We know that angle A in the triangle above is 45o, that angle B is 30o and that the length b is 2 units.
To work out the remaining angle, we need to remember that the angles within a triangle always add up to 180o.
Since we know A and B add up to 75o, the angle C must be 105o.
Now to find the length a, we can use the first part of the sine rule above. We can rearrange a/sinA = b/sinB to get a=bsinA/sinB.
Since we know A and B we can evaluate this expression to get
a=
Finally we can use the second part of the sine rule to find the length c:
b/sinB=c/sinC, so c=bsinC/sinB
That gives c=2sin(105o)/sin(30o)
which is 4sin(105o).
We can write sin(105o) as sin(150o-45o) then use the sin(A-B) rule to write this as
sin(150o)cos(45o)-cos(150o)sin(45o)
Putting in the values for the sine and cosine of these special angles gives
c=