The vertices of triangleABC are A......

Question

The vertices of triangleABC are A(1,2,3) B(-3,-6,8) and C(-5,6,-4) what is <A (angle A)?

Answer 1

If we name the angles A,B, and C and the corresponding opposite sides (for angle A, a= distance of B to C) a, b, and c, we can find the angle A from knowing all the sides and using the Law of Cosines:

b² + c² - 2bc*cos(A) = a²

A = cos^-1((a² - b² - c²)/(-2bc))

we know all the sides by doing the distance formula between points:

d = sqrt((x2-x1)^2 + (y2-y1)^2) + (z2-z1)^2)

ex: a = distance from B to C = sqrt((-3+5)^2 + (-6-6)^2 +(8+4)^2) = sqrt(4 + 144 + 144) = 17.09

Find b and c and plug into the equation for A

I hope that helps.