Law of Sines - Online Tutor, Practice Problems & Exam Prep

Intro to Law of Sines

If I gave you this right triangle on the left here and asked you to solve for the missing sides, you'd be able to do it pretty quickly given everything you've learned in the course. Because you have a hypotenuse and an angle, you could use SOHCAHTOA to find that other side, which that side is just equal to 3. Then you could use SOH CAH TOA again or the Pythagorean theorem to solve for the other side. Pretty straightforward. What if I gave you this triangle over here and asked you to do the same thing? The key difference here is that this angle is 90 degrees. It's 70. So this is not a right triangle to non-right triangle, which means that this isn't really a hypotenuse. And that means that you can't use SOHCAHTOA or the Pythagorean theorem. It just doesn't work. So unlike right triangles, you cannot solve for missing sides in a non-right triangle by using Sogutto and the Pythagorean theorem. So how do we actually solve for this? Well, what I'm going to show you in this video is we use a different equation called the law of sines. And it sounds kind of scary at first, but I'm actually going to break it down for you and show you that it's really straightforward to use. Let's go ahead and get started.

Alright. So I'm actually going to just show you the law of sines. It's an equation here. It's the sine of A over a is equal to the sine of B over b, and that's equal to the sine of C over c.

sin A a = sin B b = sin C c

So, a couple of notation systems here. So what you're going to see here is that angles in these problems are always capital letters A, B, and C, and sides are always lowercase letters. So really what we see here is that the law of sines is really just three ratios, and it's comparing ratios of angles on the top to lengths. It's called the law of sines because all the three terms involve sine, and there's a pattern here. It's always an angle over a side, A over a, B over b, C over c, and so on.

So then how do I actually solve and use this law of sines to solve for that missing variable? Basically, what happens here is you have three ratios. You're just going to try to pick two out of the three in which you know or can figure out three out of four variables. What do I mean by this? Basically, what happens here is in this problem, I know what A, C, and c are. My job is I have to pick two out of the three sort of terms in which I know three out of four variables, and I can solve that missing one. So, if I try to pick these two over here, for example, what happens is I only know one out of the four. That's definitely not going to work. I know nothing about B or b. If I try to pick these two over here, notice how it doesn't actually include my target variable, but, also, I only know two out of four variables, so I also can't use this. So, basically, because I know nothing about B, I'm just going to ignore it, and I'm just going to pick the first and the third ratio. Those are going to be my two out of three. So notice how here I have three out of four variables, and I can solve for this missing one.

Alright. So I'm just going to rewrite this. The sine of A over a is equal to the sine of C over c. Trying to solve for this little a over here, there's a couple of ways I can do this. I can cross-multiply, but really what I like to do in these problems is when you're solving for a side, you could basically just flip this fraction. This becomes a over sine of A. Whatever you do to one side, you do to the other. So this is going to be c over the sine of C. And then now what you can do is you can move this sine of A up to the top and you can just start plugging in numbers. So a over here, the reason you have to do this is because otherwise you're solving for a denominator. So if you plug in these numbers, what you're gonna get is a = c / sin(C) * sin(A). So in other words, c is equal to 6 and then sine of C, we'll see we'll see here that the sine of angle C is 70 degrees. So in other words, 6 over sine of 70 times the sine of A. So this is going to be the sine of and this is going to be 30 degrees over here.

Alright. When you actually plug in all of these numbers, what you're gonna see here is that you're gonna get a is equal to 3.19. Alright? So we just get 3.19, and that is the side. That is this side over here, 3.19. Alright? That's how you use the law of sines. Notice how the answer makes perfect sense because here in the right triangle with a hypotenuse of 6, if you just go straight down, that's only a length of 3. But in this triangle, you kind of have to go a little bit farther because this is sort of like this weird diagonal thing. So you get something that's a little bit longer than 3.19. Alright. This totally makes sense.

One last point I want to make here, by the way, is that instead of seeing capital letters A, B, and C, you may see some Greek symbols like alpha, beta, and gamma, but it works the exact same way. Alright? That's how you use the law of sines. Let me know if you have any questions, and let's get some practice.

Use the Law of Sines to find the length of side $a$ to two decimal places.



• A. 8.20
• B. 4.39
• C. 2.20
• D. 1.61

Use the Law of Sines to find the angle $B$ to the nearest tenth of a degree.



• A. 48.6°
• B. 77.2°
• C. 40.5°
• D. 35.3°

Solving SAA & ASA Triangles

So by now, we've seen how to use the law of sines to solve for missing sides in a non-right triangle. One of the most common types of problems you'll run into is where a problem will give you some angles and sides, and they'll ask you to classify and then solve a triangle by finding all of the other missing information, the other missing sides and angles. Now if that seems kind of intimidating, don't worry because what I'm going to show you in this video is that, really, to solve these problems, all we need is the law of sines that we just learned, plus another familiar equation we've seen a bunch of times before. So I'm going to walk you through a step-by-step process of how to solve these 2 specific types of triangles. But first, I want to introduce you to the 3 types or categories of triangles for which you'll use the law of sines. Let's just jump right in. Alright? So you're going to see a bunch of acronyms in your books and classes like ASA, SAA, and SSA. It's kind of just a soup of letters of ss's and a's. And, really, all these types or acronyms just have to deal with is the information that you're given at the start of a problem. The first one is called an angle side angle. This is where you were given 2 angles and the side between them. For example, you were given big A, big C, and little b. The next one is called the side angle angle. This is where you're given 2 angles and a side that's not between them, but adjacent to them. So an example of this would be big A, big C, and little c. And then finally, the last one is called the side side angle. This is where you're given, you guessed it, 2 sides and an angle. An example of this would be little b, little c, and big C. We're actually not really going to talk about this one at all in this video. We're going to talk about this one in a later video. All we're going to cover, really, is these 2 in this specific video. Alright? So without further ado, let's just jump right into our example here because the first thing it asks us to do is classify this triangle. So let's just jump right in. Alright? So I've got this information here. A equals 30, C equals 70 degrees, and then little c equals 6. Alright? Regardless of if you can't sort of tell off the bat what type of triangle it is, the first thing you're always going to do is just try to draw or attempt to sketch the triangle because then you'll be able to see what it is from there. Alright? So I'm just going to go ahead and sketch this. Your sketch doesn't have to look exactly like mine, but what I always like to do is look at the angles first because it's easiest to visualize what's going on there. I know that one of the angles is going to be 30 degrees. So what I'm going to do is I'm going to draw a little sort of leg like this, and the 30-degree angle is going to look something like that. So this is going to be 30 degrees. Alright? Now the next one, the next angle that I know is going to be 70 degrees, which is going to look something like this, but then a little bit sort of tilted in, not exactly a 90-degree angle. So it's going to look something like this. Alright? So this is going to be my 70 degrees. And I'm basically just sort of going to keep on going with those lines until they sort of intersect like this. And that's going to be my other angle. So this is sort of a sketch of my triangle. This is going to be angle A, and that's angle C based on what I've given, which means that by default, this has to be B. Right? And remember, the rule for the sides, they always have to go opposite their corresponding angles. So this is little c, which we know is 6. This is going to be little a, and this is going to be little b. So our job is to now solve for this triangle by solving for these missing variables, which we don't know over here, big B, little a, and little b. However, what we can do right now is we can actually classify this triangle because we've already drawn the sketch. If you notice here, what we've got is we've got an angle and angle and one of the sides that's not between the angles, but adjacent. So that means that this is actually exactly what this type of triangle, set. Right? So the example for this is ACC, and that's exactly what we were given inside this problem. But we just drew it out, and we kind of confirmed this. Right? So this is going to be an SAA triangle. Right? Side angle angle. Alright? So if we are given an angle side angle or SAA triangle, we're going to go ahead and stick to these steps. And I'm just going to jump right into the second one. So the easiest thing to solve for in these types of problems is actually this 3rd angle over here. How do we solve for the 3rd angle? Well, we're actually just going to go ahead and use a familiar equation that we've seen before, which is the fact that all of the angles in a triangle have to add up to 180 degrees. This is sometimes referred to as the angle sum formula. Basically, since we know these 2, we can always solve for the third one, which is by subtracting this from 180. So here's what I'm going to do. I'm going to do in step 2. I'm just going to set that A + B + C is equal to 180 degrees. And so, therefore, I can find B just by subtracting the other two angles, that I know from 180. So, this is going to be 180-70-30, which is really just -100. So therefore, B is equal to this is going to be 80 degrees. A lot of times, you'll just be able to use really quick mental math to figure this out, but you could sort of solve this out if the numbers aren't so nice. We just figured out one of our missing variables is 80 degrees. Perfect. Great. So now that we have all of the angles solved, how do we solve for the 2 missing sides, a and b? Well, we're just going to go ahead and jump to our 3rd step over here, which is we're going to use the law of sines. That's how we solve for missing sides in a non-right triangle. Alright? So we're just going to have to do this twice. And, really, in this third step, it actually doesn't matter which one you go ahead and solve for first. It doesn't matter. You could do a or b. Alright? So remember how the law of sines works? We're going to have to go ahead and set up our sine Aa equals sine Bb equals sine Cc. Remember, I'm looking for a in this specific, in this example over here. So I'm gonna have to sort of pick 2 out of these 3 ratios in which I can figure out 3 out of 4 variables. Alright? If I try to look at b, what happens is I know the angle, but I don't know the side. I know the angle A, but I don't know the side. So I can't pick these 2 because that's only 2 out of 4. So instead, I'm going to ignore this one, and I'm going to pick these 2 over here because I know ever

Classify the triangle, then solve: $A=60°,B=15°,c=6$.

• A. $SAA,a=6.69,b=22.4,C=105\degree$
• B. $ASA,a=6.69,b=22.4,C=105\degree$
• C. $ASA,a=5.38,b=1.61,C=105\degree$
• D. $SAA,a=5.38,b=1.61,C=105\degree$

An engineer wants to measure the distance to cross a river. If $B=30\degree$, $a=300$$ft$, $C=100\degree$ find the shortest distance (in $ft$) you’d have to travel to cross the river.

• A. $459.6ft$
• B. $195.8 ft$
• C. $152.3ft$
• D. $233.4ft$

Solving SSA Triangles ("Ambiguous" Case)

In the last few videos, we saw how to use the law of sines to solve these two types of triangles, ASA and SAA. Now we're going to take a look at how we can solve the last case that I mentioned, the SSA triangle, where you have two sides and an angle that isn't in between them. These types of problems are a little bit trickier because as you're solving them, you may encounter one of three possibilities: no solution, one solution, or in some cases, you might actually get two solutions. This is why your books refer to this as the ambiguous case because you don't know what type of answer you'll get until you actually start working it out. This might seem really confusing at first, but I've solved a ton of these problems. The way you solve them is pretty much the same every single time. So what I'm going to do is break down this problem for you. We're going to work it out step by step, and I'll show you a good way to get the right answer every single time. Let's just jump right in. Throughout the solution, we'll see the different possibilities that we might run into. So let's get started here.

In this problem, we're going to solve for each angle in the triangle, not just the sides. We've got \( a = 6 \), \( b = 8 \), and then \( A = 41^{\circ} \). In these types of problems, the first thing you would typically do is try to sketch the triangle out and label the given information. But with these triangles, it's harder because you have two sides and an angle, so it's kind of hard to sketch what the full triangle is going to look like because you only sort of know what one corner is going to look like. So, I don't even bother trying to sketch it in the first part here. What I try to do first is use the law of sines to find a second angle.

So, we set up a law of sines; we have \( \frac{\sin(A)}{a} = \frac{\sin(B)}{b} \). I don't have to write out the \( C \) part because I know nothing about \( C \) or \( \angle C \). So, I can solve for \( \sin(B) \) because that's the only variable in this case that I don't know. I have the other three data points. So, really, what happens is \( \sin(B) = \frac{8 \cdot \sin(41^{\circ})}{6} \). When you work this out, what you're going to get here is approximately 0.875. This is the first major milestone of this problem, which is that we have \( \sin \) of some angle is equal to some number on the right side.

If this number is greater than 1, then you don't have a solution because the sine of any angle can't exceed 1. However, in our problem, we have a number less than 1, so we continue on to the next step, using the inverse sine to solve for that angle. \( B \) is \( \sin^{-1}(0.875) \), which is approximately 61 degrees. However, there are two possible angles that will give you this sine value within the range of 0 to 180 degrees. The other angle is \( 180^{\circ} - 61^{\circ} = 119^{\circ} \). We have to consider both angles \( B_1 = 61^{\circ} \) and \( B_2 = 119^{\circ} \).

Next, to confirm the possibility of a second triangle, we add angles \( A \) and \( B_2 \): \( 41^{\circ} + 119^{\circ} = 160^{\circ} \). The sum is less than 180 degrees, indicating a possible triangle, thus two solutions are possible. You can then calculate the remaining angles using the angle sum formula \( A + B + C = 180^{\circ} \), and for \( B_1 \), \( C = 180^{\circ} - (41^{\circ} + 61^{\circ}) = 78^{\circ} \), and for \( B_2 \), \( C = 180^{\circ} - 160^{\circ} = 20^{\circ} \). Both \( C_1 = 78^{\circ} \) and \( C_2 = 20^{\circ} \) are possible solutions.

This example illustrates why the SSA case is known as the ambiguous case, as you can end up with zero, one, or two possible triangles depending on the specific angle and side measurements provided. Let's go ahead and get some practice, now that we've broken down how these solutions are derived!

Example 1

Everyone, in this example, we're going to tackle a problem where we're given a triangle with side a equals 1, side b equals 4, and angle A equals 30 degrees. If you haven't noticed, we are provided with two sides, a and b, and the angle A, which corresponds with one of those sides. This scenario is a classic example of an SSA (Side-Side-Angle) triangle. If ever in doubt, it's helpful to sketch a quick triangle, labeling side a and side b, with angle A opposite its corresponding side, confirming the SSA relationship. This setup frequently leads to the ambiguous case, where the exact nature of the triangle is unclear, but a sketch can help confirm your analysis.

Let’s proceed to solve this type of triangle without initially drawing it. The first step involves setting up the Law of Sines to determine a second angle. This can be expressed as: A a = B b We only focus on angles A and B and sides a and b due to the absence of data about angle C or side c. By rearranging and substituting the known values, the equation simplifies to: B = b * A a Substituting the numbers results in: B = 4 * 30° 1 = 2

This outcome reveals that the sine of angle B equals 2, which is impossible since sine values range from -1 to 1. Therefore, this problem does not have a solution. To understand why, consider the triangle setup: angle A is 30 degrees, side b (opposite angle A) is 4, and side a is 1. Side a is too short to complete the triangle irrespective of its orientation, which confirms why there is no solution. Hence, no further calculations are needed.

Example 2

Hey, everyone. Let's take a look at this example. We have another triangle over here in which we're given 2 sides, b and c. And then the other angle that we're given is one of the corresponding angles or one of the corresponding letters. This is, again, a side side angle triangle. Let's go ahead and stick to the first steps over here, which is we're gonna set up a law of sines so that we can get an additional angle. Let's go ahead and get started.

So, this first step over here, we're gonna set up a law of sines. Notice how, again, the only two letters involved here are b and c. So when we actually set up our law of sines, we'll completely disregard the A term because we don't know anything about A or little a. So in other words, we just have Bb equals Cc. Alright?

Again, what happens in these problems is you're gonna have big B over here, and little b and little c, so you can go ahead and solve for C. That's going to be that second angle that you find. Okay? So what we're going to do here is once we cross multiply, move this to the other side, we're going to see that sinC is equal to 2 times the sinB, which is the sin of 29 degrees divided by little b, which is 4. Now if you go ahead and work this out, what you're gonna get is you're gonna get something like 0.242. And whenever you do these problems, again, I like to hold on to a few more decimal places because otherwise, you get some rounding errors. But when you solve for this, we're gonna see here is that we get 0.242. Remember for the first step, if you ever get something that's larger than 1, you're gonna have no solution, but that's not what happened here. Right? So we actually are gonna get something that's less than 1 or equal to 1. We move on to step 2, and we're perfectly fine to continue the problem.

The second step here is we're gonna use the inverse sine to solve for 2 possible angles here. So what's going on? Well, basically, what happens is if we solve for C, remember, we can take the inverse sine. So we're just gonna take the inverse sine of 0.242, what you're gonna get here is you're gonna get the inverse sine is equal to 14 degrees. But that's not the only angle that when you plug it back into sine, you'll get this number for. So that's actually where we sort of split our problem into two parts. So, basically, what happens is that C₁ is equals to the inverse sine, which is 14 degrees, but C₂ is just equal to 180 minus C₁. Right? So in other words, it's just minus the angle that we just found over here. So it's 180 minus 14, and then C₂, therefore, is equal to 166 degrees. So we've got these two different angles over here that when you plug them both back into the sine, you could double-check this, you'll actually get back to 0.242.

Alright? So how do we figure out which one is possible? Well, actually, that leads to the third step. So after we're done with step 2, we're gonna see that we have our 2 possible angles. And by the way, this didn't happen over here, which is that the sin of the angle is equal to 1. So again, this may happen, but it's rare. We're just gonna keep going on to step number 3. So, for step number 3, what we're gonna do is for this angle here that we just calculated, the second one, we're gonna add it to the given angle that we have, and there's a really good reason why. So here's what this means, add to the given angle. The only other given angle that we have is the fact that B is equal to 29. So what does this mean? It means that if you add B+C₂, and this is just gonna equal 29+ 166 degrees, so 29 degrees plus 166 degrees. If you work this out, what you're gonna get is that this is equal to 195. But how is it possible that two angles in a triangle add up to something that's greater than 180 degrees? That's im 해결합니다