Find, correct to the nearest degree, the three angles of the triangle with the vertices d (0,1,1), e (2, 4,3) -, and f (1, 2, 1)

Well, here's one way to do it at least ...

For reference, let 'a' be the side opposite A (segment BC), 'b' be the side opposite B (segment AC) and 'c' be the side opposite C (segment AB).

Let P = (4,0) be the projection of B onto the x-axis.

Let Q = (-3,0) be the projection of C onto the x-axis.

Look at the angle QAC. It has tangent = 5/4 (do you see why?), so angle A is atan (5/4).

Likewise, angle PAB has tangent = 6/3 = 2, so angle PAB is atan (2).

Angle A, then, is 180 - atan (5/4) - atan (2) = 65.225. One down, two to go.

||b|| = sqrt (41) (use Pythagorian Theorum on triangle AQC)

||c|| = sqrt (45) (use Pythagorian Theorum on triangle APB)

Using the Law of Cosines ...

||a||^2 = ||b||^2 + ||c||^2 - 2 (||b||) (||c||) cos (A)

||a||^2 = 41 + 45 - 2 (sqrt (41)) (sqrt (45)) (.4191)

||a||^2 = 86 - 36

||a||^2 = 50

||a|| = sqrt (50)

Now apply the Law of Sines to find the other two angles.

||b|| / sin (B) = ||a|| / sin (A)

sqrt (41) / sin (B) = sqrt (50) /.9080

(.9080) sqrt (41) / sqrt (50) = sin (B)

.8222 = sin (B)

asin (.8222) = B

55.305 = B

Two down, one to go ...

||c|| / sin (C) = ||a|| / sin (A)

sqrt (45) / sin (C) = sqrt (50) /.9080

(.9080) sqrt (45) / sqrt (50) = sin (C)

.8614 = sin (C)

asin (.8614) = C

59.470 = C

So your three angles are:

A = 65.225

B = 55.305

C = 59.470