Answer:
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°
