Your exercise probably deals with the
"ambiguous case", that is, the information usually consists of two sides and a non-contained angle.
make sketches marking the given information, even though in reality the triangle may not be possible.
by the sine law:
Sin C/27 = sin 29°/13
sinC = 1.0069
this is not possible since the sine of any angle must be between -1 and +1.
So, no solution is possible
sinB/11 = sin100.1/20
sinB = .54147
angle B = 32.784
but the sine is + in quadrants I or II
so angle B could also have been 180-32.784 = 147.726°.
However, a triangle cannot have two obtuse angles, so the second case is not possible.
Find the third side and angle in the usual way.
sinB/5 = sin76/15
sinB = .3234
angle B = 18.9° or 161.1°
In this case the sum of 161.1 + 76 is already > 180, so we have to go with the solution of
angle B = 18.9°.
The rest of the question is routine.
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