Law of cosines: a2 = b2 + c2 – 2bccos(A) Find the measure of Q, the smallest angle in a triangle whose sides have lengths 4, 5, and 6. Round the measure to the nearest whole degree. 34° 41° 51° 56°
Accepted Solution
A:
a² = b² + c² – 2bccos(A) The smallest angle is across the smallest side, so 4²=5²+6²-2*5*6*cos(Q) 16=25+36-60*cos(Q) 16-36-25=-60*cos(Q) 45=60*cos(Q) cos(Q)=45/60 Q=cos⁻¹(45/60) ≈41⁰