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Math CHALLENGE...Post the hardest math SAT I questions you can find..

Hyper2400Hyper2400 Posts: 770Registered User Member
i could use the practice :)....let's see who can solve it the most efficiently...we can all learn from these efficient methods...
Post edited by Hyper2400 on
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Replies to: Math CHALLENGE...Post the hardest math SAT I questions you can find..

  • ExRunnerExRunner Posts: 731Registered User Member
    Given a triangle ABC as described below: Side AB = Side AC. Draw a line from C to side AB. call that line CD. Now draw a line from B to side AC. Call that line BE. Let angle EBC = 60 degrees, angle BCD equal 70 degrees, angle ABE equal 20 degrees, and angle DCE equal 10 degrees. Now draw line DE. The question--find what angle EDC is.

    (This can be, and is meant to be, solved geometrically, without trig. )
  • iduckyiducky Posts: 39Registered User Junior Member
    is it 10 degrees? LOL just a guess
  • ExRunnerExRunner Posts: 731Registered User Member
    No, it is not.
  • geddes97geddes97 Posts: 175Registered User Junior Member
    Is this 75?
  • elladorkesselladorkess Posts: 238Registered User Junior Member
    Is it 60 degrees?
  • iduckyiducky Posts: 39Registered User Junior Member
    oops i put my tick mark on the wrong side.. giving me the wrong isoceles!! did it again but i still thinkg i got it wrong .. 65? .. if the mid point (let's say O) is where everythign intersects and BOC = 50 thus DOE = 50 .. (my logic might be wrong but i believe triangle DOE is an isoceles) leave 130 degrees divided by 2.. 65? there must be an easier way ahha


    EDIT: Vehement answers works out for me.. did that the first time with my wrong isoceles.. but changed my method second time around =/ whoops
  • VehementVehement Posts: 957Registered User Member
    double post
  • VehementVehement Posts: 957Registered User Member
    its 70.....


    ok the full steps

    first add up bde, ebc, ecb and ecd. you will get 160.

    180 - 160 = angel a

    180 - angel a(20)/2 = ade = 80

    180 - dbe - dbc - bce = angel bdc = 30

    subtract bdc(30) and ade(80) from 180 = the answer = 70
  • xiggixiggi Posts: 21,609Registered User Senior Member
    :) ........ :)
  • ExRunnerExRunner Posts: 731Registered User Member
    It is not 70.
  • gcf101gcf101 Posts: 2,214Super Moderator Senior Member
    It'll make it so much easier to jump right to the next question if you are not interested in one before.
    thanks much!
  • VehementVehement Posts: 957Registered User Member
    Hmm you copied the question wrong then? or i seriously misread. i have figured out every angle in the triangle so i really don't know where my prob is
  • greenclothgreencloth Posts: 235Registered User Junior Member
    I messed up
  • xiggixiggi Posts: 21,609Registered User Senior Member
    Vehement, the problem is far from being a sixty seconds SAT question. It is the Gruber's challenge. It takes a ... long time to develop a solution.
  • ExRunnerExRunner Posts: 731Registered User Member
    And btw, "180 - angel a(20)/2 = ade = 80" is where you go wrong. You can't assume that.
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