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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. –  yunone Jun 5 '11 at 18:08
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3 Answers

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? –  Adam Jun 5 '11 at 20:37
    
@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? –  Isaac Jun 5 '11 at 20:39
    
I think so, sin= 12\13 cos=5/13 tan=12\5? –  Adam Jun 5 '11 at 20:43
    
@Adam: Yes, and the other 3 trig functions? –  Isaac Jun 5 '11 at 20:45
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I should be able to just reverse them so csc= 13/12 sec=13/5 and cot=5/12 –  Adam Jun 5 '11 at 20:46
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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? –  Adam Jun 5 '11 at 18:41
    
Also I don't get how to determine what side is x and what is y. –  Adam Jun 5 '11 at 18:47
    
@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 $r$? –  yunone Jun 5 '11 at 18:48
    
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. –  Adam Jun 5 '11 at 18:53
    
@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. –  yunone Jun 5 '11 at 19:03
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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 triangle –  GEdgar Jun 5 '11 at 18:20
    
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? –  Adam Jun 5 '11 at 18:23
    
@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. –  yunone Jun 5 '11 at 18:32
    
That is correct, so how do I figure out what the angle of b is? –  Adam Jun 5 '11 at 18:36
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