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.


  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.

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  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°

  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°

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

  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
    Sri Lanka

  6. This is a link to the drawing
    ΔABC is isosceles on B => x=7α

    Define A’ symmetric A by BD
    Then DA=DA’, BA=BA’, ang(BDA) = 30° => ang(BDA′) = 30° => ang(ADA′) = 60°

    DA=DA’ and ang(ADA′) = 60° => ΔADA’ is equilateral

    ΔABC is congruent to ΔABA’ (SSS) => ang(BAA′) = ang(BA'A) = x
    Therefore ang(DAA′) = x +5α =60°
    But x=7α => 7α + 5α =60 => 12α =60 => α=5° => x= 35°