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.
Let triangle ABC be isosceles with AB=AC. Angle BAC
equals 20 degrees. Let points D and E be on sides AB
and AC repectively such that angle BCD is greater than
angle CBE. What integer measures ( in degrees ) can
angles BCD and CBE take such that the measure ( in
degrees ) of angle EDC is an integer.
Let b, c, and d be the measures of angles CBE, BCD,
and EDC respectively. Let F be the intersection of
BE and CD. Applying the sine rule to triangles
EDF: EF*sin(b+c-d) = DF*sin(d)
BDF: BF*sin(80-b) = DF*sin(100-c)
CEF: CF*sin(80-c) = EF*sin(100-b)
BCF: BF*sin(b) = CF*sin(c)
Eliminating the lengths from these equations gives
sin(d) sin(b)*sin(80-c)*sin(100-c) P
------------ = ----------------------------- = ---
sin(b+c-d) sin(c)*sin(80-b)*sin(100-b) Q
solving for d we get
P*sin(b+c)
tan(d) = ----------------
Q + P*cos(b+c)
I wrote a small program in Perl and got
BCD CBE EDC
-----------------
50 20 10
50 40 30
60 30 10
60 50 30 <-- Langley's Problem
65 25 5
65 60 40
70 50 10
70 60 20 <-- This Problem
There might be round off errors in the program, so
it would be nice to have a synthetic proof or
disproof for each of these cases.
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Posted by Bractals
on 2004-11-06 14:28:55 |
Generalization
