Find the $\angle ACB$ of $\triangle ABC$.

Question

If $PC=2BP$, $\angle ABC= 45^\circ$, and $\angle APC=60^\circ$, find $\angle ACB$.

All solutions are acceptable but please try solving using reflection of point $C$ through the line segment $AP$.

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I found a solution by adding a dot named D but in the given solution which I've drawn, I didn't understand how alpha was found. Although it is the original given solution in the geometry book.

It was only pointed that $\angle ACB = \angle ADP = \frac{1}{2}(180^\circ - \angle BDP) = 75^\circ$. Please explain this for me because this solution is blurry and vague for me.

Answer 1

Sometimes, geometry consists of dropping the right line or introducing the accurate point...

Denote by $D$ the point on $AP$ such that $\angle PDC=90°$ and let $BP=x$. Note now that the triangle $\Delta PCD$ is a $30°-60°-90°$ triangle, which implies that $PD=x$.

Therefore $\Delta BPD$ is isosceles and hence $\angle PBD=30° \Rightarrow \angle DBA=15°$.

Easy angle chasing leads to the conclusion that $DC=DB=DA$, which implies that $\Delta CDA$ is also isosceles. Thus $\angle ACD=45°$. $$\therefore \angle ACB=45°+30°=75° \blacksquare$$

Answer 2

By calculating some angles and applying the sine rule we have: $$|AB|=\frac{3\sqrt{2}+\sqrt{6}}{2}\cdot x$$ Then a second application of the sine rule on $\triangle ABC$ gives: $$\frac{3x}{\sin{(135^\circ-\angle ACB)}}=\frac{\frac{3\sqrt{2}+\sqrt{6}}{2}\cdot x}{\sin{(\angle ACB)}}$$ $$\frac{3}{\sin{(135^\circ-\angle ACB)}}=\frac{3\sqrt{2}+\sqrt{6}}{2\sin{(\angle ACB)}}$$ The only solution in $0^\circ \lt\angle ACB \lt 180^\circ$ to the above equation is then: $$\angle ACB =75^\circ$$

Answer 3

Hint: Drop an altitude from $A$ to $\overline{BC}$ and then use trigonometry.