09 Jun '04 22:19>
OK my first theory was obviously flawed but heres a second attempt.
Label the DC BE intersect X
Assign a distance of 10 to BC
With this we can work out the distances of BX and CX
Now we can work out the distances of XE, CE, XD and BD
Using the distance of XE, XD and angle DXE (70) we can solve the CDE angle.
29.999999999997772
Can the poster tell me if this is the right answer??
Label the DC BE intersect X
Assign a distance of 10 to BC
With this we can work out the distances of BX and CX
Now we can work out the distances of XE, CE, XD and BD
Using the distance of XE, XD and angle DXE (70) we can solve the CDE angle.
29.999999999997772
Can the poster tell me if this is the right answer??