The vertices of a triangle are A(1,4,2) B(-2,1,2) C(2,3,-4). Find angle A angle B and angle C.
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Answered by
30
Answer:
A = 90°, B ≈ 55.46° and C ≈ 34.54°
Step-by-step explanation:
The vectors along the sides are:
AB = B - A = ( -3, -3, 0 ), BC = C - B = ( 4, 2, -6 ), CA = A - C = ( -1, 1, 6 ).
The squares of the lengths of the sides are then
|AB|² = 9 + 9 + 0 = 18
|BC|² = 16 + 4 + 36 = 56
|CA|² = 1 + 1 + 36 = 38
Notice that |AB|² + |CA|² = |BC|², so by the converse of Pythagoras' Theorem, there is a right angle at A.
For the angle at B, we have
cos B = |AB| / |BC| = 3√2 / 2√14 = 3 / 2√7
=> B ≈ 55.46°
and C = 90° - B ≈ 34.54°
Answered by
20
Answer:
Cos A=0,cos B=3/2 root 7,cos C=root of 19/28
Step-by-step explanation:
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