# $P$ 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

• As usually asked for "raw questions" without any comment of yours : have you worked on this question ? Commented Oct 7, 2019 at 16:25
• @JeanMarie I attempted to as I showed above but unfortunately I didn’t see anything else I could do Commented Oct 7, 2019 at 17:18
• I suggest producing $BC$ to $E$ so that $DE\cong BD$
– dfnu
Commented Oct 7, 2019 at 18:32

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,

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

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

• Do you have any ideas for proving it without trigno using constructions and geometry only Commented Oct 7, 2019 at 18:42
• @AkshajBansal - Just found a geometric answer. Commented Oct 7, 2019 at 19:53