Find, corrrect to the nearest degree, the three angles of the triangle with the given vertices. a(1,0), b(3,6), c(-1,4)

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°

it is contrapositive

i could not see the question too well so check yourself again too . this is because i want you to get it correct.

the ratio would be 4: 3, the four being for jackson, and the three being not for him.