In a triangle abc ab = 7 bc = 16 angle b = 110 degrees find the angles a and c.

AC² = AB² + BC² – 2 * AB * BC * cos B;

AB = 7; BC = 16; cos 110 ° ≈ – 0.34;

AC² = 7² + 16² + 2 * 7 * 16 * 0.34 = 49 + 256 + 76.16 ≈ 381.16

AC = √381.16 ≈ 19.5;

By the sine theorem:

AC / sin B = AB / sin C = BC / sin A;

sin 110 ° ≈ 0.94;

sin A = BC * sin B: AC = 16 * 0.94: 19.5 ≈ 0.77; <A ≈ 51 °;

sin C = AB * sin A: BC = 7 * 0.77: 16 ≈ 0.33; <C ≈ 29 °.