SM2 12.2-3: Given Two Sides of a Triangle, Find All Six Trig Ratios

Added: 2026-05-15

[00:00:00]Hello and welcome. Let's do some trigonometry. So  find these six trig ratios for the angle uh A for the right triangle shown below. So what's  great about this is that we're not on the standard coordinate plane. We're dealing with  just a triangle. All of these values we know are going to be positive. So we're given this  side and we're given this side and if you know two out of the three sides of a right triangle you  have a particular theorem we call that Pythagorean Theorem that'll find the third. So let's  just dive right into it. So I'd say okay well what we're missing, I'm going to call that  a. Um we're missing a leg. So if we use Pythagorean theorem which is leg squared  plus leg squared equals hypotenuse squared, then simplify everything and you're good to  go. So we have a squared, which is just a squared, plus 4 equals 64.

[00:00:51]If we want to solve for a you'll want to move  the 4 over so we have a squared equals 60. And then we undo a square with a square root. Now  remember technically you should have plus or minus square root 60 but again if it's just  a plain right triangle, right? We don't know anything about it. It's just a length a geometric  figure, it can't possibly be a negative length, and thus you're good with positive square root 60. So  that's all right. So now we have a equals square root of 60 and be careful that can simplify.  That is 6 and 10. Uh six is two and three, ten is five and two. What do they have in common?  I have a pair of two so a 2 can come out. Remember for every two that are the same one comes out and  then you have a five and a 3 left behind which multiplied together is 15. So that means your  missing side is 2 square root 15. And once you have your triangle down all of the sides, getting  your six trig ratios are not that bad. Okay. So first off you have your main three. How can we  find those ones? Acronym to the rescue, SOHCAHTOA.

[00:02:01]All right. So the S stands for sine, C cosine, T  tangent. All right. So that's going to be these three right there is going to be your SOHCAHTOA.  Now whatever isn't the main three, so that'd be this one and this one and that one. Okay. Those  are what we call the reciprocal functions where they are literally reciprocals. In other words  flipped fractions, or flipped numbers of your main three: sine, cosine, tangent. Well you're like great  that's that's nice to know they're just flips.

[00:02:31]How do I know which one's which? Well first  off the easiest one to identify is going to be tangent and cotangent. It's like pilot and  copilot they sound like they should go together because they do. So tangent cotangent those are  the reciprocals of each other. All right. So let's just fill those out really quick. So if we know  tangent is the opposite side over the adjacent side. Where I'd say okay what is our angle A? So  that means you're looking at the angle up here. If it was angle B you'd be looking at the angle down  here. Why does it matter? Well once we know that A, angle A up there is the angle we're looking for  that determines your opposite side, which directly opposite of our angle A is going to be the  segment from B to C. This is your opposite side.

[00:03:21]Okay. And then that makes that this little  segment right here from A to C your value of 2 that is literally side by side with your  angle. It is adjacent to it, so that would be your adjacent side. And then here's a right angle,  across from it is your hypotenuse and that will always be true regardless of whatever A, and which  angle you're using either A or B. Hypotenuse will always be hypotenuse. Alrighty. So now we know  which, oh that's cool, which angle we're looking at. Now I say okay, tangent is the opposite over  the adjacent. We just said that the opposite was 2 square root 15. So that means our tangent of  theta equals 2 square root 15 over the adjacent 2.

[00:04:06]Okay. Now be careful if you can simplify this  fraction simplify the fraction. All right guys two over two, what does that give you?

[00:04:18]One. That's one square root 15 or in other words  square root of 15. Do not put 2 square root 15 over 2. Simplify that down. So that means that  you'll have square root of 15 as an answer here.

[00:04:30]Okay. Now if you wanted to write the reciprocal,  well square root 15 is technically over one so the flip of that would be 1 over square root  15. Now in addition what you could also do is you could rationalize this where you can multiply  top and bottom by the square root. So square root of 15 top and bottom. 1 times square root of 15,  square root of 15. Square root of 15 times square root of 15, that's two of the same thing,  so 1 will come out and this is an equivalent answer to that. It'll accept both. Whatever  you find easiest to think about go for it.

[00:05:08]Okay. Now let's tackle our sines and cosines. How  do we know which ones are reciprocals of our sine cosine? Well how I remember this next one is I  always remember s with a c, right? Sine and cosine um are the main three so that means that sine  and cosine can't be reciprocals of each other, right? These are not reciprocals. Which means  that the s, oops let me do a different color, the s for sine its reciprocal has to start with the  c but it can't be cosine so has to be cosecant, right? Cosine starts with a c so its reciprocal  has to start with an s. So there you go. Now you know how to find all six trig functions. Let's  keep it going. So let's uh just do cosine next.

[00:05:52]So cosine is the adjacent over the hypotenuse.  Our adjacent is 2. Our hypotenuse is 8. 2 eighths is the ratio. How would we simplify that? Choose  a common number, goes into itself once, into 8 four times. So 1/4 is your cosine. The reciprocal  of 1 over 4 is 4 over 1 or in other words 4. Okay.

[00:06:17]Sine is the opposite side over the hypotenuse. The  opposite side is 2 square root 15. The hypotenuse is 8. We can simplify this further. Look at your  outside numbers, right? 2 over 8 we just said simplified to 1 over 4. So that means that your  sine is going to be the square root of 15 over 4.

[00:06:40]Because 1 times root 15 is root 15 and then over  4 on the bottom. For the-- use that: take the reciprocal to get 4 over square root of  15. If you want to rationalize you could do that, right? But those would all be correct for  your six trig ratios. Thanks for watching.