Monday, May 19, 2008

Elearn Geometry Problem 48



See complete Problem 48
Angles, Triangle, Congruence. Level: High School, SAT Prep, College geometry

Post your solutions or ideas in the comments.

6 comments:

  1. I've no idea how to solve this by plane geometry alone & would very much appreciate it if someone gave me a hint. In the meanwhile, AB=BC gives us x=7α. Now in Tr. ABC, AC/AB=sin(14α)/sin(7α)=2cos(7α)and in Tr. ABD, AD/AB=AC/AB=sin(5α+30)/sin(30). This gives us, 2sin(5α+30)=2cos(7α)
    or sin(5α+30)=sin(90-7α), which in turn, yields(5α+30)=(90-7α) or 12α=60 or α=5 deg.
    Hence x = 7α = 35 deg.
    Ajit: ajitathle@gmail.com

    ReplyDelete
  2. This comment has been removed by a blog administrator.

    ReplyDelete
  3. Solution to Problem 48:


    Let BE the perpendicular bisector of AC.
    Let F be a point on the extension of DB, so that FD⊥AF, then ∆BFA≅∆BEA.
    m∠FBA=m∠D+m∠BAD=5a+30°, and m∠CAB=m∠FAB=7a.
    Therefore: m∠FAB+m∠FBA=7a+5a+30°=90°,12a=60° and a=5°
    Since ∆ABC is isosceles, x=7a=7(5°) and x=35°

    ReplyDelete
  4. Let ΔABO be equilateral triangle. ∠AOB=60° and ∠ADB=30°, so OA=OB=OD=AB.
    AB=OA, BC=OD, AC=AD ---> ΔABC and ΔOAD are congruent.
    ---> ∠AOD=7α ---> ∠OAB=12α=60°---> α=5° ---> x=35°

    ReplyDelete
    Replies
    1. ∠AOD=7α --> ∠OAD=7α, please revise!

      Delete
  5. The key is to use the 30 degrees.

    Draw a 30-60-90 Tr. on AD, Tr. ADE, with AE perpendicular to BD.

    If BF is the perpendicular bisector of AC then easily Tr. ABE is congruent with Tr. BCF.

    So < EAB = x = 7@ and hence 12@ = 60 and so x = 7@ = 35

    Sumith Peiris
    Moratuwa
    Sri Lanka

    ReplyDelete