00:01
This problem wants us to use triangle pqr with right angle r to find sine of p, cosine of p, and tangent of p, as well as sine of q, cosine of q, and tangent of q.
00:09
And we are supposed to express each ratio as a fraction and as a decimal to the nearest hundredth.
00:14
So first to set up all of our fractions for all six of our problems, we need to remember what sine, cosine, and tangent represent in our right angle based off of our angle we're focused on.
00:24
And for our first three problems, since we're focused on angle p, we will find sine of p by remembering that sine is the opposite side of the hypotenuse side in comparison to p.
00:35
So in comparison to p, 12 is our opposite side and our hypotenuse is always across from our right angle.
00:41
So that means 37 is our hypotenuse.
00:44
So sine of p would be equal to 12 over 37.
00:50
And then for cosine of p, that's the adjacent of the hypotenuse.
00:54
So in comparison to p, adjacent would be the side next to it that's not the hypotenuse, which is 35, still over the same hypotenuse, 37.
01:02
And then for tangent of p, that will be the opposite side over the adjacent side.
01:07
So that's our opposite 12 over our adjacent 35.
01:12
And that sets up all of our fractions.
01:13
And now we're going to focus on our q angle for sine of q, cosine of q, and tangent of q, which means we're going to see a lot of the same sides used, but just in different combinations, because now we're focused on the q angle instead of p.
01:25
So sine is still opposite over our hypotenuse.
01:28
So opposite of q would be 35...