ABCD is a field on horizontal ground. AD = 163 m, BD = 329 m, CD = 152 m and angle BDC = 48°. The bearing of B from A is 151° and the bearing of D from A is 237°.

1. Find the angle between AB and AD at A using the two bearings from A.

Bearing of B from A $=151^\circ$.

Bearing of D from A $=237^\circ$.

So
$\angle BAD=237^\circ-151^\circ=86^\circ.$

2. Use the sine rule in triangle ABD to find angle ABD.

In $\triangle ABD$,

$\frac{\sin \angle ABD}{AD}=\frac{\sin \angle BAD}{BD}$

$\frac{\sin \angle ABD}{163}=\frac{\sin 86^\circ}{329}$

$\sin \angle ABD=\frac{163\sin86^\circ}{329}$

$\angle ABD=29.6^\circ \text{ correct to 1 d.p.}$

3. Convert the angle at B into the required bearing of D from B.

The bearing of A from B is the reverse bearing of B from A:

$151^\circ+180^\circ=331^\circ.$

Since $\angle ABD=29.6^\circ$,

$\text{bearing of D from B}=331^\circ-29.6^\circ=301.4^\circ.$

To the nearest degree, this is

$301^\circ.$

1. Apply the cosine rule to triangle $BDC$ using the included angle $\angle BDC = 48^\circ$.

In $\triangle BDC$,
$BC^2 = BD^2 + CD^2 - 2(BD)(CD)\cos \angle BDC$
$BC^2 = 329^2 + 152^2 - 2(329)(152)\cos 48^\circ$

2. Evaluate and round the distance appropriately.

$BC^2 \approx 64421.26$
$BC \approx \sqrt{64421.26}$
$BC \approx 253.8$
So, to 3 significant figures,
$BC = 254\text{ m}$

1. Identify the right triangle and choose the appropriate trigonometric ratio.

Let the aircraft be at point $P$, vertically above $D$.

Since $PD=245$ m and $CD=152$ m, triangle $CDP$ is right-angled at $D$.

For the angle of elevation $\theta$ from $C$,
$\tan \theta=\frac{245}{152}.$

2. Calculate the angle of elevation.

$\theta=\tan^{-1}\left(\frac{245}{152}\right)$
$\theta=58.19\ldots^\circ$

So, to 1 decimal place,
$\theta=58.2^\circ.$