Interior Point P! Find the Angle!

$(\overline{BC})^2 = (\overline{AB})^2 + (\overline{AC})^2 - 2(\overline{AB})(\overline{AC})\cos{50}$ ---> $eq. 1$

by sine law;

$\overline{AB} = \frac{\overline{AP}\sin{150}}{\sin{20}}$ ---> $eq. 2$

$\overline{AC} = \frac{\overline{AP}\sin{110}}{\sin{30}}$ ---> $eq. 3$

Substitute $eq. 2$ and $eq. 3$ in $eq. 1$

$(\overline{BC})^2 = (\frac{\overline{AP}\sin{150}}{\sin{20}})^2 + (\frac{\overline{AP}\sin{110}}{\sin{30}})^2 - 2(\frac{\overline{AP}\sin150}{\sin20})(\frac{\overline{AP}\sin110}{\sin30})\cos{50}$

$(\overline{BC}) = 1.4619022\overline{AP}$ ----> $eq. 4$

sine law on big triangle

$\frac{\sin{\angle{ABC}}}{\overline{AC}} = \frac{\sin{50}}{\overline{BC}}$ ----> $eq. 5$

subtitute $eq. 3$ and $eq. 4$ in $eq. 5$ . Then solve.

$\angle{ABC} = \boxed{80^0}$

We can assume that $AP = 1 cm$

in Triangle ABP from the law of sines we can prove that $AB=\frac{sin(150)}{sin(20)}$

in Triangle ACP from the law of sines we can prove that $AC= \frac{sin(110)}{sin(30)}$

and in Triangle ABC from the law of cosines we can get the value of $BC$

and from the law of sines we can prove that $sin(ABC)=\frac{AC \times sin(50)}{ِAB }$ and we can get it easily

This is from the 1996 USAMO .