Thursday, July 10, 2014

Geometry Problem 1027: Triangle, Double, Quadruple, Angle, Congruence

Geometry Problem. Post your solution in the comments box below.
Level: Mathematics Education, High School, Honors Geometry, College.

Click the figure below to see the complete problem 1027.

Online Math: Geometry Problem 1027: Triangle, Double, Quadruple, Angle, Congruence

7 comments:

  1. Let b= AC=AD
    We have DC=DE= 2. b. sin(2α )
    ∠(BED)=3α
    Relation in triangle AED => ED/sin(2α )= AD/sin(3α) = 2.b.sin(2α )/ sin(2α )= 2.b
    So sin(3α )= ½= sin(30) => α = 10
    So x= 90- 4α = 50

    ReplyDelete
  2. There is a geometric solution for this problem ,right?

    ReplyDelete
    Replies
    1. To Dima
      Problem 1027
      Yes, there is a geometric solution, Thanks.

      Delete
  3. We build
    AH perpendicular BC
    DF perpendicular AB

    DF=DH = 1/2ED X =50 grade

    Erina New Jersey

    ReplyDelete
    Replies
    1. Hi Erina,
      Please, send again your complete solution, there is some problem with the system.
      Thanks. Antonio

      Delete
  4. Hi Antonio

    Please publish the full solution

    Prob 1027
    We build AH perpendicular BC and DF perpendicular AB
    CH=HD=1/2DC=1/2ED
    DF=DH=1/2ED
    α =10
    X=90-4α
    X=50

    Prob 1028

    BE=ED=EC EM=EN EN =1/2 EC
    Triangle ECN < ECN =30 ECN = 90-4X -> X =15

    Prob 1029
    We note <BDF = 90
    BD perpendicular AC


    Erina Erina Jersey

    ReplyDelete
  5. We build

    1. AH perpendicular BC --> CH =HD =1/2 DC = 1/2ED
    2. DF perpendicular AB ----> DF=DH =1/2 ED
    3. triangle DEF α =10
    4. X=90-4α X = 50

    Erina New Jersey

    ReplyDelete