Points A, B, C and D are at sea level. AD = 980 km, AB = 710 km and CD = 1100 km. The bearing of B and C from A are 148° and 140° respectively. The bearing of B and D from C are 300° and 016° respectively.

1. Find the bearing of A from C using the bearing of C from A.

The bearing of C from A is $140^\circ$.

Therefore the bearing of A from C is the opposite direction:
$140^\circ + 180^\circ = 320^\circ.$

2. Find angle ACB from the bearings at C.

At C, the bearing of B from C is $300^\circ$ and the bearing of A from C is $320^\circ$.

Hence
$\angle ACB = 320^\circ - 300^\circ = 20^\circ.$

1. Find the remaining angle in triangle $ACB$.

From the bearings at $A$,
$\angle BAC = 148^\circ - 140^\circ = 8^\circ.$
From part (a),
$\angle ACB = 20^\circ.$
Hence
$\angle ABC = 180^\circ - 8^\circ - 20^\circ = 152^\circ.$

2. Apply the sine rule to find $AC$.

Using the sine rule in triangle $ACB$,
$\frac{AC}{\sin 152^\circ}=\frac{AB}{\sin 20^\circ}.$
Since $AB=710$,
$AC=\frac{710\sin 152^\circ}{\sin 20^\circ}.$

3. Calculate and round the distance appropriately.

$AC=\frac{710\sin 152^\circ}{\sin 20^\circ}$
$AC=974.58\ldots$
$AC\approx 975\text{ km}.$

1. Find angle $ACD$ using the bearings at $C$.

From part (a), the bearing of $A$ from $C$ is $320^\circ$.

The bearing of $D$ from $C$ is $016^\circ$.

Hence
$\angle ACD = 360^\circ - 320^\circ + 16^\circ = 56^\circ.$

2. Use the sine rule in triangle $ACD$ to find angle $ADC$.

Using $AC = 975\text{ km}$ from part (b), $AD = 980\text{ km}$ and $\angle ACD = 56^\circ$,

$\frac{\sin \angle ADC}{AC}=\frac{\sin 56^\circ}{AD}$

$\frac{\sin \angle ADC}{975}=\frac{\sin 56^\circ}{980}$

$\sin \angle ADC=\frac{975\sin 56^\circ}{980}$

$\angle ADC=55.6^\circ \text{ correct to 1 d.p.}$

3. Convert the angle at $D$ into the bearing of $A$ from $D$.

The bearing of $D$ from $C$ is $016^\circ$, so the bearing of $C$ from $D$ is
$016^\circ + 180^\circ = 196^\circ.$

At $D$, the angle from $DC$ to $DA$ is $55.6^\circ$.

Therefore the bearing of $A$ from $D$ is
$196^\circ + 55.6^\circ = 251.6^\circ.$

To the nearest degree,
$251.6^\circ \approx 252^\circ.$

1. Find the angle between the ship's route $AC$ and the line $AB$.

At $A$, the bearings of $B$ and $C$ are $148^\circ$ and $140^\circ$ respectively.

$\angle BAC = 148^\circ - 140^\circ = 8^\circ$

2. Calculate the perpendicular distance from $B$ to the route $AC$.

The shortest distance from $B$ to the straight line $AC$ is the perpendicular distance.

Let this distance be $d$.

$\sin 8^\circ = \frac{d}{710}$

$d = 710\sin 8^\circ$

$d = 98.801\ldots$

$d \approx 98.8\text{ km}$

1. Convert the height to kilometres and identify the trigonometric ratio.

Height of plane above sea level:
$900\text{ m}=0.9\text{ km}$
Let the horizontal distance between the plane and $C$ be $x$ km.

Since the angle of depression is $18^\circ$, the angle of elevation from $C$ to the plane is also $18^\circ$.

$\tan 18^\circ=\frac{0.9}{x}$

2. Solve for the horizontal distance.

$x=\frac{0.9}{\tan 18^\circ}$
$x=2.769\ldots$
$x=2.77\text{ km}$