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

Everyone. Welcome back. So in this practice problem, we're gonna take a look at how we can calculate the area of a triangle, but it's not actually drawn for us. All we're given is some variables. Let's check it out. So we've got 2 angles, a is 30, b is 80, and we've got that little b is equal to 10. So let's see if we can actually just draw this triangle out and then go from there. Alright? So I'm gonna start with these two angles here. I've got 1 is 30 and one is 80. So I'm gonna draw a little base like this, and I'm gonna use the fact that I know that one of the angles is gonna be 30 degrees, which is gonna look something like this. So one of the angles is gonna be 30. Let me just draw this straight up like this. And I've got that one of the angles is gonna be 80 degrees, which is gonna look something like this. It's not quite gonna be a right angle. And what really, what I'm gonna do is I'm just gonna wait for these 2 sort of lines to intersect. Alright? So that's roughly what my triangle is gonna look like. This is going to be a. This is going to be the 80 degrees, which is b, which means that this is going to be angle c. Alright? And I actually don't know what that is. What that means is that this is side a, this is side b, which we know is 10, and this is side c, again, which is unknown. Alright? So this is basically what my triangle looks like if I sketched it out. Alright? And so, really, what we're going to try to do is try to find the area. Alright? So the area, remember, is just going to be 1 half base times height, and there's sort of 3 different versions of this equation. So let's take a look at the variables that I know and figure if we could eliminate at least 2 of the equations. Alright? So if you look through these variables, what you'll see here is that you don't know little c or big c or big b. So what that means over here is that when you go to look at these 2nd and third equations, because I don't know little c, then I can't or actually little a, and I have 2 unknowns in this equation I def can't solve. What about this third equation over here? So I don't know what little a is. I do know what little b is, but I don't know what sign of big c is equal to. Again, 2 unknowns. The only one that sort of seems to make sense is gonna be this top equation over here because I have little b and I have the sine of big a, but I actually don't know what this little c over here is. Here is. So I only have one unknown to sort of solve for. Alright. So remember that this is just gonna be 1 half of a base times a height, and I'm just gonna use bc times the sine of a. I want to rearrange this equation here because, really, the base of this triangle is c. So I want to put that sort of first. So this is going to be 1 half of c times b sine a. Alright? So remember, this term over here is gonna be the base of my triangle, which means this term over here has to represent the height. Why? Let's take a look at our triangle and see if we can sort of understand why. Remember, to draw a height, you're gonna take the one of the, corners over here that's usually the the corner that's opposite of the base and just draw a dashed line that goes all the way down until it intersects, and you make a little right triangle. So you're basically sort of creating a little right triangle like this. The heights of this little triangle that you've made can be related by sohcahtoa. It's basically just the hypotenuse of this right triangle that you've made times the sine of this angle. So, in other words, it's b times the sine of a. So, in other words, it's just 10 times the sine of 30 degrees. Therefore, the height of this triangle is actually just 5. Okay? So that's the height. We already know what this is. Now, we don't actually know what this little c is. As long as we can figure that out, then we can figure out the area of this triangle. Alright? So we already have what this sort of, this other term over here is. Now I just need to go ahead and solve for a little c. How do I do that? Well, basically, what we can do here is I know that I at least know one side and its opposing angle. So if I'm trying to solve for a missing side over here, I can use the law of sines. Alright? So again, sometimes in these problems, you're going to set up your area equation, and you're going to get stuck because you don't have all the information that you need to solve. But you'll be able to unlock that variable by solving for law of sines. Alright? So what I'm gonna do here is the law of sines is I'm gonna use these 2 b terms over here because I know the sine and the angle. So So in the word, this is going to be sine of b over b equals the sine of big C over c. Alright. I'm looking for this little c over here. I have what big B and little b are. The only problem is in this problem here is that I actually don't know what big C is equal to. Alright. So how do we solve for this? Well, it might seem like you're gonna get stuck again because there's only 2 out of 4 variables here. But what happens here is when you actually know what 2 of the angles are inside of this triangle, so we can all solve for the 3rd. Right? We have these two angles over here. Solving for that third one is super easy. We've done this a 1000000 times before. This angle c is just going to be 180 minuteus the other two angles, which are 8030. So in other words, c is equal to 70 degrees. Alright? Very straightforward. So basically we figured out that this is 70 degrees over here. And now what we can do is now we know that this is equal to 70. We can set up our equation. We've done this a bunch of times before. We sort of flip these two fractions. This is c over sine of big C is equal to b over sine of big B. And this is really just if you cross multiply, this is going to be 10 divided by the sine of big B, which is 80 times the sign of big C, which we just figured out, is 70. When you go ahead and plug this in, what you're going to get is 9.54. All right. So Remember, we're not quite done yet here. All we've done is we figured out for this little c over here, and now we can actually just sort of plug this into our area equation. So the area, putting all this stuff together, is going to be 1 half of 9.54 times the heights, which is just 5. You could replug all this stuff in, but we did that over here. This is 10 sin 30, which equals 5. So if you work this out, what you're going to get here is an area of 23. And I'm just going to round this to the nearest tenth. It's 85. So this is going to be 23.9, and that is going to be the area of this triangle. Alright? So that is the area of this big triangle that we've created. Let me know if you have any questions, and I'll see you in the next one.

13. Non-Right Triangles

Area of SAS & ASA Triangles

Precalculus

13. Non-Right Triangles

Area of SAS & ASA Triangles

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Multiple Choice

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

$44.8m^2$

$11.9m^2$

$26.2m^2$

$23.9m^2$

Verified step by step guidance

Identify the given values: angle A = 30°, side b = 10 m, and angle B = 80°.

Use the fact that the sum of angles in a triangle is 180° to find angle C. Calculate C = 180° - A - B.

Apply the Law of Sines to find side a. The Law of Sines states: $ \frac{a}{\sin A} = \frac{b}{\sin B} $. Rearrange to solve for a: $ a = b \cdot \frac{\sin A}{\sin B} $.

Use the formula for the area of a triangle when two angles and a side are known: $ \text{Area} = \frac{1}{2} \cdot a \cdot b \cdot \sin C $.

Substitute the known values into the area formula and simplify to find the area of the triangle.

4:2m

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Find the area of the triangle: A=30°A=30\degreeA=30°, b=10mb=10mb=10m, B=80°B=80\degreeB=80°.

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