# Video: Using the Sine Rule to Solve Ambiguous Questions

Learn how to apply the sine rule to a range of nonright triangle problems in which the question has some ambiguity such that there may be no feasible answers or more than one feasible answer. We consider the unit circle to find alternate solutions.

11:56

### Video Transcript

In this video, we’re gonna look at how to use the sine rule to answer questions in which there is more than one possible interpretation of the question given the information. For example, if a triangle 𝐴𝐵𝐶 is described as having side 𝐴𝐵 with length five centimetres and side 𝐵𝐶 with length four centimetres, and the measure or size of angle 𝐵𝐴𝐶 is forty-five degrees, then there are two possible situations. Side 𝐵𝐶 could be in two different orientations and each gives quite a different answer to the question: Find the measure of angle 𝐴𝐶𝐵.

Now first remember that the sine rule states in any given triangle, the ratio of the length of a side and the sign of its opposite angle is the same for any side. So if we have triangle with vertices 𝐴, 𝐵, and 𝐶 and we label the sides like this, then the ratio 𝑎 over sin 𝐴 gives the same result as 𝑏 over sin 𝐵, which also gives the same result as 𝑐 over sin 𝐶. Although sometimes it’s more convenient to put those ratios the other way up, so sin 𝐴 over 𝑎 is equal to sin 𝐵 over 𝑏 is equal to sin 𝐶 over 𝑐. Now we’re gonna use this sine rule in this video to answer some ambiguous questions.

Now going back to the question we started looking at, we can see the side length 𝑎 is four centimetres and side length 𝑐 is five centimetres. And also the measure or size of angle 𝐵𝐴𝐶 is forty-five degrees. Now given that the angle we’re looking for is angle 𝐶 there, we can use this part of the sine rule: sin of angle 𝐴 over 𝑎 is equal to sin of angle 𝐶 over 𝑐. So sin forty-five over four is equal to sin 𝐶 over five. Well I can rearrange this by multiplying both sides by five. So that’s sin 𝐶 is equal to five times sin forty-five over four. And five sin forty-five over four is five root two over eight. So 𝑐 is the inverse sin of five root two over eight, and my calculator tells me that that, to two decimal places, is sixty-two point one one degrees, so there’s our answer, well or is it? That’s one of our answers. Remember we said there were two possible answers. If we think back to our original scenarios, it looks like we’ve solved this question here, so we’ve still got to think about what would the angle 𝐵 if side 𝐵𝐶 came in back towards 𝑎 rather than out away from 𝑎. Now thinking back to a unit circle, sin of 𝜃 here is equal to the length of the opposite side in that triangle divided by the length of the hypotenuse.

Well the length of the opposite side is the height of that triangle and the hypotenuse is one, so the sin 𝜃 is the height divided by one. So sin 𝜃 is the height of that triangle. And in this case, that is a positive amount here. But there’s an equivalent triangle over here. So if I have an angle of 𝜃 the same here, they will have the same height. So in other words, there are two values of 𝜃 between zero and a hundred and eighty degrees which give triangles of that particular height which have the same sine value. And they are gonna have angle sizes of 𝜃 and whatever this measure is here. So it’s a hundred and eighty minus 𝜃.

Now in our case, we said 𝜃 was sixty-two point one one degrees. So in our case, this other angle here is gonna be a hundred and eighty minus sixty-two point one one degrees, which is a hundred and seventeen point eight nine degrees. So now we found our two possible answers: sixty-two point one one degrees or one hundred and seventeen point eight nine degrees. So the information we were given in the first place was ambiguous and it lead to two possible scenarios. So in the first case, we could have had an angle there of sixty-two point one one degrees and in the second case there it could have been a hundred and seventeen point eight nine degrees.

Okay let’s have a look at another question. In triangle 𝐴𝐵𝐶, the measure of angle 𝐴, in other words, the size of angle 𝐴 is seventy-two degrees. The length 𝐴𝐶 is fifty centimetres and the length 𝐵𝐶 is ten centimetres. If such a triangle exists, find the measure of angle 𝐵 correct to two decimal places. Okay well there’s a bit of a clue in this question. It says if such a triangle exists, find the measure of an angle 𝐵 correct to two decimal places. So first of all, we have to check if it’s possible for that triangle to exist.

So first, let’s do a sketch. Now it doesn’t matter that this is dramatically not to scale. We can still do our calculations based on what we’ve written down. So we’ve just written in the side names here, we’ve got 𝑎, little 𝑎, is ten centimetres and little side 𝑏 is fifty centimetres. Now we’ve got the measure of angle 𝐴 and the length of side 𝑎. We’ve got the length of side 𝑏. And although we haven’t got the measure of angle 𝐵 there, but we can use the sine rule on angles 𝐴 and 𝐵.

And remember that means that the sine of angle 𝐵 divided by the length of side 𝑏 is equal to the sine of angle 𝐴 divided by the length of side 𝑎. So that means that the sin of angle 𝐵 divided by fifty is equal to the sin of seventy-two degrees divided by ten. Now if I multiply both sides by fifty, the fifty is cancelled on the left-hand side, so that leaves me with sin 𝐵 is equal to fifty sin seventy-two divided by ten. And my calculator tells me that fifty sin seventy-two over ten is equal to four point seven five five two eight two five eight one and so on. But the sine of an angle must be between negative one and positive one, so this is impossible. Such a triangle cannot possibly exist.

If we try to work out sine to the minus one of all this lot, our calculator would give us an error. The numbers we were given in the question would lead to sine of an angle giving an answer of over four which is impossible. Sine of an angle must always be negative one and one, so such a triangle cannot exist. That’s our answer.

Let’s look at one more question then. In triangle 𝐴𝐵𝐶, the measure of angle 𝐴 is forty-eight degrees. Side 𝑎 is nineteen centimetres. Side 𝑐 is twenty-one centimetres. If such a triangle exists, find the possible measures of the other angles and the possible lengths of side 𝑏, all correct to one decimal place. So again, let’s start off by doing a sketch. And again, we haven’t done this to scale, so it doesn’t necessarily that accurate. Now we know the size of angle 𝐴 and the size of side 𝑎. We know the size of side 𝑐, but we don’t know the size of angle 𝐶.

So we can use this form of the sine rule: sin 𝐶 over 𝑐 is equal to sin 𝐴 over 𝑎. So filling in those unknown values, sin 𝐶 over twenty-one is equal to sin forty-eight over nineteen. Now we can multiply both sides by twenty-one. So the twenty-one cancels on the left-hand side and we get sin 𝐶 is equal to twenty-one sin forty-eight over nineteen. Now that means that 𝑐 is equal to the inverse sine of all this lot. I’m gonna put that into the calculator. This is what comes out: fifty-five point two two two two three two zero one. We’ve only been asked for one decimal place. So we’ve got 𝐶 is equal to fifty-five point two degrees. That’s our first answer.

But remember, this side 𝐵𝐶 could have been orientated like that or it could have been oriented like that, so there’s another possible value here. We found this answer here, but there’s also another version of the answer here. So again, thinking back to our unit circle, we found this answer here. But there’s another solution that gives the same height of the triangle, the same sin 𝜃 value between zero and a hundred and eighty and it’s this angle here. So that’s a hundred and eighty minus fifty-five point two. So that gives us two possible different values for 𝑐.

Now for the rest of the question, we’ve got to be quite careful to keep the right combination of values together. You can’t just mix and match these throughout the rest of question. You’ve got to keep the right angles with the right angles. Now if 𝐶 is equal to fifty-five point two degrees, then angle 𝐵 here is equal to a hundred and eighty minus forty-eight minus fifty-five point two, because angles in a triangle sum to a hundred and eighty degrees. And that works out to be seventy-six point eight degrees. But if 𝐶 was a hundred and twenty-four point eight degrees, then 𝐵 is gonna be a hundred and eighty minus this forty-eight minus the hundred and twenty-four point eight. And that would give us an answer for 𝐵 of seven point two degrees. So we’ve got two different versions of the triangle with two different corresponding values of 𝐵 and 𝐶.

Now to get a nice accurate answer for the length of side 𝑏 and round to one decimal place is actually quite tricky. We could use the cosine rule or we could use the sine rule with two possible different values for 𝑏. Well let’s stick to the sine rule for now, but we’ll keep a few extra places of accuracy in our calculation for angle 𝐵. Now remember in our calculator, we had all these decimal places. Now we don’t need to keep all of these, we-we will do in this particular question just to keep the most accuracy that we can. And we’ve said that 𝐶 was fifty-five point two two two two three two o one and so on. And if we use that figure for calculating the measures of angle 𝐵, we’d have get-we’d have got seventy-six point seven seven seven seven six seven nine nine and seven point two two two two three two zero one. So they’re the values we gonna use in the next part of the calculation.

Then going back to our sine rule, 𝑏 over sin 𝐵 equals 𝑎 over sin 𝐴. Now these values, 𝑎 and sin 𝐴, I was given in the question itself, so these are as accurate as they can possibly be. And we’re using the most accurate version of the answer we can for sin 𝐵. So when I work this out, I’ve got 𝑏 over sin seventy-six point all that lot is equal to nineteen over sin forty-eight. So if I multiply both sides by sin seventy-six point all that lot, I get nineteen sin seventy-six point all that lot over sin forty-eight, which my calculator tells me to one decimal place is twenty-four point nine centimetres. So when 𝐶 is fifty-five point two degrees and 𝐵 is seventy-six point eight degrees, then the length of side 𝑏 is twenty-four point nine centimetres. So just quickly let’s do the calculation again with the other versions.

So we’re using the same values for the length of side 𝑎 and the measure of angle 𝐴, but now we’re using a size of seven point two two two two three two o one for angle 𝐵. So 𝑏 over sin seven point two that lot is equal to nineteen over sin forty-eight. And when we rearrange that, we get 𝐵 is equal to three point two to one decimal place. So when 𝐶 is a hundred and twenty-four point eight degrees and 𝐵 is seven point two degrees, then to one centimetre the length of side 𝑏 would be three point two centimetres.

So the last thing that remains is for us to make it absolutely clear which measurements go with which measurements. And so we’re writing out our answer like this when 𝐶 is fifty-five point two degrees, then 𝐵 is seventy-six point eight degrees. And the length of side 𝑏 is twenty-four point nine centimetres all correct to one decimal place. And when 𝐶 is a one hundred twenty-four point eight degrees, then 𝐵 would only be seven point two degrees, and the length of side 𝑏 would only be three point two centimetres. And it’s worth just writing in the level of accuracy that you’ve given your answers to.