geometrytriangles
Considering the attached image, is that possible to compute for $\angle x$ in the isosceles triangle below, with no further known values?
$\\$I have found $\angle B$ and $\angle C$ are $50^o$. So $\angle B$ is divided by $\overline{BP}$ to $10^o$ and $40^o$. But I can't find how the $\angle C$ is divided by $\overline{CP}$. Also its clear that $\overline{BP}$ is perpendicular to leg $\overline{AC}$.
I saw the following solution may years ago:
On side $AD$ construct in exterior equilateral triangle $ADE$. Connect $BE$.
Then $AB=AC, AE=BC, \angle BAE=\angle ABC$ gives $\Delta BAE =\Delta ABC$ and hence $AB=BE$.
But then $$AB=BE, BD=BD, DA=DE \Rightarrow ADB =EDB$$
Hence $\angle ADB=\angle EDB$. Since the two angles add to $300^\circ$ they are each $150^\circ$. Then $\angle ABD + \angle ADB+ \angle BAD=180^\circ$ gives $ABD=10^\circ$.
The problem evidently intends the base to be the third side that is not necessarily equal to the other two.
The height of the perpendicular, then, is $15$, so the area is indeed $$\frac12\cdot 30 \cdot 15 = 225$$
You have assumed that the base and one of the other sides constitute the pair of equal sides.
I think your interpretation isn't incorrect, but is not what the problem intended.
Best Answer
Construct equilateral triangle $ABX$ as in the picture.
Note that the concave angle $BXA$ equals $300^\circ = 2\cdot 150^\circ = 2\angle BPA$, hence $P$ lies on the circle with center $X$ and radius $XA=XB$. It follows that $\angle PXB = 2\angle PAB = 40^\circ$.
On the other hand, $AB=AX=AC$, so $A$ is the circumcenter of triangle $BXC$, hence $\angle CXB = \frac 12 \cdot \angle CAB = \frac 12 \cdot 80^\circ = 40^\circ$.
Since $\angle PXB = 40^\circ = \angle CXB$, points $X, P, C$ are collinear.
Since $AC=AX$, we have $\angle PCA = \angle AXP = 2\cdot \angle ABP = 20^\circ$, hence $$\angle APC = 180^\circ - \angle CAP - \angle PCA = 180^\circ - 60^\circ - 20^\circ = 100^\circ.$$