Point inside a triangle

From the angles given, we can find that $\angle APB = 150^\circ$ , $\angle APC = 110^\circ$ , and $\angle CPB = 100^\circ$ .

Let $AB=1$ , then by sine rule on $\triangle ABP$ :

$\begin{aligned} \frac {AP}{\sin 20^\circ} & = \frac {BP}{\sin 10^\circ} = \frac {AB}{\sin 150^\circ} = \frac 1{0.5} = 2 \\ \implies AP & = 2 \sin 20^\circ \\ BP & = 2 \sin 10^\circ \end{aligned}$

Using sine rule again on $\triangle ACP$ :

$\begin{aligned} \frac {AC}{\sin 110^\circ} & = \frac {AP}{\sin 30^\circ} = \frac {2 \sin 20^\circ}{0.5} \\ \implies AC & = 4 \sin 20^\circ \sin 110^\circ \end{aligned}$

Let $\angle ACB = \theta$ , then $\angle ABC = 130^\circ - \theta$ . Using sine rule on $\triangle ABC$ :

$\begin{aligned} \frac {\sin \theta}{\color{#3D99F6}AB} & = \frac {\sin (130^\circ-\theta)}{\color{#3D99F6}AC} & \small \color{#3D99F6} \text{Note that }AB=1 \text{ and }AC = 4 \sin 20^\circ \sin 110^\circ \\ \sin \theta & = \frac {\sin (130^\circ-\theta)}{\color{#3D99F6}4 \sin 20^\circ \sin 110^\circ } & \small \color{#3D99F6} \text{By } \frac {\cos (A-B) - \cos (A+B)}2 = \sin A \sin B \\ & = \frac {\sin 130^\circ \cos \theta - \cos 130^\circ \sin \theta}{\color{#3D99F6}2 (\cos 90^\circ - \cos 130^\circ)} \\ & = \frac {\sin 130^\circ \cos \theta - \cos 130^\circ \sin \theta }{-2 \cos 130^\circ} \\ & = - \frac {{\color{#3D99F6}\tan 130^\circ} \cos \theta}2 + \frac {\sin \theta}2 & \small \color{#3D99F6} \text{Note that }\tan (180^\circ - x) = - \tan x \\ & = {\color{#3D99F6}\tan 50^\circ} \cos \theta \\ \tan \theta & = \tan 50^\circ \\ \implies \theta & = 50^\circ \end{aligned}$

Therefore, the largest angle of $\triangle ABC$ is $\angle ABC = 180^\circ - 50^\circ - 50^\circ = \boxed{80^\circ}$

This is a problem (#5) from USA Math Olympiad 1996: Solution (as obtained from - https://www.normalesup.org/~kortchem/olympiades/Problemes/mc96-97-01feb.pdf): Let D be the reflection of A across the line BP. Then triangle APD is isosceles with vertex angle ∠APD = 2(180 − ∠BPA) = 2(∠P AB + ∠ABP) = 2(10 + 20) = 60°, and so is equilateral. Also, ∠DBA = 2∠P BA = 40. Since ∠BAC = 50°, we have DB ⊥ AC. Let E be the intersection of DB with CP. Then ∠PED = 180 − ∠CED = 180 − (90 − ∠ACE) = 90 + 30 = 120 ° and so ∠PED + ∠DAP = 180°. We deduce that the quadrilateral APED is cyclic, and therefore ∠DEA = ∠DPA = 60°. Finally, we note that ∠DEA = 60° = ∠DEC. Since AC ⊥ DE, we deduce that A and C are symmetric across the line DE, which implies that BA = BC. Then ∠PCB =50-30=20° and ∠CBP=180-20-100=60°. This makes ∠ABC the largest at 80°.