In a △ ABC, AB = 11 and AC = 60. Find sin B, if ∠A = 90°

The correct answer is $\frac{60}{61}$

By Pythagoras theorem,
In a △ ABC, AB² + AC² = BC²

∴ BC = √(AC² + AB²) = √(3600 + 121) = 61

sin B = $\frac{Perpendicular}{Hypotenuse}$ = $\frac{AC}{BC}$ = $\frac{60}{61}$