Question

How do you solve triangle ABC, given B=41°, c=7, a=10?

Answer 1

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

and `sin A/a = sin B/b.`

`b^2 = 100 + 49 - 140(0.75) = 149 - 105 = 44 -> b = 6.63`

`sin A/ 10 = sin (41)/6.63` ->

`sin A = (10(0.66))/6.63 = 1 -> A = 90` deg

C = 90 - B = 90 - 41 = 49 deg

Answer 2

Hope this helps!