hey guys, anyone know how to solve for theta? I'm pretty

Just looking at it, I don't think you need any trig stuff, it looks like it's mostly adding up angles and such.

>>204484
Riddle me this: How can you use trig functions without a side length.

Listen to >>204494 and remember that if you add the angles of a triangle up you get 180°. Also keep in mind that this picture is horribly not to scale as they say in the picture.

>>204503
You can attempt trig things, but it would be absolutely a waste of time and retarded.

File: Screenshot_2016-10-13-22-46-00.png (468KB, 1080x1920px) Image search: [Google]

468KB, 1080x1920px

it was fun

>>204612
△ABC = 200°?
Center △ = 90°?

I got stuck until I realized that triangles aren't the only rule you need to look at. Also remember that the sum of angles on a 4-sided figure add up to 360, and that makes it possible to determine the angle.

File: problem1-large.gif (12KB, 330x887px) Image search: [Google]

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20°

>>204612
Angle A can not be 40
It is 20
Calculate as total angle of triangle is 180

>>204612
obviously wrong
>>204655
can't be, your triangle ABG has total interior angles of 190

Bump. I think i solved it, one moment

>>205258 (You)
I'm this anon, so here goes:

First, we can easily deduce that (angles are in degrees):
△BDE = 170 - Ɵ
△CDE = 130 - Ɵ
△DEC = 30 + Ɵ
△BDC = 40

From the sinus law, we have:
In triangle BDE, sin Ɵ/BD = sin(170 - Ɵ)/BE = sin 10/DE
In triangle DEC, sin 20/DE = sin(130 - Ɵ)/EC = sin (30 + Ɵ)/DC
In triangle BDC, sin 60/BD = sin 80/DC = sin 40/BC


From the first relation, we deduce BD = sin Ɵ*DE/sin 10
From the second relation, we deduce DC = sin (30 + Ɵ)*DE/sin 20
From the third relation, we have BD = sin 60*DC/sin 80


Relations 1 and 2 give us:
sin Ɵ*DE/sin 10 = sin 60*DC/sin 80


We then replace DC, we get:
sin Ɵ*DE/sin 10 = sin 60*sin (30 + Ɵ)*DE/(sin 20*sin 80)


We can simplify by DE, then we have to solve an equation with one unknown (Ɵ):
sin Ɵ/sin 10 = sin 60*sin (30 + Ɵ)/(sin 20*sin 80)


We use the relation sin(a+b) = sin a*sin b + cos a*cos b, we get:
sin Ɵ/sin 10 = sin 60*[sin 30*sin Ɵ + cos 30*cos Ɵ]/(sin 20*sin 80)


The rest is really tedious but we can easily deduce theta. I'll let you do this part yourself :)