Question

Can you please explain how to write the ratios for sin x and cos x? When angles are x, y, and z. The hypotenuse being 12 and is between x and y. The other leg being x to z and 5. The last leg being z to y and sqrt 119.

Answer 1

Your triangle has acute angles X and Y, and right angle Z.

For an acute angle A in a right triangle:

The sine is the ratio of the opposite leg to the hypotenuse.
sin A = opp/hyp

The cosine is the ratio of the adjacent leg to the hypotenuse.
cos A = adj/opp

The hypotenuse of a right triangle is the side opposite the right angle. It is the longest side of a right triangle. There is only one hypotenuse in a triangle, so there is no confusion with the hypotenuse.

The two sides that form the right angle are called the legs. Each leg is opposite an acute angle. The legs may or may not be congruent to each other, but each leg is always shorter than the hypotenuse. Since there are two legs, we need to be able to distinguish them. If you take an acute angle as your angle of interest, the leg that is part of the angle is called the adjacent leg. The other leg is the opposite leg. Adjacent leg and opposite leg are relative terms. They depend on the acute angle you are considering.

For your triangle, if you look at angle X, then the adjacent leg is side XZ. The opposite leg for angle X is side YZ.

Using the ratios mentioned above for sine and cosine, you get:

sin X = opp/hyp = sqrt(119)/12

cos X = adj/hyp = 5/12