Find the Angle

[Thud&Blunder]

The isosceles triangle ABC has AB = AC and angle BAC = 20 degrees.

 

The point D lies on AC such that angle DBC = 60 degrees.

The point E lies on AB such that angle ECB = 50 degrees.

 

Find angle EDB.

[Thud&Blunder]

Here's a diagram.

[Callipygous]

this is really bothering me. it seems like there is enough information. geometry got me a bunch of info, but im still missing the 4 crucial angles. algebra seems to just go in circles, and i cant think of a way to apply trig to it...

 

maybe i should have paid more attention in intermediate algebra. : P

[reverse]

Um, I think I see a way to do it.

 

did you want a clue, maybe we can work together on it.

[Callipygous]

sure

[reverse]

Well, I haven’t thought it right through,

But my idea is this.

 

There are two overlapping triangles, one facing left and one facing right.

Each of the two triangles has only one unknown angle.

 

The two original triangles are locked into a relationship.

 

You can calculate the two unknown angles with the aid of Pythagoras.

 

Relationship between the two unknown angles will give me that answer,

or at least the answer when subtracted from 180 degrees.

 

Any use to you?

[Callipygous]

your gonna have to tell me which triangles your talking about.

 

 

there are 4 angles in the whole thing that i cant find. marked w, x, y and z.

[Dapthar]

i have it uploaded, cant find a way to include it.

Click the 'Manage Attachments' button below the 'Submit Reply' button, and select your attachment.

[uncool]

sin(FBE)/sin(FEB) = EF/BF

sin(FCD)/sin(FDC) = DF/CF

sin(FBC)/sin(FCB) = CF/BF

sin(FED)/sin(FDE) = DF/EF

So:

 

sin(FED)/sin(FDE) = DF/EF = (CF*sin(FCD)/sin(FDC))/(BF*sin(FBE)/sin(FEB)) = (sin(FEB)*sin(FCD)*sin(FBC))/(sin(FBE)*sin(FDC)*sin(FCB)) = (sin(40)*sin(20)*sin(50))/(sin(30)*sin(50)*sin(60))

 

FED = 110-x

sin(110-x)/sin(x) = (sin(40)*sin(20)*sin(50))/(sin(30)*sin(50)*sin(60))

sin(110)*tan(x) - cos(110) = (sin(40)*sin(20)*sin(50))/(sin(30)*sin(50)*sin(60))

tan(x) = ((sin(40)*sin(20)*sin(50))/(sin(30)*sin(50)*sin(60)) + cos(110))/sin(110)

Find x.

-Uncool-

[Callipygous]

havent been able to find my graphing calc recently... : (

[reverse]

I might be on the wrong track,

but I see only a total of three overlapping transparent triangles.

 

The other angles are an illusion.

 

Do you get what I mean?

[Callipygous]

not at all.

[dan19_83]

Callipygous, using your diagram above i have come up with four eqaution as follows:

 

40 + X + Y = 180

50 + W + Z = 180

70 + Y + Z = 180

20 + X + W = 180

 

using these four equations it should then be possible to work out the values for W, X, Y and Z.

I'll let you do the calculations!!

[Callipygous]

Callipygous' date=' using your diagram above i have come up with four eqaution as follows:

 

40 + X + Y = 180

50 + W + Z = 180

70 + Y + Z = 180

20 + X + W = 180

 

using these four equations it should then be possible to work out the values for W, X, Y and Z.

I'll let you do the calculations!![/quote']

 

 

thats all well and good, except your working with the exact same relationships all the way around.

 

in normal systems that kind of thinking works great, but all the numbers are the same in this one (its always related to 180 the same way). the end result is that when you do normal system solving methods your variables cancel out and you end up with things like 140=140.

 

which is what i meant about the algebra going in cirlces.

[dan19_83]

ok, had to figure that one out the hard way!! took me half an hour of calculations to figure out that you were right.

 

here's an idea, this is just an idea though so it might be way off the mark.

The triangle that is enclosed by the two angles of 50 degrees and the corner of the main triangle is an isoleces (?!) triangle.

could some trigo be used here to get your 4 unknown angles?

just a thought.

[reverse]

not at all.

 

Ummmmm...

 

Ok,

Halfway between point A and point B there is another point.

 

please call it point D

 

and.

 

Halfway between point A and point C there is another point.

 

Please call it point E.

 

Please remove the LINE that goes from point D to point E.

 

Now throw your brain at it.

[Callipygous]

if you remove that line you also remove the angle we are looking for.

[reverse]

Correct,

Please do it anyway .

there is no penalty.

You can put it back later, after you have looked at the triangles in a new light.

[Callipygous]

ok... i dont understand what your getting at. when you remove that line you remove the only parts of the problem im still trying to figure out. maybe be a little more specific?

[reverse]

That's ok.

I'm not sure what I'm getting at either.

so with the line DE removed, can you now see only three overlapping triangles?

[Callipygous]

there are still more than 3 in the picture, but i think i can see which three you would be talking about.

[reverse]

Yes!

 

The three big ones.

 

Can you use Pythagoras to work out the Two unknown angles?

I mean.

s=o/h

c=a/h

t=o/a

[Callipygous]

you mean the two big angles on either side?

 

140 and 130.

 

basic geometry will get you that.

[reverse]

Yes do it that way,

why make life hard.

 

So now are you able to put those two angles and your removed DE line back on your diagram.

[Callipygous]

and.... accomplish nothing.

 

i was fully aware of this fact the entire time. its doesnt help. all that tells you is the 4 angles total 270.