How can I solve this Laws of Sines problem? 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.  
 A: 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
A: This might help:
${AD}\cdot{\sin(\angle DAC)}=
CD=\displaystyle{BD}\cdot{\sin(\angle DBC)}$
A: Let $d$ be the measure of angle $BDC$.  Then you have $BC=88.3 \sin d, AC=BC+5.26=91.2 \sin(d+2.8)$  Expand the sum, plug $BC$ into the last and you should be able to get to something that looks like $e \sin d + f \cos d$, which you can combine.
A: To compute directly an angle or a length, you must have at least 2 legnth and an angle or 2angle and a length, and then use the law of Sines. For the moment you have 1 angle and 1 length
The idea is to use the other triangle (on which you have or may have full information) to gain the missing information. You could compute the angle ABD within the first triangle, and then deduce the angle CBD on the second one. And then you have the necessary information to use the law of Sines.
A: You can use law of cosine http://www.transtutors.com/math-homework-help/laws-of-triangle/ to find AB = 5.26
Then you can use pythagoras theorem
BC * BC + CD *  CD = 88.3 * 88.3
and
CD * CD + (BC + 5.26) * (BC + 5.26) = 91.2 * 91.2
Solve for BC using above 2 equations
