Current time:0:00Total duration:3:15

0 energy points

# 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.

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, let me put another label there Let me put another label there just for fun Let' call this！ listen, you can call this a ray cause 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！ they give us these two angles right over here, these adjacent angles, they give it to us in terms of x And what I wanna 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, you could！ we could turn this into a line CD, so it's not just a ray anymore, so it just keeps going on and on in both directions The fact they'd given us a parallel line tells us that maybe we can use a little！ we know about transversals and parallel lines to figure out some of the angles here And you might recognize, 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 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+10 right over here, and its corresponding angle is right down here This would also be x +10 And if this is x + 10, then you have a vertical angle right over here that would also be x + 10 Or you could say that you have alternate interior angles that would also be congruent Either way, these base angles wanna be x+10 Well, it's an isosceles triangle so you're 2 base angles are going to be congruent So if this is x+10, then this is going to be x+10 as well And now we have the 3 angles of a triangle expressed as as functions of, expressed in terms of x So when we take their sum, then they'd be equal to 180 and then we can actually solve for x We get 2x + x + 10 + x + 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 + 10 and another +10, so that gives us a +20, is equal to 180 And we can subtract 20 from both sides of that And we get 4x 4x is equal to 160 Divide both sides by four, and we get x is equal to 40 And we're done! We figured out what x is And we can actually figure out what these angles are This is x+10, then you have 40 + 10 This right over here is going to be a 50 degree angle This is 2x, so 2 * 40 This is an 80 degree angle It doesn't look like it, the way I've drawn it And that's why you should never take anything 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 2 base angles right over here are also going to be are also going to be 50 degrees So you have 50 degrees, 50 degrees, and 80, they add up to 180 degrees