x+y+2=0
x+2=-y
-x-2=y
2x+5y-3=0
2x-3=-5y
(-2x+3)/5=y
-x-2 = (-2x+3)/5
-x-2 = -2x/5+3/5
-3x/5 = 13/5
x = -13/3 (this is the point where the two equations are equal)
Now, assuming you want the angle of the interior part of the point:
Let f(x) = -x-2 and g(x) = (-2x+3)/5
f(-2) = 0
g(1.5) = 0
(calculated some zero's as "starting points" for the "sides" of the triangle)
Let x2 = -13/3 and y2 = f(-13/3) = g(-13/3) = 7/3
Let x1 = -2 and y1 = f(-2) = 0
Use distance formula to find the distance between the two points:
d = Sqrt((x2-x1)²+(y2-y1)²)
d = Sqrt((-13/3-(-2))²+(7/3-0)²)
d = Sqrt(49/9+49/9)
d = Sqrt(98/9) = 3.29983164553722
let's call this side A
Now,
Let x2 = -13/3 and y2 = f(-13/3) = g(-13/3) = 7/3
Let x1 = 1.5 and y1 = g(1.5) = 0
d = Sqrt((-13/3-1.5)²+(7/3-0)²)
d = Sqrt(34.0277777777777+49/9)
d = 6.28269227499024
let's call this side B
We know the distance from x=-2 to x=1.5... 3.5 - call this side C
Now, using the law of cosines
Now, we want the angle opposite side C:
cos(c) = (A²+B²-C²)/(2*A*B)
cos(c) = 0.919145030018056
c = 23.1985905136485°
Or to be more precise, 23 degrees, 11 minutes, and 54.9 seconds
I'm not sure that your teacher would prefer you to do it this way - but I can't remember any other shorter ways of doing it right off the top of my head.
