P.o.t.W. #32 Solution
equal areas: \displaystyle\frac{1}{2}x\cdot x\sin2{{\theta}}=\frac{1}{2}x\cdot x\sin{{\theta}} hence, \sin2{{\theta}}=\sin{{\theta}} , 0^{\circ}<{{\theta}}<90^{\circ} \sin2{{\theta}}=\sin{{\theta}}\;\;\;\;\Rightarrow\;\;\;\;2\sin{{\theta}}\cos{{\theta}}=\sin{{\theta}} \sin{{\theta}}=0 when {{\theta}}=0^{\circ} or 180^\circ which cannot occur; so, divide by \sin{{\theta}} since \sin{{\theta}}\ne0...