Angles of parallelogram : I'm stumped!

[rahidz2003]

Here's the question, it sounds so simple, but I can't figure it out
Does it even have an answer?

ABCD is a parallelogram. AC and BD meet at point E.
Angle AED is 60 degrees.
Angle BAE is 40 degrees.

What is the measure of angle DBC?

I've figured out all the angles except for DBC, ACB, BDA, and CAD, I need help!

 

[Opalg]

There doesn't seem to be a neat geometric approach to this one. It looks as though you have to attack it with some heavy-duty trigonometry.

Use the sine rule in the triangle ABE. That will tell you the ratio AE/BE. But AE=EC, so you can then use the sine rule again, in the triangle BEC. If \(\displaystyle \alpha\) denotes the angle DBC, then the angle ECB is \(\displaystyle 120^{\circ}-\alpha.\) The sine rule gives you a trigonometric equation for \(\displaystyle \alpha\). I get the answer to be \(\displaystyle \displaystyle\alpha = \arctan\left(\frac{\sqrt3}{4\cos20^{\circ} - 1}\right)\approx 32.122^{\circ}.\)

 

[rahidz2003]

There doesn't seem to be a neat geometric approach to this one. It looks as though you have to attack it with some heavy-duty trigonometry.

Use the sine rule in the triangle ABE. That will tell you the ratio AE/BE. But AE=EC, so you can then use the sine rule again, in the triangle BEC. If \(\displaystyle \alpha\) denotes the angle DBC, then the angle ECB is \(\displaystyle 120^{\circ}-\alpha.\) The sine rule gives you a trigonometric equation for \(\displaystyle \alpha\). I get the answer to be \(\displaystyle \displaystyle\alpha = \arctan\left(\frac{\sqrt3}{4\cos20^{\circ} - 1}\right)\approx 32.122^{\circ}.\)

Eww, that's not a pretty number

Thanks for the equation and help though, I really appreciate it!