ABCD is a field on horizontal ground. AD = 168 m, BD = 325 m and CD = 151 m.
The bearing of B from A is 155° and the bearing of D from A is 235°.
Angle BDC = 48°.

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

Bearing of B from A $=155^\circ$.

Bearing of D from A $=235^\circ$.

So,
$\angle BAD=235^\circ-155^\circ=80^\circ.$

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

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

$\frac{\sin \angle ABD}{168}=\frac{\sin 80^\circ}{325}$

$\sin \angle ABD=\frac{168\sin80^\circ}{325}$

$\angle ABD=30.6^\circ \text{ correct to 1 d.p.}$

3. Find the bearing of D from B using the reverse bearing of A from B.

Bearing of A from B is the reverse bearing of B from A:
$155^\circ+180^\circ=335^\circ.$

At B, the angle between $BA$ and $BD$ is $30.6^\circ$.

Hence bearing of D from B is
$335^\circ-30.6^\circ=304.4^\circ.$

To the nearest degree,
$304^\circ.$

1. Find the missing side $BC$ using the cosine rule in triangle $BCD$.

In $\triangle BCD$,

$BC^2 = BD^2 + CD^2 - 2(BD)(CD)\cos \angle BDC$

$BC^2 = 325^2 + 151^2 - 2(325)(151)\cos 48^\circ$

$BC = 250.5\text{ m} \text{ approximately}$

2. Calculate the total length of fencing needed around triangle $BCD$.

Perimeter of $\triangle BCD$:

$BD + CD + BC = 325 + 151 + 250.5$

$= 726.5\text{ m} \text{ approximately}$

3. Find the total cost and round to the nearest $100 dollars$.

Cost of fence:

$726.5 \times 85 = 61752.5$

Correct to the nearest $100 dollars$,

$61752.5 \approx 61800$