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

Jun 9, 2015

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

#### Explanation:

Use the trig identity:${b}^{2} = {a}^{2} + {c}^{2} - 2 a c . \cos B$

${b}^{2}$ = 4624 + 12100 - 14960cos B = 16724 - 11029.66 = 5694.33
b = 75.46

$\sin C = \frac{c . \sin B}{b} = \frac{110 \left(0.68\right)}{75.46} = \frac{74.31}{75.46} = 0.98$
C = 80 deg

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