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

$\frac{12}{60}$

$\frac{12}{61}$

$\frac{61}{60}$

$\frac{60}{61}$

Solution

The correct answer is

$\frac{60}{61}$

Explanation

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

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

sin B =

$\frac{Perpendicular}{Hypotenuse}$ =

$\frac{AC}{BC}$ =

$\frac{60}{61}$