Question 968332
A=(5,2) 
B=(7,5) 
C=(8,4). 
<pre>
{{{drawing(400,3200/11,-2,9,-2,6,graph(400,3200/11,-2,9,-2,6),

triangle(5,2,7,5,8,4),locate(5,2,"A(5,2)"),locate(7,5.5,"B(7,5)"),locate(8,4,"C(8,4)") )}}}

This could be done with lengths of sides and the law of cosines but it
is more easily done with angles between lines using slopes.

All three of those angles appear to be acute (less than 90°)

The acute angle between two lines is given by 

{{{tan(theta)}}}{{{""=""}}}{{{abs((m[1]-m[2])/(1+m[1]m[2]))}}} 

&#8736;A is the angle between AB and AC.

We find the slope of AB and call it m<sub>1</sub>

{{{m[1]}}}{{{""=""}}}{{{(y[2]-y[1])/(x[2]-x[1])}}}

{{{m[1]}}}{{{""=""}}}{{{(5-2)/(7-5)}}}

{{{m[1]}}}{{{""=""}}}{{{3/2}}}

We find the slope of AC and call it m<sub>2</sub>

{{{m[2]}}}{{{""=""}}}{{{(y[2]-y[1])/(x[2]-x[1])}}}

{{{m[2]}}}{{{""=""}}}{{{(4-2)/(8-5)}}}

{{{m[2]}}}{{{""=""}}}{{{2/3}}}

Substitute these in:

{{{tan(A)}}}{{{""=""}}}{{{abs((m[1]-m[2])/(1+m[1]m[2]))}}} 

{{{tan(A)}}}{{{""=""}}}{{{abs((3/2-2/3)/(1+expr(3/2)expr(2/3)))}}} 

{{{tan(A)}}}{{{""=""}}}{{{abs((3/2-2/3)/(1+1))}}}

{{{tan(A)}}}{{{""=""}}}{{{abs((3/2-2/3)/2)}}}

Multiply top and bottom by LCD of 6

{{{tan(A)}}}{{{""=""}}}{{{abs((9-4)/12)}}}

{{{tan(A)}}}{{{""=""}}}{{{abs(5/12)}}}

{{{tan(A)}}}{{{""=""}}}{{{5/12}}}

{{{A}}}{{{""=""}}}{{{"22.61986595°"}}}

Notice that instead of calling the angle &#8736;A, here 
it is called &#952;<sub>A</sub>. 

Find the other two angles the same way.

Edwin</pre>