Answer:
45°, 45°, 90°
Step-by-step explanation:
Find the vectors to represent each sides
AB =<3-1, 6-0>=<2,6>
AC = < -1-1, 4 - 0 > = < -2, 4>
Magnitude of the vectors
AB = √(2²+6²) = 6.32
AC = √ ((-2)² + 4²) = 4.47
cosθ = vector of AB × vector AC / ( Product of the magnitude of AB and AC) = 2 × (-2) + (6×4)/ (6.32×4.47) = 20 / 28.2504
θ = arcos(20 / 28.2504 ) approx = 45°
Magnitude of the vectors
BA =<1-3,0-6>=<-2,-6>
BC =<-1-3,4-6>=<-4,-2>
Magnitude of the vectors equals
BA = √((-2)² + (-6)²) = 6.325
BC = √((-4)² + (-2)²) = 4.4721
cosθ = (-2×-4) + (-6 ×-2) / (6.325 × 4.4721) = 20 / 28.286
θ = arcos (20 / 28.286 ) = 45°
Magnitude of the vectors
CB =<3--1, 6-4>=<4,2>
CA=<1--1,0-4>=<2,-4>
Magnitude of the vector =
CB = √(4² + 2²) = 4.4721
CA = √(2² + (-4)²) = 4.4721
cosθ = (4×2) + (2×-4) / (4.4721×4.4721) = 0
θ = arcos 0 = 90°
