Find, correct to the nearest degree, the three angles of the triangle with the given ven
A(1, 0, -1), B(4, -3,0), C(1, 2, 3)
o
CAB =
O
LABC =
O
LBCA =

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9514 1404 393

Answer:

∠CAB = 86°

∠ABC = 43°

∠BCA = 51°

Step-by-step explanation:

This can be done a couple of different ways (as with most math problems). We can use the distance formula to find the side lengths, then the law of cosines to find the angles. Or, we could use the dot product. In the end, the math is about the same.

The lengths of the sides are given by the distance formula.

9514 1404 393

Answer:∠CAB = 86°

∠ABC = 43°

∠BCA = 51°

Step-by-step explanation:This can be done a couple of different ways (as with most math problems). We can use the distance formula to find the side lengths, then the law of cosines to find the angles. Or, we could use the dot product. In the end, the math is about the same.

The lengths of the sides are given by the distance formula.

AB² = (4-1)² +(-3-0)² +(0-(-1)) = 16 +9 +1 = 26

BC² = (1-4)² +(2-(-3))³ +(3-0)² = 9 +25 +9 = 43

CA² = (1-1)² +(0-2)² +(-1-3)² = 4 +16 = 20

From the law of cosines, …

∠A = arccos((AB² +CA² -BC²)/(2·AB·CA)) = arccos((26 +20 -43)/(2√(26·20)))

∠A = arccos(3/(4√130)) ≈ 86°

∠B = arccos((AB² +BC² -AC²)/(2·AB·BC)) = arccos((26 +43 -20)/(2√(26·43)))

∠B = arccos(49/(2√1118)) ≈ 43°

∠C = arccos((BC² +CA² -AB²)/(2·BC·CA)) = arccos((43 +20 -26)/(2√(43·20)))

∠C = arccos(37/(4√215)) ≈ 51°

The three angles are …

∠CAB = 86°∠ABC = 43°∠BCA = 51°_____

Additional commentThis sort of repetitive arithmetic is nicely done by a spreadsheet.