smc

Notice that $6^2+8^2=10^2.$ That makes us want to construct a right triangle. Rotate $△APC$ $60^{\circ }$ about A. Note that $△PAC\cong △P^{\prime }AB$, so $$\angle P^{\prime }AP=\angle PAB+\angle P^{\prime }AB=\angle PAB+\angle PAC=60^{\circ }.$$ Therefore, $△AP{P}^{\prime }$ is equilateral, so $P^{\prime }P=8$, which means $\angle P^{\prime }PB=90^{\circ }.$ Let $\angle B{P}^{\prime }P=\alpha .$ Notice that $\cos \alpha =\frac{8}{10}=\frac{4}{5},$ and $\sin \alpha =\frac{3}{5}.$ Applying the Law of Cosines to $△APC$ (remembering $\angle APC=\angle A{P}^{\prime }B$): $$\begin{array}{rl}A{C}^2& =10^2+8^2-2\cdot 10\cdot 8\cdot \cos (60^{\circ }+\alpha )\\ & =164-160(\cos 60\cos \alpha -\sin 60\sin \alpha )\\ & =164-160(\frac{2}{5}-\frac{3\sqrt{3}}{10})\\ & =164-16(4-3\sqrt{3})\\ & =100+48\sqrt{3}.\end{array}$$ We want to find the area of $△ABC$, which is $$A{C}^2\cdot \frac{\sqrt{3}}{4}=25\sqrt{3}+36\approx \boxed{79}.$$