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Suppose ABC is a right triangle with sides of lengths a b and c and the right angle at C find the unknown side length using the Pythagorean Theorem, and then find the values of the six trig function for angle B.

11) a= 5 b= 12

c= 13 so how do I find the angles? I know I have 90 degree + a + b = 180 but that is it.

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    In a right triangle, the values for the six standard trig functions of an angle can be computed as the ratios of certain sides.2011-06-05

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In general, when we talk about a triangle and we've named the vertices with capital letters (your $\triangle ABC$ has vertices $A$, $B$, and $C$), we name the lengths of the sides opposite each vertex with the lower-case version of the letter labeling that vertex (the side opposite $B$ in your triangle is called $b$, which is the side with endpoints $A$ and $C$).

Since you've talked about $x$, $y$, and $r$ in other comments, I think you may be working with trigonometry on a coordinate plane. Let's try drawing a picture of the situation. We're talking about finding a trig function of angle $B$, so we want to put $B$ at the origin. We want the right angle, which is at $C$ to be on the $x$-axis. Since $a=5$ is the length of the side between $B$ and $C$, let's put $C$ at $5$ on the $x$-axis. With a right angle at $C$, point $A$ will be directly above or below $C$. Since $b=12$ is the length of the side between $A$ and $C$, let's put $A$ at $(5,12)$, directly above $C$.

Here's the picture so far:

diagram as described above

Given this picture, do you know what the $x$, $y$, and $r$ are for finding the trig functions of angle $B$?

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    This makes more sense, I think x would be 5, r would be 13 and y would be 12?2011-06-05
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    @Adam: Yes. This is sometimes called "standard position" for the angle $B$—drawing it with a right triangle where the right angle is somewhere on the $x$-axis. Can you find the six trig functions of $B$ now?2011-06-05
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    I think so, sin= 12\13 cos=5/13 tan=12\5?2011-06-05
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    @Adam: Yes, and the other 3 trig functions?2011-06-05
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    I should be able to just reverse them so csc= 13/12 sec=13/5 and cot=5/122011-06-05
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    @Adam: That's right. Now, if you have $\triangle DEF$ with right angle at $D$, $e=7$, and $f=24$, what is $\sin F$?2011-06-05
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Use the Law of Cosines. (Technically, you could also use the Law of Sines, but given your starting data the Law of Cosines would be easier to use).

Edit: Or, if you didn't want to use dynamite when a pick-axe would be better, just use the basic trig functions and solve for the ratio of the sides.

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    sort of overkill in a *right* triangle2011-06-05
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    I can't read the wikipedia article, it doesn't make sense. Although if I am correct its as simple as stating that all 90 degree angles are sin=1 tan= undefined and cos=0 correct?2011-06-05
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    @Adam, I don't think $B$ is the right angle in this problem. The right angle would most likely be denoted $C$, and $B$ is opposite $b$, not $c$, if my speculation is correct.2011-06-05
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    That is correct, so how do I figure out what the angle of b is?2011-06-05
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In a right triangle, $\sin$ and $\cos$ are paricularly easy to calculate, since in this case $\sin B$ is just the ratio of the opposite over the hypotenuse, and $\cos B$ is the ratio of the adjacent over the hypotenuse. To find $\tan B$, $\cot B$, $\csc B$, $\sec B$, just express them in terms of $\sin B$ and $\cos B$ and simplify.

I assume $B$ is the angle opposite $b$, so if you draw your triangle out, you should see $\sin B=\frac{12}{13}$. The rest follows quite like this.

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    I don't get why I can't just use the values of x y and r aren't they all defined in this triangle?2011-06-05
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    Also I don't get how to determine what side is x and what is y.2011-06-05
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    @Adam, you don't need to find the angle $B$ to calculate the value of the trig functions as the argument, which is unnecessary for this problem. If you want to know, $B=\arcsin(12/13)\approx 67.38^\circ$, but I don't think it's much help for this problem. And what are $x$, $y$, and $r$?2011-06-05
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    What is an arcsin? x is 12 because it is the adjacent, 13 is r because it is the hypotenuse and 5 is y because it is the opposite. What are the answers then? Am I wrong? I wish the idiots who wrote this book would take some consideration and put at least some of the answers in the book.2011-06-05
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    @Adam, forget I mentioned $\arcsin$ at this point. Actually, $x$ is 5, since the leg $a$ is adjacent to the angle $B$, and $y$ is 12 since leg $b$ is opposite $B$. The adjacent and opposite legs are dependent on which angle you're referring to in a triangle, there isn't a fixed adjacent leg and opposite leg.2011-06-05
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    So sin is 12/13 ?2011-06-05
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    @Adam Yes, $\sin B=12/13$. Remember to specify what angle you're putting into the sine function.2011-06-05
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    How do I determine what side as A B and C also? Is it somehow related to the angles? Also what is angle B and how is that determined?2011-06-05
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    @Adam The uppercase letter is used to denote the angle opposite the side with the lower case letter. So angle $B$ is opposite $b$.2011-06-05
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    Opposite clockwise or counter clockwise? Like I can get angle C because that is 90 so the big long line across from it is line c but how do I fimnd ab and AB2011-06-05
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    @Adam Opposite means the third leg of the triangle that doesn't form the angle. Angle B is formed by $a$ and $c$, so $b$ is the opposite leg.2011-06-05
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    I don't get that at all really. Do I just need to memorize these values as well? Pretty frustrating, guess I need to write it down.2011-06-05
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    Okay I understand this, but how do I find sin?2011-06-05
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    Remember, sin B is opposite over hypotenuse. Once you know which angle B is, (you don't have to know it's value, just where it is in your triangle), then find which leg is opposite, and which leg is the hypotenuse. Then take their ratio to find sin B.2011-06-05
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    I am trying to solve this problem but I am having trouble defining the line and angles again. I know that C is 90 degrees, and C is made up of A and B and C is across from the angle C. How do I know where A and B go though since they can go on either side of angle C?2011-06-05
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    Generally, $A$ will be opposite the side $a$, so $A$ will be made up of $b$ and $c$. Likewise, $B$ will be opposite $b$, and thus is made up of $a$ and $c$. Try drawing a picture, it helps a lot.2011-06-05
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    I am drawing it, not helping. I can get angle C and line C but the other two are a toss up since nothing defines them for certain.2011-06-05
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    Just put down $a$ and $b$, it doesn't matter. Then $A$ and $B$ will be determined by which sides you chose to be $a$ and $b$.2011-06-05
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    But A and B are defined by numbers, so I get different answers. Like if I made a on the left side I get sin = 5/13 or 12/13 if I go it the other way.2011-06-05
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    You're already given that $a=5$ and $b=12$, so that will be the same whichever way you decide to name the legs of your triangle. The angles are then determined by which leg is which, i.e., $A$ is opposite $a$, and $B$ is opposite $b$.2011-06-05
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    So how do I know which to label a and which b?2011-06-05
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    It doesn't matter. Just draw a right triangle with hypotenuse $c$, and right angle $C$. You then have two legs. Just choose one and call it $a$. So $a=5$. Then the angle opposite $a$ is $A$. The remaining leg is $b$, with opposite angle $B$. If you had chosen to instead call this leg $b$, then the opposite angle would be $B$. Draw this twice, once naming the leg $a$, and once naming it $b$, and finish the drawing. You'll notice at the end that no matter what you do, the triangle ends up the same if you just flip it over. In the end, it's the same triange.2011-06-05
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    I think I get it, if I flip the triangles around in circles and stuff they are the same. The values will always be the same though? As long as it is orientated on angle b?2011-06-05
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    Kind of yes, the triangle can be flipped and turned around but the legs and the angles are all the same relative to each other. Since all the legs and angles move around in the same way, the values will always be the same, so it doesn't matter how you label really.2011-06-05