Hyper2400

i could use the practice ....let's see who can solve it the most efficiently...we can all learn from these efficient methods...

ExRunner

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. )

iducky

is it 10 degrees? LOL just a guess

ExRunner

No, it is not.

geddes97

Is this 75?

elladorkess

Is it 60 degrees?

iducky

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

Vehement

double post

Vehement

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

xiggi

........

ExRunner

It is not 70.

gcf101

It'll make it so much easier to jump right to the next question if you are not interested in one before.
thanks much!

Vehement

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

greencloth

I messed up

xiggi

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.