-angled triangle $ACD$

In right-angled triangle $ACD$,
$\sin \angle ADC = \frac{AC}{AD} = \frac{8}{10}.$

3. Calculate the angle

$\angle ADC = \sin^{-1}\left(\frac{8}{10}\right) = 53.130\ldots^\circ$

Therefore,
$\angle ADB = 53.1^\circ \text{ correct to 1 decimal place}.$

1. Find $BC$ using the tangent ratio in right-angled triangle $ABC$

In right-angled triangle $ABC$,

$\tan 32^\circ = \frac{AC}{BC}$

$\tan 32^\circ = \frac{8}{BC}$

$BC = \frac{8}{\tan 32^\circ} = 12.802\ldots\text{ km}$

2. Find $CD$ using Pythagoras' theorem in right-angled triangle $ACD$

In right-angled triangle $ACD$,

$AD^2 = AC^2 + CD^2$

$10^2 = 8^2 + CD^2$

$CD^2 = 100 - 64 = 36$

$CD = 6\text{ km}$

3. Add the collinear distances to find $BD$

Since $B$, $C$ and $D$ lie on a straight line,

$BD = BC + CD$

$BD = 12.802\ldots + 6 = 18.802\ldots$

$BD = 18.8\text{ km} \quad \text{(3 s.f.)}$

1. Identify the angle $\angle ADE$ using the straight line $CDE$

Since $C, D$ and $E$ lie on a straight line,
$\angle ADE = 180^\circ - \angle ADC.$

2. Use the cosine relationship for supplementary angles and the right triangle $ACD$

In right-angled triangle $ACD$,
$AD=10,\quad AC=8.$
Hence
$CD=\sqrt{10^2-8^2}=6.$
So
$\cos \angle ADC=\frac{CD}{AD}=\frac{6}{10}=0.6.$
Therefore
$\cos \angle ADE=\cos(180^\circ-\angle ADC)=-\cos \angle ADC=-0.6.$

1. Convert the given angle from the east direction to a bearing from north

At $B$, the line $BC$ points due east. Since $\angle ABC=32^\circ$, the line $BA$ is $32^\circ$ south of east.

Bearing of $A$ from $B$:
$90^\circ+32^\circ=122^\circ$

1. Use the equidistance condition to find the horizontal distance $CE$

From earlier, in right-angled triangle $ACD$,
$CD=\sqrt{10^2-8^2}=6\text{ km}.$
Since $D$ is equidistant from $C$ and $E$,
$DE=DC=6\text{ km}.$
Hence
$CE=CD+DE=6+6=12\text{ km}.$

2. Apply Pythagoras' theorem to find $AE$

Since $AC\perp CE$, triangle $ACE$ is right-angled.
$AE^2=AC^2+CE^2$
$AE^2=8^2+12^2=64+144=208$
$AE=\sqrt{208}=14.4\text{ km} \quad \text{(to 3 s.f.)}$

1. Identify the right triangle and the horizontal and vertical distances from A to E

From earlier parts, $CD=6\text{ km}$. Since D is equidistant from C and E,

$DE=CD=6\text{ km}$.

Hence

$CE=CD+DE=6+6=12\text{ km}$.

Also, $AC=8\text{ km}$, so from A to E, the eastward distance is $12\text{ km}$ and the northward distance is $8\text{ km}$.

2. Use trigonometry to find the angle east of north

Let $\theta$ be the bearing angle of E from A, measured clockwise from north.

In the right-angled triangle,

$\tan \theta = \frac{12}{8}$

$\theta = \tan^{-1}\left(\frac{12}{8}\right)$

$\theta = 56.3^\circ$

3. Write the angle as a three-figure bearing

Bearings are written as three figures, measured clockwise from north.

Therefore the bearing of E from A is

$056.3^\circ$