Monday, May 19, 2008

Elearn Geometry Problem 44



See complete Problem 44
Angles, Triangle. Level: High School, SAT Prep, College geometry

Post your solutions or ideas in the comments.

6 comments:

  1. Tr. ABC gives us, AC/AB = sin(x)/sin(2α)
    Tr. ACD gives us, AC/CD = sin(3α)/sin(2α)
    But AB=CD, so sin(x)/sin(2α)= sin(3α)/sin(2α)
    Thus, sin(x)=sin(3α)or x=3α. Now in Tr. ABC,
    3α+4α+2α =180 or α =20 deg. and x=3α=60 deg.
    Ajit:ajitathle@gmail.com

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  4. The solution is uploaded to the following link:
    https://docs.google.com/open?id=0B6XXCq92fLJJWHh3X19VYWFYNjQ

    Sir, My i send my solution in tamil language as a copy, so that my students can easily understand the way.
    With regards,
    PradheepKannah.

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  5. Let AD meet BC at E. Drop perpendiculars DH and DJ from D to AC and AB respectively. Drop another perpendicular BFG from B to AF to meet the bisector of < BAD at F and AD at G

    Since D is the incentre of Tr. ABC < ABD = x/2 = 90-3@. Since < ABG = AGB = 90-@, < DBG = 2@ and so BD = BG.

    Now Tr. s CDH, ABF & AFG all being congruent ASA, BG = BD = 2r where r is the inradius. But JD also = r so in right Tr. BDJ, BD = 2DJ hence this Tr. is 30-60-90.

    So x/2 = 30 and x = 60

    Sumith Peiris
    Moratuwa
    Sri Lanka

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