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Finding angles in isosceles triangles (example 2)

Sal combines what we know about isosceles triangles and parallel lines with the power of algebra to solve the angles of an isosceles triangle. Created by Sal Khan.

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Video transcript

So what do we have here? We have a triangle, and we know that the length of AC is equal to the length of CB. So this is an isosceles triangle, we have two of its legs are equal to each other. And then they also tell us that this line up here, they didn't put another label there. Let me put another label there just for fun. Let's call this, you could even call this a ray because it's starting at C, that line or ray CD is parallel to this segment AB over here, and that's interesting. Then they give us these two angles right over here, these adjacent angles. They give it to us in terms of x. And what I want to do in this video is try to figure out what x is. And so given that they told us that this line and this line are parallel, and we can turn this into line CD, so it's not just a ray anymore, so it just keeps going on and on in both directions. The fact that they've given us a parallel line tells us that maybe we can use some of what we know about transversals and parallel lines to figure out some of the angles here. And you might recognize that this right over here, this line-- let me do that in a better color. You might recognize that line CB is a transversal for those two parallel lines. Let's let me draw both of the parallel lines a little bit more so that you can recognize that as a transversal, and then a few things might jump out. You have this x plus 10 right over here, and its corresponding angle is right down here. This would also be x plus 10. And if this is x plus 10, then you have a vertical angle right over here that would also be x plus 10. Or you could say that you have alternate interior angles that would also be congruent. Either way, this base angle is going to be x plus 10. Well, it's an isosceles triangle. So your two base angles are going to be congruent. So if this is x plus 10, then this is going to be x plus 10 as well. And now we have the three angles of a triangle expressed as functions of-- expressed in terms of x. So when we take their sum, they need to be equal to 180, and then we can actually solve for x. We get 2x plus x plus 10 plus x plus 10 is going to be equal to 180 degrees. And then we can add up the x's. So we have a 2x there plus an x plus another x, that gives us 4x. 4 x's. And then we have a plus 10 and another plus 10, so that gives us a plus 20, is equal to 180. And we can subtract 20 from both sides of that, and we get 4x is equal to 160. Divide both sides by 4, and we get x is equal to 40. And we're done. We've figured out what x is, and then we could actually figure out what these angles are. If this is x plus 10 then you have 40 plus 10, this right over here is going to be a 50-degree angle. This is 2x, so 2 times 40, this is an 80-degree angle. It doesn't look at it the way I've drawn it, and that's why you should never assume anything based on how a diagram is drawn. So this right over here is going to be an 80-degree angle, and then these two base angles right over here are also going to be 50 degrees. So you have 50 degrees, 50 degrees, and 80, they add up to 180 degrees.