Question

If Asinθ = B, then prove that A > Bsinθ

Answer 1

sin(t)<1 right
now
say x=sin(t)
Ax=B
=> Ax^2=B sin(t)
=>A=(B sin(t))/x^2
as x^2<1
A> B(sin(t))

Answer 2

Thats right. But I think its easier.
A >= A sin t = B >= B sin(t)
voila