If you can remember your perfect squares you'll
remember that 15² is 225 and that is exactly twice
the area of 112.5m². So a square which is 15m on a
side has exactly twice that area.
Therefore since the triangle is isosceles it can
only be half of a 15m×15m square cut across the
diagonal. So its vertex angle is the interior
angle of a square and therefore is 90°.
If that's too much for you, then you can work it
out this way. Draw ΔABC
Draw the line CE, which is the median, vertex angle bisector and
altitude from the vertex to the base (in green). Label it h
and label each of the equal halves of the base AE=BE=x.
Area = 
base·height
The base is 2x, so
121.5 = (2x)(h)
121.5 = xh
And by the Pythagorean theorem,
x²+h² = 15²
x²+h² = 225
So you have this system to solve
x²+h² = 225
121.5 = xh
Remove the decimal by multiplying
the second equation by 2
225 = 2xh
2xh = 225
Make a perfect square trinomial by subtracting
that equation from both sides of
x²+h² = 225
to make it a perfect square trinomial:
x²-2xh+h² = 0
(x-h)(x-h) = 0
(x-h)² = 0
x-h=0
x=h
That tells us that congruent right triangles
ΔACE and ΔBCE are isosceles and the base
angles of an isosceles right triangle
are 45° each, so
∠ACE = ∠BCE = 45° and
∠ACB = ∠ACE+∠BCE = 45°+45° = 90°
Edwin
