$P$ is a point inside triangle $ABC$ such that $\angle PBC=30°,\angle PBA=8°$ and $\angle PAB=\angle PAC=22°$. Find $\angle APC$, in degrees. $P$ is a point inside triangle $ABC$ such that $\angle PBC=30°,\angle PBA=8°$ and $\angle PAB=\angle PAC=22°$. Find $\angle APC$, in degrees.
If you continue the $AP$ so that it meets $BC$ at $D$ you get that angle $DBP$ = $BPD$ therefore triangle $BDP$ is isosceles but I don’t know what to do next, hints and solutions would be appreciated
Taken from the 2009 IWYMIC 

 A: Let $\angle APC = \alpha$ and use the sine rule for the triangles ABP and ACP with the common side AP,
$$\frac{BP}{CP}=\frac{\sin(158-\alpha)}{\sin 8} \tag{1}$$
where $\angle ACP = 180 - 22 - \alpha = 158 - \alpha$ is used. Also, use the sine rule for the triangles BPD and CPD with the common side DP,
$$\frac{BP}{CP}= \frac{ \sin(\alpha-60)}{\sin 30}\tag{2}$$
where $\angle PCD = \alpha - \angle PDC = \alpha - (30+\angle BPD) = \alpha - 60$ is used. Combine (1) and (2) to get
$$\sin(158-\alpha)=2\sin 8 \sin(\alpha -60)$$
Rewrite both sides as,
$$\cos(\alpha-68)=\cos(68-\alpha)-\cos(\alpha-52)$$
which yields $\cos(\alpha - 52) = 0$, or $\alpha = 90+ 52 = 142$. Thus,
$$\angle APC = 142^\circ$$
A: 
Here is a geometric derivation:
1) Construct CQ $\perp$ AD and connect QD. 
2) Because the right triangles AQX and ACX share the side AX and $ \angle QAX = \angle CAX=22$, they are congruent. In turn, the triangles QDX and CDX are congruent, which yields 
$$QD = DC\tag{1}$$
3) $\angle PBD = \angle BPD = 30$ makes the triangle PBD isosceles and,
$$ \angle PDC = \angle PDQ = \angle QDB = 60; \>\>\>\>\>BD = DP\tag{2}$$
4) Because of (1) and (2), the triangles BDQ and PDC are congruent, which yields,
$$\angle CPD = \angle QBD = 38$$
As a result,
$$\angle APC = 180 - 38 = 142^\circ$$
