1. ## Angles of Triangle

In the figure, Angle DAB=20 degrees and AB = AD = CD. Find the value of x and y. x being angle ADB, y being angle BDC.

All help is appreciated. DCB is 20 degrees right?

2. You need to attach the figure or we cannot tell what DCB is.

Since AB = AD, $\displaystyle \triangle ABD$ is isosceles so the base angles are the same.

Also, the angles of a triangle add up to $\displaystyle 180^o$

Now see if you can continue.

3. Hello, Rocher!

We still need more specifics.
Do we have a parallelogram?

In the figure: .$\displaystyle \angle DAB = 20^o,\;AB = AD = CD$
Find: .$\displaystyle x = \angle ADB,\:y = \angle BDC.$
According to your description, CD can be anywhere . . .
Code:
                              D
*
*  x *    *
*         *   y     *
*              *              *
*                   *                  * C
*  20°                   *
*  *  *  *  *  *  *  *  *  *  *
A                             B

4. Here is the figure:

Nevermind.

5. Originally Posted by Rocher
In the figure, Angle DAB=20 degrees and AB = AD = CD. Find the value of x and y. x being angle ADB, y being angle BDC.

All help is appreciated. DCB is 20 degrees right?
This diagram is messed up somehow. Every statement you made above has something wrong with it!

Angle x is ABD, not ADB. Angle y is CAD, not BDC. You have AB = AD = AC, not AB = AD = CD. DCB is a straight line segment, not an angle. And though this is not a critical point to make, angle DAB doesn't look anywhere near 20 degrees on your diagram.

-Dan

6. Heh, I'm a retard. Drew the wrong figure, and now its time for class. Oh well, thanks guys.