# Thread: I need help on how to find variables for Isosceles Triangles

1. ## I need help on how to find variables for Isosceles Triangles

Here is what I'm dealing with:

I know y would be 49, but I don't know how to find the value of x. I'm a beginner at Geometry, so hopefully someone can explain this to me in a way that I understand.

2. Originally Posted by ladyk
Here is what I'm dealing with:

I know y would be 49, but I don't know how to find the value of x. I'm a beginner at Geometry, so hopefully someone can explain this to me in a way that I understand.
The angles opposite the congruent sides are congruent. (y is not 49)

In classical (Euclidean) geometry, the sum of the measures of the angles in a triangle equals $180^{\circ}$

Use these two facts to solve for x and y.

3. Ok, so I meant to put y would be 46, and therefore x would be 88. That's what I got

4. Originally Posted by ladyk
Ok, so I meant to put y would be 46, and therefore x would be 88. That's what I got
x is not 88. You found the angle for the bigger triangle. What about the one whose measures are x, 90, 46?

5. OK, so y is 46, and x is 90. Why would x be 90? I want to understand why that is.

6. I labeled your triangle.

Consider triangle BCD. I mentioned previously that the sum of the measures of the angles in a triangle equals 180. Therefore:
x+46+90=180

What you did was x+y+46=180, which is incorrect. The measure of angle B is not x. As you can see, there is an angle bisector that divides angle B into 2 congruent angles(<CBD and <DBA)

7. Thank you so much Chop Suey. When it's explained simply, you realize how easy it actually is. Would x=44 though?

8. Yes.

9. Thank you!