Point P on side BC of triangle ABC such that PC=2BP. Find ACB if ABC=45º, APC=60º.
I can't solve this one. Tried some stuff but can't work it out. Can this be done using just simple geometry (like angle calculations and properties of equal triangles,etc.)?
Well, trigonometry works, I think.
By Sine Law $$\frac{\sin \widehat{BAC}}{BC}=\frac{\sin\widehat{ABC}}{AC}$$ $$\frac{\sin \widehat{PAC}}{PC}=\frac{\sin\widehat{APC}}{AC}$$ so $$\frac{\sin \widehat{BAC}}{\sin\widehat{PAC}}=\frac{\sqrt{3}}{\sqrt{2}}$$ and obviously $\widehat{BAC}=\widehat{PAC}+15^\circ$. Let us set $\widehat{BAC}=x$ and $\widehat{PAC}=y$, then $$\sin x=\sin (y+15^\circ)=\frac{1+\sqrt{3}}{2\sqrt{2}}\sin y + \frac{\sqrt{3}-1}{2\sqrt{2}}\cos y$$ so $$\frac{\sin x}{\sin y}=\frac{1+\sqrt{3}}{2\sqrt{2}} + \frac{\sqrt{3}-1}{2\sqrt{2}}\cot y$$ from which you obtain $$\cot y=1$$ hence $y=45^\circ$. Therefore $\widehat{ACB}=75^\circ$.