In right $\triangle ABC$ with $\angle B = 90^\circ$, $BC = 6$ and $AB = 8$, then $\sin A = ?$

With $\angle B=90^\circ$, $BC=6$ is opposite to $A$ and the hypotenuse $AC=\sqrt{6^2+8^2}=10$, so $\sin A=\dfrac{BC}{AC}=\dfrac{6}{10}=\dfrac35$.