WEBVTT
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Let's look at how we can solve an oblique triangle. An oblique triangle is a triangle that does not have
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a right angle, so we don't have the Pythagorean theorem to look at, and we don't have the sine is opposite
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over the hypotenuse and that sort of thing. So we have other tools to use, and one of the tools we can use is the
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law of sines. In any triangle you have six parts, and you have these ratios that will always be true. The
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sine of alpha is to a as the sine of beta is to b as the sine of gamma is to c, or you can use the reciprocal.
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Now, one thing that students tend to have trouble with is when do they use the law of sines. So I'm going
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to show you a technique that you can use. If you set up a little chart with your sides and your angles listed and
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then you check off or fill in the material that you're given, the values that you're given, then you'll know.
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Anytime you have one side and the angle opposite that side you have a ratio. You can set up one of these
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ratios, and then you can find the other missing parts. Anytime you have a side and the opposite angle you
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can use the law of sines. There are actually three cases. We have the angle-side-angle, side-angle-angle, and
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side-side-angle. These are the cases that you use the law of sines to solve the triangle.
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Let's solve a triangle using the law of sines. First thing you always want to do is you want to organize your
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data. Set up a little chart here. You're going to find that you'll be able to solve these much more efficiently
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because you'll immediately have everything in order. You can see what you have, what you can find
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quickly, and we'll go from there. So what we have here is we know that c is equal to 65. We're given that
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alpha is equal to 46 degrees and beta is equal to 20 degrees. Now, you might say, well, I don't have a
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ratio here, so how am I going to solve using the law of sines because you have to have a ratio. But we have
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two angles, and anytime you have two angles in a triangle you find the third angle very quickly because
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the sum of the angles is 180 degrees. So we're going to know that gamma is going to be 180 degrees minus
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46 degrees minus 20 degrees. So we have an exact value for gamma. It's 114 degrees. So that's one of our
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missing parts. So now we have that ratio that we said we needed to use the law of sines. So we're going
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to find sides a and b by setting up the ratio a is to the sine of 46 degrees. Now, remember we use what we're
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given, so we have that 65 is to the sine of 114 degrees. So a is equal to 65 times the sine of 46 degrees
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divided by the sine of 114 degrees. We'll calculate that in a minute, and we're going to do the same thing
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for side b. So we have this. Now we know that b is to the sine of 20 degrees as 65 is to the sine of 114 degrees,
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so we can solve for b. [Side] b is 65 times the sine of 20 degrees divided by the sine of 114 degrees. Now we can
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find a decimal approximation for these two, and we solve our triangle. Alright, make sure again that you
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have it in degree mode because if you've been working in radian mode and degree mode you might want
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to make sure that you double check that because you could get seriously wrong answers. Divide by the sine
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of 114 and we get that a is approximately 51.18, and now we'll find b. So multiply 65 times the sine of 20
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divided by the sine of 114, and oops, I got the wrong answer there. I can see that. I got sine sine in there,
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so we'll redo it. So we have 65. Part of the thing is in any kind of math you're going to get the wrong
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answer. The trick is to know when you've got the wrong answer and check your answers again. So it's not
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that we don't make mistakes. It's just knowing when you do. Alright, so this one works out better. This is
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approximately 24.34. Now I want to point out something in a triangle. I want you to notice this because
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this can help you in other triangles when you're solving problems make sure you've got your data
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organized. You'll notice that the largest side is always going to be opposite the largest angle. The
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smallest side is opposite the smallest angle. So you can kind of give a visual check there and make sure
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that you've written it down. You might find the right answer and then write it in the wrong place. So
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if you kind of just do a visual check of that, you'll see that you've got everything as it should be.
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So now we have solved that triangle.
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Now we're going to look at something called the ambiguous case within the law of sines, and this is when
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you have side-side-angle. You have an angle, the side opposite that angle, and one other side. If you set it
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up in your little chart, if you look at this, and you have the three sides and the three angles. If you're given
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one side and the angle opposite that and either of the other two sides, it doesn't matter, and this can be in
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any order, then that's what we call the ambiguous case. Now we call it that because you might go back and
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remember about inverse sines. We know that on your calculator if you enter sine inverse of 1/2 it's going
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to give you an answer of 30 degrees, but you have to remember that the sine of 150 degrees is also equal
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to 1/2. Your calculator's not going to tell you that, but 30 degrees and 150 degrees could both be angles in
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a triangle. They're both less than 180 degrees. So this creates what we call the ambiguous case. Now the
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ambiguous case has three outcomes possible. You can have zero triangles, you can have one triangle, or
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you can have two triangles, but I'm going to show you how you can find that out very quickly. We're going
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to continue with this technique, and now we'll solve an example using the law of sines, ambiguous case.
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Alright, as we always do, we're going to take the triangle and the information we're given, and we're
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going to put it in our chart. So we're told that a is equal to 3, b is equal to 2. We don't have a value for c.
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We're given that alpha is equal to 40 degrees, and then we have to find beta and gamma. Now, this is the
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ambiguous case. Remember, how do you know it's the ambiguous case? You're given one side and the
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angle opposite that side and one of the other two sides. So now, we look at this and we say, "Alright,
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what's the first thing I can find?" Well obviously I don't have any way of finding c or gamma, but I can use
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the law of sines to find beta. So I'm going to set up my ratio, and we're going to find beta. I know that the
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sine of beta is to 2, which is side b, as the sine of 40 degrees is to 3. Alright, now I'm going to solve for beta.
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This will take a little algebra here. We know that the sine of beta is 2 times the sine of 40 degrees divided
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by 3, so beta is sine inverse of 2 times the sine of 40 degrees divided by 3. So we're going to find that value,
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and be careful when you put this in your calculator. Make sure you put parentheses around this whole
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expression because otherwise you would definitely get a wrong answer. So we have sine inverse of 2
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times the sine of 40 divided by 3, and we get 25 point. Well I guess we'll just round it off to one decimal,
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so that would be 25.4 degrees, so beta is approximately 25.4 degrees. Now, I know I have at least one
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triangle because I found beta. So the question is how do I know if there's one triangle, just this one, or if
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there are two triangles. So here's what you do. Remember that the sine of 25.4 degrees is the same as the
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sine of the supplement to 25.4 degrees. If I find the supplement of 25.4 degrees, so we have 180 minus 25.4,
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that's 154.6 degrees. So the sine of 154.6 degrees is the same number in here. So the question is, okay, can
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that work in this triangle? Well, what would happen if I substituted 154.6 degrees in for this? Let me just
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erase this here, and let's put it here, and let's see what would happen. So I'm saying, ok, I found the other
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angle and I'm good to go. But now I look and I see, oops, I've got 154.6 degrees plus 40 degrees is greater
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than 180 degrees, and the sum of all of them can only be 180 degrees. So we know that that is not going to
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be the case and that the only value for beta in this particular set of data that we're given is 25.4 degrees,
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and so now we know we only have one triangle and we're going to find the other two missing parts.
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Well we can't find c. Remember a squared plus b squared is not equal to c squared, so we're going to find
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gamma using the fact that the sum of the angles of a triangle is 180. So gamma, actually we'll put
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approximately because we're approximating this. So gamma is approximately 180 minus 40 minus 25.4,
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so that's 114.6 degrees. Now we can find the third side, c. Now remember, don't use beta to find c because
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you've already approximated here. You've already approximated here. Use what you're given, more
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exact values. You'll get a better approximation. You're less likely to make a roundoff error. So we know
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that c is to the sine of 114.6 degrees as 3 is to the sine of 40 degrees, so we can solve for c now. So c is
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3 times the sine of 114.6 degrees divided by the sine of 40 degrees. So we'll calculate that, and we get
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4.24. Now let's go back and double check. Remember we talked about the fact that the largest angle is
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opposite the largest side, and we see that this is the largest angle, the largest side, and that the ratios hold.
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So now we have found our triangle. We've discovered that in this particular set of measurements
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there was only one triangle, so we have solved this oblique triangle.
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Let's take a look at another example of the ambiguous case. How do I know it's the ambiguous case?
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Because I have one side, a given angle opposite that side, and one other side. So the first thing that we're going to
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do is we're going to try to find alpha because that's all really that I can do. So I'm going to set up my
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trig ratios for the law of sines. The sine of alpha is to 2 as the sine of 60 degrees is to 1, so the sine of alpha
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is 2 times the sine of 60 degrees. Now you might already see what's happening here, but just in case
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you don't, or it's an angle you're maybe not familiar with, alpha is sine inverse of 2 times the sine
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of 60 degrees. This is a good chance to review one of our exact value angles. Sine of 60 degrees is the
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square root of 3 over 2. So what am I asking here? I'm asking, "What angle has the sine square root of 3?"
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No angle has the sine square root of 3. Remember sine is always between 1 and negative 1, so this angle
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does not exist. It does not exist and so there are zero triangles given this particular set of measurements.
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Now let's do an example with the ambiguous case that has two triangles, and I want to show you how you
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can discover that there are in fact going to be two triangles. Just from looking at this data you can't tell.
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We're going to continue just like we did in the past. The first thing we're going to do is we're going to
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find this first angle gamma. So we know, from the law of sines, that the sine of gamma is to 10 as the sine
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of 35 degrees is to 6. So we have that the sine of gamma (oops I wrote alpha, not gamma). The sine of gamma
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is 10 times the sine of 35 degrees divided by 6, so gamma is sine inverse of 10 times the sine 35 degrees
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divided by 6. So now we're going to find this first gamma. Again, you have to be careful when you're
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putting in your values. Make sure you put parentheses around that numerator. So we're going to have
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sine inverse of 10 times the sine of 35 divided by 6, and we get a value of 72.9 degrees. Let's make that
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approximate. Alright, 72.9 degrees. So we know that we have this triangle. Gamma is approximately
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72.9 degrees. Now remember there are two angles less than 180 degrees that have this same sine, so we're
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going to find, we'll call this one gamma one, so we're going to find gamma two. So gamma two is 180 degrees
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minus 72.9 degrees, and that's going to be 107.1 degrees. We'll just do one decimal place. So here's the question.
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Can we have two triangles? Well, if instead of 72.9 degrees I wrote in 107.1 degrees, the question is
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could that work? Yes, it can because 35 plus 107.1 degrees is less than 180 degrees. Now I know I have
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two triangles, so we're going to make another chart here with our data. So we'll put back here, we have that
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this is 72.9, and now this is going to be triangle one, and we're going to have triangle two. So we know that
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a is 6. I'm going to call that b one and this beta one and that gamma one just to keep myself straight so
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I don't get confused. So this will be b two. [Sides] a and c have the same values, so that doesn't change.
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Alpha is 35 degrees. This will be beta two, and this will be gamma two which we've already found. So that's going
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to be approximately equal to 107.1 degrees. Now we have our two triangles set up, and it's really not going
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to be very difficult to solve for these missing parts. We can find these two angles by subtracting from 180
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degrees. So we know that beta one is approximately 180 degrees minus 35 degrees minus 72.9 degrees.
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Alright, so we have 180 minus 35 minus 72.9, and that's going to be 72.1 degrees. They're kind of close,
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and we'll talk about what happens there in a minute. So then beta two is approximately 180 degrees
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minus 35 degrees minus 107.1 degrees, and let's see what that is. That's 37.9, so this is approximately
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37.9 degrees. We have found everything but these two sides. Now I want to show you something here,
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and this is kind of a way you can... You know there's always ways that you can check yourself and maybe
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not know exactly, but you can tell if you're kind of going in the right track. Remember we talked about the
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fact that the largest angle is opposite the largest side, and they kind of have these ratios. You'll notice that
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these two angles are really close in value, so if I do my work correctly, I would expect that these two
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sides should be really close in value, and you've got the same thing down here. This is 35. This is 37, not
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quite as close as they are, but that's going to kind of give you an idea that b two is going to be just a little
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bit bigger than 6, and this one is going to be just a little bit less than 10. So that will kind of help you when
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you're doing your calculations. Make sure you don't put something in wrong. So let's find b one. [Side] b one is
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to the sine of 72.1 degrees, now remember once again use what you're given not what you find, so as
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6 is to the sine of 35 degrees. So b one is 6 times the sine of 72.1 degrees divided by the sine of 35 degrees.
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Alright, so let's see what that is. 6 times the sine of 72.1 divided by the sine of 35, and that's 9.95, and that's
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what I expected. I wanted it to be close. Now we'll find b two. Again, we're going to use what we're given,
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not what we've found, so we set up that b two is to the sine of 37.9 degrees as 6 is to the sine of 35 degrees.
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So b two is 6 times the sine of 37.9 degrees divided by the sine of 35 degrees. Let's calculate that.
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6 times the sine of 37.9 divided by the sine of 35 is 6.43. Again, that's what I expected. I wanted those to
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be close because the angles are close, so that's just kind of a visual check you can give yourself.
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So now we have solved the law of sines case where we have two triangles.