Thread: 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?

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

Since AB = AD, is isosceles so the base angles are the same.

Also, the angles of a triangle add up to

Now see if you can continue.

Hello, Rocher!

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

In the figure: .
Find: .

According to your description, CD can be anywhere . . .

Code:

                              D
                              *
                          *  x *    *
                      *         *   y     *
                  *              *              *
              *                   *                  * C
          *  20°                   *
      *  *  *  *  *  *  *  *  *  *  *
      A                             B

Here is the figure:

Nevermind.

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

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