OK, let's try with no figure. We have an isosceles triangle sides s,s, and b.
Opposite b is angle t.
Draw the altitude h to bisect t. We have two right triangles, legs b/2 and h, hypotenuse s. The angle opposite b/2 is t/2 so
sin(t/2) = (b/2)/s = b/2s
So we arrived at the first part,
b = 2s sin(t/2)
The area of a triangle with sides s,s and included angle t is
A = (1/2) s² sin t