Angles of Triangle

• Dec 21st 2007, 08:28 PM
Rocher
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?
• Dec 22nd 2007, 12:52 AM
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.
• Dec 22nd 2007, 04:36 AM
Soroban
Hello, Rocher!

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

Quote:

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```
• Dec 22nd 2007, 08:40 AM
Rocher
Here is the figure:

Nevermind.
• Dec 22nd 2007, 08:49 AM
topsquark
Quote:

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
• Dec 22nd 2007, 08:51 AM
Rocher
Heh, I'm a retard. Drew the wrong figure, and now its time for class. Oh well, thanks guys.