Sign, sin, sine

Refer to the diagram above for the solution.

Let $E$ be the point where the perpendicular produced from $C$ intersects $AB$ .

Let $F$ be the point where the perpendicular produced from $A$ intersects $BC$ .

Cut a perpendicular from $A$ to the line $DC$ . Let $G$ be the intersection of the perpendicular from $A$ and line $DC$ .

Observe that $\angle AGC = \angle AEC = \angle GCE = \angle GAE =90^{\circ}$

Thus we can conclude that $AECG$ is a rectangle. $\Longrightarrow AE=GC, AG=EC$

$AB=DC$

$AB-AE=DC-CG$

$EB=DG$

Also, $DA^2+AP^2=821^2$

$DA^2+(AE^2+EP^2)=821^2$

$EP^2+EB^2=700^2$

$DA^2-EB^2+AE^2=821^2-700^2$

$DA^2-DG^2+AE^2=121\cdot 1521$

$AG^2+AE^2=121\cdot 1521$

$EC^2+AE^2=121\cdot 1521$

$AC^2=121\cdot 1521$

$AC=\sqrt{121\cdot 1521}$

$AC=11\cdot 39$

$AC=429$