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?

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- Dec 21st 2007, 08:28 PMRocherAngles 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 AMbadgerigar
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 AMSoroban
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: .$\displaystyle x = \angle ADB,\:y = \angle BDC.$

**anywhere**. . .Code:`D`

*

* x * *

* * y *

* * *

* * * C

* 20° *

* * * * * * * * * * *

A B

- Dec 22nd 2007, 08:40 AMRocher
Here is the figure:

Nevermind. - Dec 22nd 2007, 08:49 AMtopsquark
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 AMRocher
Heh, I'm a retard. Drew the wrong figure, and now its time for class. Oh well, thanks guys.