Answer:
Since we know that ΔPQR is a right triangle, we can also asume that:
sin R = cos P = 3/5
So the answer is (d).
* This formular can also be applied to other right triangles.
In a right triangle, sine of one acute angle will always be equal to cosine of the other acute angle.
And we can check this by actually finding cos P using the lengths of the sides, by calculating PR first:
PR = √(PQ² + RQ²) = √(12² + 16²) = 20
=> cos P = PQ/PR = 12/20 = 3/5
