$P$ 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.

Question

$P$ 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

Answer 1

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,

$$\angle APC = 180 - 38 = 142^\circ$$

Answer 2

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, $$\angle APC = 142^\circ$$