Triangle ABC is isosceles with AB=AC. Point D is on side AB such that angle BCD is 70 degrees.
Point E is on side AC such that angle EBC is 60 degrees.
Angle ABE equals 20 degrees, and angle DCE equals 10 degrees.
Find angle EDC. Justify your answer.
(In reply to re(2): Just following the rules
"The two angles at the left have to sum to 140. The two at the right have to sum to 150, and one of these is the angle EDC, which we are looking for."
But there are other constraints. For example, if you made angle DEB equal to 60 degrees, ED would be parallel to CB, which is clearly impossible as D must be higher than E (as it rises 70 degrees at BCD while E is up at only a 60 degree angle EBC).
By the law of sines CE/CB = sin 60 / sin 40, and DB/CB = sin 70 / sin 30, so DB/CE = sin 70 sin 40 / (sin 30 sin 60).
But DB/ED must equal sin DEB / sin 20, and CE/ED must equal sin EDC / sin 10, so DB/CE = sin DEB sin 10 / (sin EDC sin 20).
So sin DEB sin 10 / (sin EDC sin 20) = sin 70 sin 40 / (sin 30 sin 60). In other words, sin DEB/sin EDC = sin 20 sin 40 sin 70 / (sin 10 sin 30 sin 70). Not every pair of DEB and EDC that add up to 130 (completing the triangle with a 50-degree angle at F) fit this bill.
Posted by Charlie
on 2004-11-04 19:17:21