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

Angle at vertex  A : 59.2° B: 61.7° C : 59.1°

Step-by-step explanation:

1. First find the length of the vectors  

AB = B-A = (4-1, 3-0, 0-1) = (3, 3, -1)

AC = C-A = (1-1, 4-0, 3-1) = (0, 4, 2)

BC = C-B = (1-4, 4-3, 3-0) = (-3, 1, 3)

2. Find magnitude of the vectors

|AB| = √(3^2+3^2+〖(-1)〗^2 )=√19  

|AC| = √(0^2+4^2+2^2 )=2√5

|BC| = √(〖(-3)〗^2+1^2+3^2 )=√19

3. Find the angles between them

cos θ = (a.b)/(|a||b|) > θ = arc cos ((a.b)/(|a||b|))

AB, AC : θ =arc cos (AB.AC)/(|AB||AC|) = arc cos ( (3.0+3.4+ -1.2)/(√19  x 2√5)=  10/(2√95) ) =59.136, which is approximately 59.14°

AB, BC : θ =arc cos (AB.BC)/(|AB||BC|) = arc cos( (3.-3+3.1+ -1.3)/(√19  x √19)=  (-9)/19 ) = 118.27°

Because of the direction of BC is pointing relative to AB this is the angle outside the triangle, and we should find the supplementary angle.

180-118.27 = 61.73°

If we used –(BC) = CB in the formula (just negate the numerator) we would have gotten the correct angle on first try.

The 3rd angle should be what’s left after subtracting from 180°;

180-61.73-59.14 = 59.13 °

You can confirm using the formula again :

AC, BC : θ =arc cos (AC.BC)/(|AC||BC|) = arc cos( (0.-3+4.1+ 2.3)/(√19  x 2√5)=  10/(2√95) = 59.13°

Rounding everything up so they add up to 180 degrees we have Angle at vertex  A : 59.2° B: 61.7° C : 59.1°

a: 1/6, j: 1/4

step-by-step explanation: you are welcome

r and t.

both are 64 degrees.