Nagwa uses cookies to ensure you get the best experience on our website. Learn more about our Privacy Policy.

Video: The Law of Sines: The Ambiguous Case

Lauren McNaughten

Learn about the ambiguous case of the law of sines and why it leads to two possible solutions for other lengths and angles in a nonright triangle. Follow an example in a right-angled triangle and another in a nonright triangle.

10:43

Video Transcript

In this video, we’re going to look at the ambiguous case of the law of sines. Suppose you’re given some information about a triangle such as the lengths of two of its sides and the size of one of its angles. The ambiguous case occurs when that information doesn’t define a unique triangle, but in fact it’s possible to draw more than one triangle using the information given. More specifically, it occurs when the angle that you’re given is not included between the two sides.

So let’s look at an example then to clarify what we mean. I have here a triangle in which I have the lengths of two sides, five centimetres and six centimetres, and the size of one angle. And you’ll see that that angle is not included between the two sides, so it’s not between the sides of six centimetres and five centimetres. I’m asked to calculate the size of angle 𝐴.

So I’m gonna try to answer this problem then using the law of sines, so I need to recall what it tells me. Remember then that it tells me that the ratio between each side and the sine of its opposite angle is equal. Now I’ve chosen to write it in this format here where the sines of the angles are in the numerator because this particular question asked me to calculate an angle, so it will be slightly more straightforward to use this version than it will be to use the version where the sides are in the numerator.

So I’m going to write down then this law of sines using the information in this question. So I will have that sine of angle 𝐴 over six is equal to sine of fifty over five. This gives me an equation that I can solve in order to work out 𝐴. So the first step is to multiply both sides of this equation by six. And in doing so, I have that sine of 𝐴 is equal to six sin fifty over five. Now to work out angle 𝐴, I need to do sine inverse of both sides.

So I have that 𝐴 is equal to sin inverse of six sin fifty over five. Now I’m gonna use my calculator to evaluate this. And when I do, it tells me 𝐴 is equal to sixty-six point eight one seven or sixty-seven degrees to the nearest degree.

So, brilliant! Fantastic! We’ve finished the question you might think. But have a look at the diagram and have a look specifically at angle 𝐴. Now this diagram isn’t supposed to be deliberately misleading, and what you notice is that angle 𝐴 is an obtuse angle. So there’s no way that angle 𝐴 can be equal to sixty-seven degrees; it has to be somewhere between ninety and one hundred and eighty degrees.

So what’s gone wrong? Well nothing in the maths that we’ve done so far is incorrect. All of those steps that we followed are perfectly fine. It’s just those steps haven’t actually found us angle 𝐴 in this particular triangle because there’s another triangle we can draw that would use this same information.

If I extend the base of the triangle out in this direction here, and if I were then to take my compass and put it on point 𝐵 and set it to five centimetres and then draw a little arc of points, you see that actually it crosses this line again. And actually there’s another point on this baseline here that is still five centimetres away from 𝐵, and I’ll label that then as 𝐴 dash.

So I have two triangles I have triangle 𝐴𝐵𝐶, the original triangle, and I have triangle 𝐴 dash 𝐵𝐶. And both of those triangles have the same information in them. So in both of those two triangles, it’s true that the measure of angle 𝐶 is fifty degrees, 𝐵𝐶 is six centimetres, and either 𝐴𝐵 or 𝐴 dash 𝐵 depending on which triangle you’re talking about is equal to five centimetres. So this collection of information doesn’t uniquely specify a single triangle.

What all of this means then is that the angle we’ve just calculated, sixty-seven degrees, isn’t actually angle 𝐴. It’s angle 𝐴 dash which we can see is an acute angle. So we have this angle here. However, we can use this to work out angle 𝐴. If you look at the triangle formed by 𝐴, 𝐵, and 𝐴 dash, you’ll see that that’s an isosceles triangle because it has those two sides of the same length, both five centimetres.

What that means then is that the other base angle must also be equal to sixty-seven degrees. And then we can work out angle 𝐴 that we’re looking for by doing a hundred and eighty minus sixty-seven as you’ll see those two angles sit on a straight line together.

So subtracting sixty-seven from a hundred and eighty gives us a value of one hundred and thirteen degrees, and this is then the size of the angle 𝐴 that we were looking for. This is illustrative of a general property of the sine ratio, which is that for an angle 𝜃 between zero and a hundred and eighty degrees, sine of 𝜃 is in fact equal to sine of a hundred and eighty minus 𝜃. This relationship always holds true for supplementary angles.

So in this question, we applied the law of sines correctly and we got an answer for angle 𝐴, but then we saw that our answer couldn’t actually be the correct value given the context of the diagram and the fact that angle 𝐴 is an obtuse angle. If we hadn’t been given the diagram to refer to and instead we’ve just been given the list of information in green, then either these two values for 𝐴 would be valid. So 𝐴 could either be sixty-seven degrees or a hundred and thirteen degrees, and we’d have to provide two possible answers to this problem.

This problem tells us 𝐴𝐵𝐶 is a triangle where the measure of angle 𝐵 is a hundred and ten degrees, side 𝑏 is sixteen centimetres and 𝑐, side 𝑐, is twelve centimetres. How many possible solutions are there for the other lengths and angles? Now the inclusion of the word possible here tells me there a couple of situations that could exist. It could be that the information I’ve got doesn’t actually describe a triangle at all and it’s not possible to draw a triangle that fulfills those requirements.

Or it could be that that information describes a unique triangle, in which case is only one possible solution for the other lengths and angles. It could also be that this is an example of the ambiguous case of the law of sines and in fact there are two possible solutions for the other lengths and angles.

So I’m gonna start off with a sketch of what triangle 𝐴𝐵𝐶 could look like. Now as I said, it may be the case that it doesn’t exist at all or it may be the case that there’s more than one triangle I could draw. But I’ll start off with just one idea. So triangle 𝐴𝐵𝐶 could look like this, and I’ve put all the relevant information onto my diagram. Now I’m gonna start off by trying to calculate angle 𝐶, and I’m going to use the law of sines for that.

Remember this is what the law of sines tells me, that the ratio between the sine of each angle and the opposite side is constant throughout the triangle. So I’m going to use information I know, which is side 𝑏 and angle 𝐵 and also side 𝑐. And its angle 𝐶 I’m looking to calculate. So substituting this information gives me then that sine of a hundred and ten over sixteen is equal to sin 𝐶 over twelve. Now I’m looking to solve this equation to work out angle 𝐶, so I’m gonna multiply both sides by twelve.

And I’ve just swapped the order of the two sides around here. But it tells me that sine of 𝐶 is equal to twelve sin hundred and ten over sixteen. In order to work out angle 𝐶 then, I need to use the inverse sine function. And this tells me that angle 𝐶 is equal to sine inverse of twelve sin a hundred and ten over sixteen. Now I’ll use my calculator to evaluate this.

And doing so, I get that 𝐶 is equal to forty-four point eight one zero nine or forty-five degrees when rounded to the nearest degree. So this tells me then that there is at least one possible solution for the other lengths and angles. I’ve worked out angle 𝐶 is forty-five degrees, so I could work out angle 𝐴 by subtracting the forty-five degrees and the hundred and ten degrees from a hundred and eighty, which is the angle sum of a triangle. And then consequently, I could apply the law of sines again to work out the size — of side 𝑎.

So now the question is are there actually two possible solutions. Is this an example of the ambiguous case of the sine rule when there’s another possible value for 𝐶 and hence another possible value for 𝐴 and so on? So if you recall the way that we work out the other possible value of the angle is we subtract this from a hundred and eighty. And if I were to do that, I would get that 𝐶 is equal to a hundred and thirty-five degrees.

Now the question is is this possible. Is it ok for 𝐶 to be equal to a hundred and thirty-five degrees? And to determine whether that’s the case, we need to look back at the original triangle. We already knew angle 𝐵 in the triangle was a hundred and ten degrees. This means then that it can’t possibly be the case that angle 𝐶 is a hundred and thirty-five degrees because if I add those two together it would exceed a hundred and eighty degrees, which is the angle sum for a triangle.

What this means then is that angle 𝐶 must be that forty-five degrees that I originally worked out. And therefore, there is a unique solution to this problem. So there’s only one possible solution for the lengths of the sides and the sizes of the angles.

So this check is essential if you think there is a second possible solution you must look at the other angles in the triangle and confirm that it doesn’t exceed the angle sum. In summary then, we’ve seen what the ambiguous case of the law of sines is. We’ve seen how and why to rises. And we’ve seen how to determine a second possible solution if one exists.