If in a triangle ABC, angle $ABC = \frac{\pi}{2}$ and sides AB=12 and BC = 5, then the value of sinC + cosA is:

Correct Answer: (A) $\frac{24}{13}$

In right-angled triangle ABC with $\angle$ B = 90^$\circ$, AB = 12 and BC = 5, the hypotenuse AC = $\sqrt{12^2 + 5^2}$ = 13 by Pythagoras theorem. Then $\sin$ C = $\frac{AB}{AC}$ = $\frac{12}{13}$ and $\cos$ A = $\frac{AB}{AC}$ = $\frac{12}{13}$, so $\sin$ C + $\cos$ A = $\frac{12}{13}$ + $\frac{12}{13}$ = $\frac{24}{13}$.