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Today we're working on the review for the chapter 17 18 and 19 test that's coming up this week question number one asks us to identify the sine cosine tangent cosecant secant and cotangent of the reference angle theta so remember it's helpful to start by circling your reference angle and then for this one we're finding all six of them so I'm going to go ahead and label the five is opposite or across from the nightthe the theta that's indicated here so for this angle twelve is adjacent to that and then the thirteen is my hypotenuse so labeling those on the picture will help me when I go through and set up these ones over here remember I'm going to use the acronym sohcahtoa to help me set them up so from angle theta sine would be opposite over hypotenuse and so that's why the five goes in the top and the thirteen goes in the bottom cosine would be adjacent over hypotenuse so 12 over 13 and then tangent would be opposite over adjacent so five over twelve cosecant is the flip of sine secant is the flip of cosine and then cotangent is the flip of tangent and so you just take these three fractions that you find over here and you flip them question number two on the review gives me some information so it says the sine of angle theta is equal to seven over eight and so this is my reference angle theta a triangle helps you with this picture or trying drawing a triangle helps you solve this problem it's not a required step but it's recommended so sine is opposite over hypotenuse and so no matter which way you draw your triangle the seven should be opposite or across from whichever angle you label is theta and then eight should be on your hypotenuse which is the side of the triangle that's opposite from the 90-degree box we're being asked to find cosine of theta and so remember with sohcahtoa for sine we used opposite in hypotenuse but for cosine we need to use adjacent and hypotenuse so we need to solve for this side of the triangle over here that's saying to do that I'm going to use the Pythagorean theorem seven can go in for your a or for your B but your C squared the eight has to go in for your C because that's your hypotenuse and so to solve this I'll minus the 49 for both sides and so going through and solving it amounts to solving this when I take 64 take away 49 gives me 15 and then to solve for that missing side of the triangle I'll take the square root of both sides so I'll get that the missing side of the triangle is the square root of 15 and you do want to leave it in its square root you're not going to take it out from under its square root I was given sign but I wanted to find cosine and so for this one the cosine of angle theta remember cosine is adjacent over hypotenuse so the 15 comes over here the square root of 15 and that becomes my adjacent side and so for cosine I'm gonna have the square root of 15 in the top and then 8 in the bottom so adjacent over my hypotenuse will be the answer for that question questions over 3 & 4 have the same...
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