Question 387508
find, correct to the nearest degree the three angles of a triange with vertices A(0,3,3) B(-2,1,1) C(1,5,-1)
<pre>
The line segment AB is parallel to the vector < -2-0,1-3,1-3 > = < -2,-2,-2 >

The line segment AC is parallel to the vector < 1-0,5-3,-1-3 > = <  1, 2,-4 >
 
The line segment BC is parallel to the vector < 1-(-2),5-1,-1-1 > = <  3, 4,-2 >

We use this formula

{{{cos(theta)}}}{{{""=""}}}{{{u*v/(abs(abs(u))abs(abs(v)))}}}

Angle A is the angle between AB and AC, which is the same angle
as the angle between the vectors < -2,-2,-2 > and <  1, 2,-4 >

Their dot product is 

< -2,-2,-2 >@<  1, 2,-4 > = (-2)(1)+(-2)(2)+(-2)(-4) = -2-4+8 = 2

The norm of < -2,-2,-2 > is {{{sqrt((-2)^2+(-2)^2+(-2)^2)=sqrt(4+4+4)=sqrt(12)}}} 

The norm of < 1,2,-4 > is {{{sqrt((1)^2+(2)^2+(-4)^2)=sqrt(1+4+16)=sqrt(21)}}}

Substituting in the formula for the cosine,

{{{cos(A)}}}{{{""=""}}}{{{2/(sqrt(12)sqrt(21))}}}=.1259881577

So A = 82.76217579° or 83° to the nearest degree.

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Find the other two angles the same way.

Edwin</pre>