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

Registered User Posts: 770 Member
edited July 2005
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

## Replies to: Math CHALLENGE...Post the hardest math SAT I questions you can find..

• Registered User Posts: 731 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. )
• Registered User Posts: 39 Junior Member
is it 10 degrees? LOL just a guess
• Registered User Posts: 731 Member
No, it is not.
• Registered User Posts: 175 Junior Member
Is this 75?
• Registered User Posts: 238 Junior Member
Is it 60 degrees?
• Registered User Posts: 39 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
• Registered User Posts: 957 Member
double post
• Registered User Posts: 957 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

• Registered User Posts: 25,432 Senior Member
:) ........ :)
• Registered User Posts: 731 Member
It is not 70.
• Registered User Posts: 2,529
It'll make it so much easier to jump right to the next question if you are not interested in one before.
thanks much!
• Registered User Posts: 957 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
• Registered User Posts: 235 Junior Member
I messed up
• Registered User Posts: 25,432 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.
• Registered User Posts: 731 Member
And btw, "180 - angel a(20)/2 = ade = 80" is where you go wrong. You can't assume that.
This discussion has been closed.