there are two angles whose sine is 0.895...
Quad I and Quad II
Given a = 31.5, b = 51.8, A = 33 degrees, find the value of c in triangle ABC.
bsinA = 51.8sin33 = 28.21 -> bsinA < a < b so there are two solutions
Using sine law:
sinB/b = sinA/a
sinB/51.8 = sin33/31.5
sinB = 51.8 * sin33/31.5
sinB = 0.895628635
B = arcsin(0.895628635)
B = 63.6
C = 180 - 33 - 63.6 = 83.4
c/sinC = a/sinA
c/sin83.4 = 31.5/sin33
c = sin83.4 * 31.5/sin33
c = 57
The second solution is 29, but not sure how to solve for it.
1 answer