The lengths of the sides of a triangle are 180, 186, and 174. Find the measures of the angles of the...?

If :

a = 180

b = 186

c = 174

Angle A = arccos((b^2 + c^2 - a^2) / 2bc)

Angle A = arccos((186^2 + 174^2 - 180^2) / (2*186*174))

Angle A = arccos(32472 / 64728)

Angle A = arccos(0.5016685)

Angle A = 59.9 degrees

Angle B = arccos((a^2 + c^2 - b^2) / 2ac)

Angle B = arccos((180^2 + 174^2 - 186^2) / (2*180*174))

Angle B = arccos(28080 / 62640)

Angle B = arccos(0.44827586)

Angle B = 63.4 degrees

Angle C = arccos((a^2 + b^2 - c^2) / 2ab)

Angle C = arccos((180^2 + 186^2 - 174^2) / (2*180*186))

Angle C = arccos(36720 / 66960)

Angle C = arccos(0.548387)

Angle C = 56.7 degrees

Proof :

59.9 + 63.4 + 56.7 = 180 :ok

Note : "arccos" is the inverse of the function cosine. It gives the angle related to the cosine. On the Windows calculator, check the box "Inv" before the "cos" key.