 All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars  perplexus dot info  Measure that angle IV (Posted on 2004-11-02) 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.

 See The Solution Submitted by Brian Smith Rating: 3.6667 (9 votes) Comments: ( Back to comment list | You must be logged in to post comments.) Generalization | Comment 15 of 21 | ` `
`Let triangle ABC be isosceles with AB=AC. Angle BACequals 20 degrees. Let points D and E be on sides ABand AC repectively such that angle BCD is greater thanangle CBE. What integer measures ( in degrees ) canangles BCD and CBE take such that the measure ( indegrees ) 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 ofBE 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, soit would be nice to have a synthetic proof ordisproof for each of these cases.`
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 Posted by Bractals on 2004-11-06 14:28:55 Please log in:
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