Answer:
∠B = 22.5°, ∠C = 25.5°, c = 40.6
Step-by-step explanation:
an angle and 2 sides are given. It just so happens that the angle is across from one of the sides. Using the Law of Sines to find the missing angle B first.
[tex]\frac{sin(132)}{70} =\frac{sinB}{36}[/tex]
36 sin(132) = 70 sinB and
[tex]\frac{36sin(132)}{70} =sinB[/tex] and
sinB=0.3821887674
B=[tex]sin^-^1[/tex](0.3821887674)
B = 22.5
Now, find C using the Triangle Angle-Sum Theorem .
C = 180 - 132 - 22.5
C = 25.5 degrees
Using the Law of Sines again with angle A and side A and angle C:
[tex]\frac{sin(132)}{70} =\frac{sin(25.5)}{c}[/tex]and
[tex]c=\frac{70sin(25.5)}{sin(132)}[/tex]
c=40.6
It is the third option there. Locate the third B and that's where you solution starts.