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 big A is 40 degrees, big C is 60 degrees, and little 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 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 12⋅b⋅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. I have sin(a)/a=sin(c)/b, and this equals sin(c)/c. Alright. 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, you know, 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 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°-A-C. However, whenever you know 2 out of 3 angles, you can always easily solve for that other one. So this is going to be 180°-40°-60°, and this ends up equaling 80°. 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=3.5, and the reason you need that is because 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 going to 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.
Table of contents
- 0. Review of College Algebra4h 43m
- 1. Measuring Angles39m
- 2. Trigonometric Functions on Right Triangles2h 5m
- 3. Unit Circle1h 19m
- 4. Graphing Trigonometric Functions1h 19m
- 5. Inverse Trigonometric Functions and Basic Trigonometric Equations1h 41m
- 6. Trigonometric Identities and More Equations2h 34m
- 7. Non-Right Triangles1h 38m
- 8. Vectors2h 25m
- 9. Polar Equations2h 5m
- 10. Parametric Equations1h 6m
- 11. Graphing Complex Numbers1h 7m
7. Non-Right Triangles
Area of SAS & ASA Triangles
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