This is a homework question that was set by my teacher, but it's to see the topic our class should go over in revision, etc.

I have calculated $AB$ to be 5.26m for part (a). I simply used the law of cosines and plugged in the numbers.
part (b) is the question I have been trying for quite a while. I tried to do the law of sines, to no avail. To calculate $BC$ I need the angle opposite that which I do not have (or know how to work out.) The triangle $BDC$ has a right-angle, but this does not work as $sine(90) = 1$
A step in the right direction would be more beneficial than a full answer.
steps to solve this problem
1)
use cosine formula in $\Delta ABD$ to find side $AB$.
$AB^2=88.3^2+91.2^2-2\times88.3\times 91.2\times \cos2.8^\circ$
$AB=5.257$m
2)
find $\angle ABD$ in above triangle using same cosine formula.
$\cos \angle ABD=\dfrac{5.257^2+88.3^2-91.2^2}{2\times 5.257\times 88.3}$
$\angle ABD=122.069^\circ$
3)
Now find out $\angle BDC=\angle ABD-90^\circ=32.069^\circ$
4)
Now we can use simple trigonometry to solve
$\dfrac{BC}{DC}=\tan 32.069^\circ\implies DC=\dfrac{BC}{\tan 32.069^\circ}$
and in $\Delta ACD$
$\dfrac{AB+BC}{DC}=\tan (32.069^\circ+2.8^\circ)$
$\dfrac{5.257+BC}{DC}=\tan 34.869^\circ$
${5.257+BC}={DC}\tan 34.869^\circ$
${5.257+BC}=\dfrac{BC}{\tan 32.069^\circ}\tan 34.869^\circ$
$BC=46.879$m and $AB=5.257$m