Given $\triangle ABC\ :\ A(0,0),B(b,0)$. Point $C$ on the line $AC\ :\ y=\tan 75^{\circ} \cdot x$ and the line $BC\ :\ y=-\sqrt{3}(x-b)$, $C(\frac{\sqrt{3}(\sqrt{3}-1)b}{4},\frac{\sqrt{3}(\sqrt{3}+1)b}{4})$. Point $T(\frac{(1-\sqrt{3})b}{2},\frac{\sqrt{3}(\sqrt{3}+1)b}{2})$. Point $M(\frac{(1-\sqrt{3})b}{4},\frac{\sqrt{3}(\sqrt{3}+1)b}{4})$. Because $TC=CB, TM=MA$, is $MC // AB$. Slope of the line $BM\ :\ m_{BM}=-1$, so $\angle BMC =45^{\circ}$. Please change "angle wanted, 75-50-45 triangle," in to "angle wanted, 75-60-45 triangle,"