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### Course: ತರಗತಿ 7 ಗಣಿತ (ಭಾರತ) > Unit 6

Lesson 4: Two special triangles: Equilateral and isosceles# 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. ಸಾಲ್ ಖಾನ್ ರವರು ರಚಿಸಿದ್ದಾರೆ.

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ಇನ್ನೂ ಪೋಸ್ಟ್ಗಳಿಲ್ಲ.

## ವೀಡಿಯೊ ಪ್ರತಿಲಿಪಿ

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.