slick TMUA geometry!

Ajouté : 2026-06-05

[00:00:00]If you're looking to study maths, computer science, or economics at any of the UK's top universities, you have to do the TMUA. This problem actually comes from the AMC, but this is an amazing resource for additional problems to help you prepare for the TMUA. Let's have a look. In triangle ABC with integer side lengths, cos of A is 11 over 16, cos of B is 7 over 8, and cos of C is minus a quarter. What is the least possible perimeter for triangle ABC? And we've got five options here. Pause the video now and give this problem a go for yourself. I'm going to dive right into the solution here.

[00:00:31]So, we've got the cosines of each of the angles, and so because cosine is a one-to-one function between 0 and 180°, we could in theory just work out what these angles are individually by taking the inverse cosines here.

[00:00:44]But, none of these are pretty nice numbers, and obviously we don't have a calculator here. So, that might not help us too much.

[00:00:51]And we're trying to work out that something to do with the perimeter. So, we want the side lengths of our triangle. So, how can we go from cosines to side lengths? Well, we might think to use the cosine rule, but then you have squares of side lengths in there. So, we want something that doesn't really involve squares of side lengths. Well, we could think about the sine rule.

[00:01:13]Well, the sine rule is quite nice cuz it doesn't involve side length squared, it just involves side lengths. And we can obviously work out sine of A, sine of B, and sine of C using our good old friend sine squared theta plus cos squared theta is 1.

[00:01:28]Okay.

[00:01:29]So, we can work out sine A, sine B, sine C. Notice that I'm I'm hesitant to actually go ahead and calculate these just yet because I know it's going to take a couple of minutes to calculate these. Let's just imagine we did have sine A, sine B, and sine C.

[00:01:41]Well, what we could do then is we say that the side lengths ABC, we could say A to B to C, we know that the ratio is the same as the ratio of sine A to sine B to sine C.

[00:01:53]And then, well, we will know each of these three numbers, so we'll know what this ratio is, and then we could just perhaps find the smallest multiple which ensures that lowercase A, B, and C are integers, and then that will give us our answer. So, this seems like a pretty good method. Let's go ahead and try and work this out. So, if cos A is 11 over 16, well, we know that sin A is going to be plus or minus the square root of 1 minus cos squared A, so 11 squared over 16 squared, like so, which is going to be 1 over or plus or minus 1 over 16 times the square root of 16 squared minus 11 squared. So, 16 plus 11 times 16 minus 11, like so.

[00:02:32]And just simplifying this plus or minus 1 over 16 times 16 plus 11, that's 27, and if we square root that, that's 3 root 3.

[00:02:40]And then 16 minus 11, that's root 5, so that will just be root 3 times root 5 is root 15. So, we get plus or minus 3 over 16 times root 15. And because cos of A is positive, we know that A must be between 0 and 90, so it's an acute angle, so we're going to be taking the positive version of sin. In fact, we actually know sin will always be a positive value between 0 and 180, so in fact for all of these we can just take the positive root. Okay, sin of B, same thing, we're going to get that this is just positive square root of 1 minus 7 over 8 squared, like so, which again is going to be 1 over 8 times 8 squared minus 7 squared, which can be 8 plus 7 times 8 minus 7, so that's going to be 1 over 8 times root 15. And this is a good sign because we've got a root 15 in both of these, so when we look at the ratios, those will cancel out. And lastly, when we look at sin of C, this is going to be root 1 minus a quarter squared, like so, which is going to be 4 squared minus 1 squared, so 15 over 16 square rooted. I wrote that correctly. Yes, and that's of course 1 over 4 root 15. And so therefore, this ratio that we had up here, this is just equal to sin A which was 3 over 16 * root 15.

[00:03:57]Uh to sin B which is 1 over 8 * root 15.

[00:04:00]And the last one was then 1 quarter. So, if I just multiply through by 16 here, this is 3 to 2 to 4. So, therefore, my the lowest kind of choice of side lengths A, B, and C to ensure that uh we get the least possible perimeter whilst ensuring A, B, and C are perimeters would be A is 3, B is 2, and C is 4, like so.