Question

How do you solve the triangle given B=72°30', a=105, c=64?

Answer 1

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

b^2 = 11025 + 4096 - 6720.(0.3) = 15121 - 2016 = 13105

`b = 114.48`

Use the trig identity : `sin A/a = sin B/b`

`sin A = (a.sin B)/b = (105 (0.95))/114.18 = 0.87` -> A = 60.89 deg

C = 180 - 72.30 - 60.89 = 46.81 deg