Given a triangle with a = 9, b = 11, A = 31°, what is (are) the possible length(s) of c? Round your answer to two decimal places.
A) 16.42 or 2.44
B) 6.61
C) 14.21
D) 16.42 or 3.41
You can use the Law of Cosines to figure this. a² = b² +c² -2bc×cos(A) c² -(2b×cos(A))c +b² -a² = 0 c² -18.8577c +40 = 0 . . . . . . . substituting the given values Then we can use the quadratic formula to find c. c = (18.8577 ±√(18.8577² -160))/2 c = (18.8577 ±13.9861)/2 ≈ {2.44, 16.42}
