Area of SAS & ASA Triangles - Online Tutor, Practice Problems & Exam Prep

To calculate the area of a triangle, use the formula $(1/2)bh$, where b is the base and h is the height. For non-right triangles, find height using the sine function: $h=c\sin \left(A\right)$. In cases with missing sides, apply the Law of Sines to determine unknown values, ensuring all necessary information is available for area calculation.

Calculating Area of SAS Triangles

Everyone, you may remember that to calculate the area of any triangle, you'll always need the height. You may recognize the one-half base times height formula. It works even for non-right triangles. As long as you're given the height and base, you can find the area. But some questions will give you angles and sides, and you won't actually be given the height. But don't worry because in these types of problems, I'm going to show you how you can find it by using the sine of a known angle. So, what I'm going to show you in this video is that we're going to work out the areas of these two triangles, and they're exactly the same. We're just going to use different information to get there, and then I'll show you three new equations that you'll need to solve for the area. Let's get started here.

In this first example, we will have one-half if we have the base and the height of this triangle, then we could just use the familiar one-half base times height formula to find the area. So, really, what this becomes here is one-half. The base of this triangle is 8, and the height of this triangle is just equal to 3. When you work this out, this ends up being 12. Pretty straightforward. That's the area of this triangle. By the way, it doesn't necessarily have to be b as the base. You could have a or c. All these things could be potentially different letters. But the area of this is 12.

What about this triangle over here? The key difference is that we actually don't have that height that's already drawn for us, but the principle is still the same. We're still just going to use one-half of the base times height. What's the base of this triangle? Well, actually, it's the same thing as the triangle on the left. The base really is just b. So I'm just going to write that over here. What about the height of this triangle? Well, here's where it's a little bit different because whenever you want to draw the height of a triangle, you're just going to take the base that you've drawn, and you're basically going to draw a perpendicular line up to one of the corners. And this effectively sort of turns this into a right triangle. So, what is the value of this height over here, which I'll write as h? Well, the idea here is that if you have the sides, then basically what you can do is turn this into a right triangle, and we have an angle and the hypotenuse of this triangle so we can use sohcahtoa. So, the sine of this over here, we're going to use the sine of angle a which is equal to h over the hypotenuse, which is c. If I rearrange for this, what I can find is that h=c⋅sin⁡(a). So, basically, what happens here is that if I have the hypotenuse and angle, I can figure out the height, and this effectively becomes the height of my triangle.

So, really what happens is that the base becomes the letter b, and the height of this triangle actually becomes c⋅sin⁡(a). So b is the base and then c⋅sin⁡(a) is the height. Alright. So let's see actually what this works out to be. This is going to be area equals one-half, and I'm done my base is still b exactly like it was on the left. And then now what happens is instead of 3, I'm just going to plug in c⋅sin⁡(a), which is 6 times the sine of 30 degrees. When you work this out, what you're actually going to get is that this actually equals 12 exactly like what it did on the left side. So the areas of these two triangles are exactly the same. This actually makes total sense because if you solve for the height of this triangle, this is really just 6 times the sine of 30, which is actually just equal to 3. So using different information, we found the height of this triangle is still 3, and, therefore, the areas are going to be exactly the same. That's really all there is to it. Alright. Now, remember that the positions of these variables will be in different places throughout your triangles.

So there are sort of three different variations of this formula, kind of like how there was for the law of cosines. I'm just going to go ahead and show you those. Sometimes you'll have BC sin A. Another variation that you might see is AC sin of big B. And the last one you'll see here is AB sin C. So what you'll notice here is that the angle is always going to be the last the third letter that is not in these other two over here. And it's always two lowercase letters and the sign of an angle over here. Alright. Now, remember that all of these things will be multiplied, so you can shuffle these letters around, but this is the idea behind finding the area of a non-right triangle. Thanks for watching, and let's get some practice.

Calculating Area of SAS Triangles Example 1

Welcome back, everyone. So in this problem, we're going to try to calculate the area of this triangle. Let's get started here. Remember that at the end of the day, an area is just 1/2 times a base times a height. So area is equal to 12×b×h, and the base of this triangle appears to be b=8. What about the height of this triangle? Remember that we're going to try to draw from one of the corners down to where it intersects the base. The problem is if I try to do this here and try to figure out this height, there's a piece of this hypotenuse of this triangle that still keeps going up, and I'm losing all of this area if I try to plug it into 12b×h. Alright? So it turns out that we can't really do this, so I kind of want to explain what's going on in these triangles. But first, let's actually just take a look at the variables. We know that little b=8, little a=6, and the big C=120°. We know that there are sort of three versions of the area formula of a triangle, and usually what happens in these problems is that as you look through your variables, 2 of them won't make sense. Let's take a look here. We have 12b×c×sin a. The problem is I don't know what a is in this problem or little c. So I have 2 unknowns. That equation is not going to work. The second one is going to be 12a×c×sin b. Same problem. I don't know what c, little c is, or the big B. Both of these variables or angles are unknown. The only one that makes any sense at all is using this one over here because I actually know what all three of these values are, a, b, and c. But I want to show you why this works in this equation. So we're going to have 12a×b×sin C. Remember, area is always a base times the height. What's the base in this problem? It's really just that little b. So what I'm going to do is I'm actually just going to switch the variables around in this equation. This is going to be 12b×a×sin C. Alright? But I want to show you why that ends up being the height in this problem. Alright? So we already know that we can't sort of draw an a a little dash line straight up and have that as the base. But what we can do is we can go all the way to the corner over here of b and then basically just draw this line straight down even if it sticks out past the base of my triangle. Because what I can do is I could basically just pretend that the base of this triangle sort of keeps on moving on going on and sort of ongoing straight. So, really, this height over here is actually the height of my triangle. And notice that what we've done is we've made a smaller right triangle that's sort of as an extension of my obtuse triangle and the height of this right triangle is the same as the height of the obtuse triangle that I'm trying to find the area of. What is the height of this triangle related to the variables that I know? Well, I've got the hypotenuse, which is a. And just using SOHCAHTOA, this h would just be a×sin C. What is this angle? Well, it's basically just if we're assuming that this line here is going to be supplemental, it's going to be 180 degrees, then this is really just going to be 180 degrees minus a 120, which ends up just being 60 degrees. So this angle over here is 60 degrees. Therefore, the height would just be a×sin 60°. In other words, it would really just be 6 times the sine of 60. Alright? Now it might seem like you have to do all this work to figure out this angle over here, but, actually, you don't. Because what I'm going to show you here is that you could have done all this work to figure out this angle was 60, or you can actually just use 6 times the sine of the original angle C that you were given, which is a 120. What I want you to do is remember an old identity that you may have learned, back in a previous chapter, which is the sine of any angle is equal to the sine of 180 minus that angle. So an example of this in this problem would just be so for example, the sine of 120 degrees is the same thing as the sine of 180 minus a 120, which is the sine of 60. So you don't have to actually do all of this work to figure out what the sine of this sort of smaller right triangle is. Because what I've shown you here is that the sine of this 120, which was your original angle inside of your obtuse triangle, is actually the same as the sine of that supplementary angle. Alright? In fact, you can actually plug these things into your calculator, and what you'll see is that 6 times the sine of 60 ends up being 5.2, and 6 times the sine of 120 also ends up being 5.2. So what I'm trying to tell you in these problems here, what I'm really, what I'm trying to say is that you actually it might seem like you have to do all of this extra work to figure out the supplementary angle, but you actually don't. We could have just gone straight into using this equation, a b times the sine of C, because taking the sine of this angle with this hypotenuse is the same as taking the sine of this angle to figure out the height of this triangle. Okay? That's really what's going on here. So, really, I'm just going to take the area over here, and this is going to be 1/2. I know the base is 8. And what's the height? I just calculated the height of this triangle as 5.2. So this is going to be, sorry. This is going to be sin C, which ends up being 5.2. And the area of this triangle ends up being 20.8 units. Alright. And that is the final answer. That is the area included inside of this obtuse triangle. Alright. Thanks for watching, and I'll see you in the next one.

Calculating Area of ASA Triangles

What we've seen in the last couple of videos is that to calculate the area of a triangle when you're not given the heights, we find it by using the sine function. Now in some triangles, like the one we're going to work out here, we actually are not given all of the information that we need to plug into our area formulas. For example, this is an ASA triangle, meaning we have 2 angles and a side between them, but all of the area formulas require that we know 2 sides out of 3. So it seems like we're stuck here, but what I'm going to show in this video is that there's nothing to worry about. Because whenever this happens, we can always use the other formulas that we've learned, the law of sines and cosines, to find any missing information to plug into our area equations. Let's just jump into this problem here so I can show you it's really not so bad. Alright. Let's get started.

So we have a triangle over here. We have A is 40 degrees, C is 60 degrees, and b is 4. Alright? I've already drawn this here just to make it a little bit simpler. But, basically, the idea here is that I want to orient this triangle. So I've got my base over here, and then I can draw the height, which is going to be over here on this side. We've seen something similar to this. Alright? So what is the height of this triangle? Basically, if I make this little right triangle like this, the height, using SOHCAHTOA, is just going to be the hypotenuse of this triangle times the angle over here. In other words, it's just going to be c∙sin(A). The problem is I actually don't know what c is. Alright? So what happens is in order to go from my area formula, my area formula is going to be 1/2. The base of this triangle here is going to be b, which I know, and then the height of this triangle is going to be c∙sin(A). When you start looking at these variables, you know the b and you know the big A, but you don't know c. So in order to calculate the area, I need to actually go find this little c first. How do I do that? Well, this is a triangle in which I know I have some angles and sides here. So in order to find c, what I can do is I could just use the law of sines. Right? That's what we solve for missing sides. So what I'm going to do here is I'm going to set up my law of sines sin(A)/a ≡ sin(B)/b ≡ sin(C)/c. Remember, I'm looking for this little c over here. So what I'm going to have to do is I'm going to have to pick 2 out of these 3 ratios in which I know 3 out of 4 variables.

Let's go ahead and sort of check out which variables I know. I know little b, and I know big A, and I also know big C. So it might look here like I'm kind of stuck again because I only have 3 out of the 4 out of the 6 possible variables. But remember here that whenever you have 2 angles like A and C, you can always solve for that missing angle by using the angle sum formula. So what I'm gonna do here is, I'm just gonna go right up here and I'm gonna say that A+B+C = 180°. So therefore, B is just equal to 180 minus the other two angles. However, whenever you know 2 out of 3 angles, you can always easily solve for that other one. So this is gonna be 180 minus 40 minus 60, and this ends up equaling 80 degrees. So this angle over here is 80. And the reason we need that is we need this to basically figure out another variable so we can set up our law of sines. So now that I know what this B is, if you take a look here, the only sort of two ratios I can use are these last two ones over here. This is where I have 3 out of 4 variables. So, I'm going to pick these 2 out of 3 ratios, and we've seen how to solve for something like c before. Basically, what I'm going to do is I'm going to flip these variables. I've got c/sin(C) = b/sin(B). So, therefore, c is equal to I'm just gonna plug in everything now. I've got b is equal to 4 and then the sine of big B, which I just figured out that angle is going to be the sine of 80. Then I'm going to multiply this by big C, which goes up here, which is going to be the sine of 60 degrees. When you go ahead and plug this in, what you should get for little c is you should get 3.5. Now, remember, that's not your answer. That's not the area of this triangle. That's really just this third or this one of these sides over here that you were missing, the c here equals 3.5, and the reason you need that is that you need that to figure out the height of this triangle. So now that we've figured out c, we have everything we need to solve the problem because we figured this out. Okay. So this is just going to be one half of b, which is 4, times c, which I just figured out was 3.5 times the sine of angle A, which is 40 degrees. When you plug all this stuff into your calculator carefully, what you should find for the area is that the area is equal to about, 4.5. So this is gonna be 4.5, and that is the area of this triangle. Alright? So thanks for watching, and let me know if you have any questions.

Find the area of the triangle: $A=30\degree$, $b=10m$, $B=80\degree$.

• A. $44.8m^2$
• B. $11.9m^2$
• C. $26.2m^2$
• D. $23.9m^2$

Here’s what students ask on this topic:

How do you find the area of a triangle using the SAS (Side-Angle-Side) formula?

To find the area of a triangle using the SAS formula, you need two sides and the included angle. The formula is:

$\frac{1}{2}ab\sin C$

Here, $a$ and $b$ are the lengths of the two sides, and $C$ is the included angle. This formula leverages the sine function to calculate the height indirectly.

What is the formula for the area of a triangle when given two angles and the included side (ASA)?

For an ASA triangle, you can use the Law of Sines to find the missing side and then apply the SAS formula. First, find the third angle using:

$A+B+C=180°$

Then, use the Law of Sines to find the missing side:

$\frac{a}{\sin }=\frac{b}{\sin }$

Finally, apply the SAS formula:

$\frac{1}{2}ab\sin C$

How do you use the Law of Sines to find missing sides in a triangle?

The Law of Sines states:

$\frac{a}{\sin }=\frac{b}{\sin }=\frac{c}{\sin }$

To find a missing side, set up a proportion using the known sides and angles. For example, if you know $a$, $A$, and $B$, you can find $b$:

$\frac{a}{\sin }=\frac{b}{\sin }$

Rearrange to solve for $b$:

$b=\frac{a\sin B}{\sin }$

What is the significance of the sine function in finding the area of non-right triangles?

The sine function is crucial for finding the height of non-right triangles when the height is not directly given. By using the sine of a known angle, you can calculate the height indirectly. For example, if you know the hypotenuse $c$ and angle $A$, the height $h$ can be found using:

$h=c\sin A$

This height can then be used in the area formula:

$\frac{1}{2}bh$

How do you find the area of a triangle when given two sides and a non-included angle (SSA)?

For an SSA triangle, you can use the Law of Sines to find the missing angle and then apply the SAS formula. First, use the Law of Sines to find the missing angle:

$\frac{a}{\sin }=\frac{b}{\sin }$

Once you have the missing angle, use the SAS formula to find the area:

$\frac{1}{2}ab\sin C$