$\triangle ABC$ is isosceles with $\measuredangle BAC =80^\circ$. Find the measure of $\measuredangle APC$

Question

I have already found some angles, but I am not too sure on how to go on to the smaller triangles of $APB$ and $APC$. How do I keep going and what is the answer?

Answer 1

Apply the sine rule to the triangles ABP an ACP,

$$\frac{AP}{AB} = \frac{\sin10}{\sin(250-\theta)},\>\>\>\>\> \frac{AP}{AC}=\frac{\sin20}{\sin\theta}$$

where $\angle APB = 250 - \theta$. Given that AB = AC,

$$\sin10\sin\theta = \sin20\sin(250-\theta)$$

Simplify with $\sin20=2\sin10\cos10$,

$$\sin\theta = 2\cos10\sin(\theta-70)=\sin(\theta-60)+\sin(\theta-80)$$

Rearrange $$\sin(\theta-80) = \sin\theta -\sin(\theta-60)=2\sin30\cos(\theta-30)=\sin(120-\theta)$$

which yields the solution $\theta = 100$.

Answer 2

Using Cosine rule in $\Delta ABC$ find BC in terms of AB. Use Sine rule in triangle PBC find PC in terms of BC finally using Sine rule in triangle APC we get $\angle APC=100°$ For detailed solution I have done in below link you can check. https://www.mathsdiscussion.com/discussion-forum/topic/find-angle-apc/?part=1#postid-98

Answer 3

Draw $BM$ such that $\angle ABP=\angle PBM=10°$.
Now $BP$ becomes the bisector of $\angle ABM$. By angle chasing we can mark all possible angles:
$\angle BMP=60°$ and since $\Delta BMC$ is isosceles too, $AM$ is the bisector of $\angle BAC$ so $\angle MAC=40°$, $\angle AMC=120°$ and $\angle AMP=60°$.
Because $\angle BMP =\angle AMP = 60°$ therefore $MP$ is the bisector of $\angle AMB$.
In $\Delta AMB$, two of its bisectors $BP$ and $MP$ intersecting at $P$ therfore $AP$ is the bisector of $\angle BAM$. Thus $\angle PAM = 20°$ and $\angle APM=100°$

Answer 4

Another possible path

Draw the right-angled triangle $\triangle BQC$, in such a way that $\angle QBC = 20^\circ$. Let $R$ be the midpoint of $BQ$.

1. Show that $\square BQCP$ is cyclic.
2. Since $CR$ is median to hypotenuse $BQ$, it must be $CR \cong RQ$. Hence $R$ is the center of $\square BQCP$'s circumcircle.
3. Therefore $\triangle BPR$ is equilateral, and the triangles $\triangle BRC$ and $\triangle RCQ$ are isosceles.
4. It now suffices to show that $\triangle APB \cong \triangle APR$ (SSS criterion), to demonstrate that $\angle APR = 150^\circ$, hence $\angle APC = 100^\circ$.