Will Angle Chasing Work?

We have $\angle ABD = 30^{\circ}$ and $\angle ACD = 60^{\circ}$ . Let the angle bisector of $\angle ACD$ intersect $BC$ at $P$ . Then, $\angle ACP = \angle ABP = 30^{\circ}$ , so $ABCP$ is cyclic. Since $CP$ and $BD$ are both angle bisectors of $\triangle ACD$ , they intersect at the incenter of $\triangle ACD$ , which is $P$ . Thus, $\angle CAP = \angle CBD = 10^{\circ}$ and $\angle BCA = 60^{\circ}$ . Finally, $x^y = 60^{10}$ , which has digit sum $\boxed{36}$ .

B is the intersection of the external angle bisector of angle CAD and the angle bisector of angle ADC. Thus B is the centre of the excircle opposite D. Thus BC is the external angle bisector of angle ACD, and angle BCA is 60°. Thus

$x=60, y=10, x^{y}=604661760000000000$ which has sum of digits 36.

Draw angle bisector of $\angle{BCD}$ meeting BD at E

Now.

Since $\angle{ABE} = \angle{ACE} = 30$

Therefore ABCE is cylic

Also since angle bisectors of $\angle{D}$ and $\angle{ACD}$ meet at E.

Therefore E is the incentre of $\triangle ADC$

So,

$\angle{FAE} = 10$

This gives,

$x = ( 180 - 90 ) - 30 = 60$

So

$y = 10$

Now,

$x^y = 60^{10} = 6^{10} \times 10^{10} = 60466176 \times 10^{10}$

Therefore ,

Sum of digits = $\boxed{36}$

AXD = 180 - (DAC + BBA) = 110 (Angles in a triangle make 180)

BXC = AXD = 110 (Opposite Angles)

AXB = 180-BXC = 70 (Angles in a straight line)

CXD = AXB = 70 (Opposite Angles)

ABD = 180 - (CAB + AXB) = 30 (Angles in a triangle make 180)

I can see 2 methods for getting reming 2 angles :

Construction or Sin Rule (Both require a random length for starting point which is fine as it is only angles we care about)

I went with construction and got BCA = 60 & DBC = 10

60^10 = 6^10 * 10^10

6^10 * 10^10 = 60466176 * 10000000000

==> 6+0+4+6+6+1+7+6 +(10*0)= 36