Question

How do you find the area of the triangle given a= 68, c= 110, B= 42.5 degrees?

Answer 1

Solve triangle given B = 42.5, a = 68 and c = 110

Explanation:

Use the trig identity:`b^2 = a^2 + c^2 - 2ac.cos B`

`b^2` = 4624 + 12100 - 14960cos B = 16724 - 11029.66 = 5694.33
b = 75.46

`sin C = (c.sin B)/b = (110(0.68))/75.46 = 74.31/75.46 = 0.98`
C = 80 deg

A = 180 - b - c = 180 - 42.5 - 80 = 57.5deg