## The Best answer for Find, correct to the nearest degree, the three angles of the triangle with vertices A(1, 0, −1), B(5, −3, 0),and C(1, 2, 5)?

First find expressions for the vectors that make the triangle:

Vector AB = <5 - 1, -3 - 0, 0 + 1> = <4, -3, 1>

Vector BC = <1 - 5, 2 + 3, 5 - 0> = <-4, 5, 5>

Vector AC = <1 - 1, 2 - 0, 5 + 1> = <0, 2, 6>

Now using dot product:

Vector AB dotted with Vector AC = |AB|*|AC| * cos(∠CAB )

(4)(0) - 3(2) + 1(6) = √(9 + 16 + 1) * √(4 + 36) * cos(∠CAB )

∠CAB = 90 degrees

Vector BC dotted with Vector BA = |BC|*|AB| * cos(∠ABC)

16 + 15 - 5 = √(16 + 9 + 1) * √(16 + 50) * cos(∠ABC )

∠ABC = 51 degrees

Therefore since all angles add up to 180 degrees:

∠BCA = 180 - 90 - 51 degrees = 39 degrees

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