Two sides of triangle ABC and an angle opposite to the third side are given.

Two sides of triangle ABC and an angle opposite to the third side are given. Find the other two corners and the third side if BC = 16 cm, AC = 10 cm, angle C = 80.

Let’s omit the height BО. Consider triangle BOC and find the angle OBC.

Cos C = OC / BC;

OC = BC * Cоs 80 °;

OC = 16 * 0.1736 = 2.8 cm.

OBC angle = 180 ° – (90 ° – 80 °) = 10 °.

Consider a triangle BOC.

AO = AC – OC = 10 – 2.8 = 7.2 cm.

tg A = BO / AO = 15.8 / 7.2 = 2.2;

angle A = 66 °.

ABO angle = 180 ° – (90 ° – 66 °) = 24 °.

Find angle B.

Angle B = angle ABO + angle OBC = 24 ° + 10 ° = 34 °.

Find the AB side.

Sin A = BO / AB;

AB = BO / Sin A = 15.8 / 0.9135 = 17.2 cm.

Answer: angle A = 66 °, angle B = 34 °, AB = 17.2 cm.