Find, correct to one decimal place, the size of the smallest angle of the triangle which has sides of length 3,5,7

Answer 1

a^2 = b^2 + c^2 - 2bc cosA

b^2 = a^2 + c^2 - 2ac cosB

c^2 = b^2 + a^2 - 2ba cosC

where a, b, and c are the sides of the triangle, and A, B, C are the angles opposite the sides.

Plug a, b, c = 3, 5, 7 into these equations and solve for the angles A, B, C. The smallest of these angles is the answer

Answer 2

you can save some work by recognizing that the smallest angle is opposite the shortest side (law of sines)