Triangle Problem

My math teacher gave us a geometry problem as a sort of fun extracurricular thing and I found it interesting so maybe the mathy people here will enjoy it.

Triangle ABC is isoceles, angle BAC = 20 degrees
Angle EBC is 50 degrees.
Angle DCB is 60 degrees.
Find theta.

Here's a crappy diagram done in paint:

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27.43 degrees? sec
12.88 degrees? nvm I'm obviously retarded
nvm I think I got it, ~12.9*

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ABC = ACB = 80 degrees due to isosceles.
Name the intersection of BE and DC "F".
BFC = 70 degrees due to triangle
BFD = 110 degrees complementary.
BDF = 40 degrees due to triangle
BEC = 50 degrees due to triangle
20 + ADE + DEA = 180 triangle (1)
40 + theta + ADE = 180 line (2)
theta + 70 + DEF = 180 triangle (3)
50 + FED + DEA = 180 line (4)
This is 4 equations in 4 unknowns so it's solvable if they're linearly independent.
DEA = 160 - ADE (1)
FED = 130 - DEA = ADE - 30 (4)
theta = 140 - ADE (3 with DEF subbed in) - nope, they're not linearly independent.

That's about how far I got. It's obviously possible to solve it using trig - let BC = 1, then you can calculate the lengths EC, BD, and CD, resulting in the DEC triangle having two known side lengths and one known angle, which is enough to figure out the rest. But that seems rather unpleasant to actually go through.

Answer is 30, but when I tried to solve it numerically, I would get zero, and when I solved it using matrices, I would get 55, so I don't know what's up with that.

Well, it's less than 61 degrees I think, that's all I know. No clue how to solve the upper half of the triangle.

Answer is 30, but when I tried to solve it numerically, I would get zero, and when I solved it using matrices, I would get 55, so I don't know what's up with that.

Since I got a conflicting answer, I suppose it's time I show my work:

 Spoiler

- Prove that triangle BEC is isoceles, (angle BEC = 50*) -> BC = CE
- Law of Sines -> sin(theta)/CE = sin(DCA)/AD (with DCA = 20*) -> sin(theta) = sin20*BC/AD (1)
- Prove Triangle ADC is isoceles at D -> AD = DC -> need DC
- Law of Sines, part 2 -> sin80/DC = sin(BDC)/BC -> DC (= CE) = BC * sin80/sin40 (2)

(1) & (2) => sin(theta) = sin20*(BC/(BC * sin80/sin40)) = sin40*sin20/sin80 = 0.223
=> theta = arcsin(0.223) = ~13*

I'll go dig a hole now. Nevertheless, I think it's possible to find AD and AE with the above direction (in terms of BC's length).. then proceed to find angle ADE and AED (2 equations with 2 unknowns) then theta.

Since I got a conflicting answer, I suppose it's time I show my work:

 Spoiler

- Prove that triangle BEC is isoceles, (angle BEC = 50*) -> BC = CE
- Law of Sines -> sin(theta)/CE = sin(DCA)/AD (with DCA = 20*) -> sin(theta) = sin20*BC/AD (1)
- Prove Triangle ADC is isoceles at D -> AD = DC -> need DC
- Law of Sines, part 2 -> sin80/DC = sin(BDC)/BC -> DC (= CE) = BC * sin80/sin40 (2)

(1) & (2) => sin(theta) = sin20*(BC/(BC * sin80/sin40)) = sin40*sin20/sin80 = 0.223
=> theta = arcsin(0.223) = ~13*

I'll go dig a hole now. Nevertheless, I think it's possible to find AD and AE with the above direction (in terms of BC's length).. then proceed to find angle ADE and AED (2 equations with 2 unknowns) then theta.

Thing is, I got 30 from degrees from actually drawing the triangle to scale and measuring it. Using just geometry and algebra isn't enough to solve it, and I'm too lazy to use trig. I'll leave that part up to someone else.

Since I got a conflicting answer, I suppose it's time I show my work:

 Spoiler

- Prove that triangle BEC is isoceles, (angle BEC = 50*) -> BC = CE
- Law of Sines -> sin(theta)/CE = sin(DCA)/AD (with DCA = 20*) -> sin(theta) = sin20*BC/AD (1)
- Prove Triangle ADC is isoceles at D -> AD = DC -> need DC
- Law of Sines, part 2 -> sin80/DC = sin(BDC)/BC -> DC (= CE) = BC * sin80/sin40 (2)

(1) & (2) => sin(theta) = sin20*(BC/(BC * sin80/sin40)) = sin40*sin20/sin80 = 0.223
=> theta = arcsin(0.223) = ~13*

I'll go dig a hole now. Nevertheless, I think it's possible to find AD and AE with the above direction (in terms of BC's length).. then proceed to find angle ADE and AED (2 equations with 2 unknowns) then theta.

Thing is, I got 30 from degrees from actually drawing the triangle to scale and measuring it. Using just geometry and algebra isn't enough to solve it, and I'm too lazy to use trig. I'll leave that part up to someone else.

Trig method, not sure why it isn't getting that 30 degree answer? Did this independently of Kalovale.
DC/sin(80) = BC/sin(40)
EC = BC (since EBC is isosceles)
DCE = 20 degrees
DE^2 = DC^2 + EC^2 - 2DC*EC*cos(20)
DE/sin(20) = EC/sin(theta)
Now, working back:
sin(theta) = EC/DE*sin(20)
sub in for EC and DE
sin(theta) = BC*sin(20)/sqrt(DC^2 + BC^2 - 2DC*BC*cos(20))
keep going to get all BCs
sin(theta) = BC*sin(20)/sqrt((BC*sin(80)/sin(40))^2 + BC^2 - 2*(BC*sin(80)/sin(40))*BC*cos(20))
then we have BC/sqrt(BC^2) which factors out
sin(theta) = sin(20)/sqrt(sin^2(80)/sin^2(40) + 1 - 2*sin(80)*cos(20)/sin(40))
multiply a sin^2(40) into the bottom
sin(theta) = sin(20)*sin(40)/sqrt(sin^2(80) + sin^2(40) - 2*sin(80)*cos(20)*sin(40))

This gives theta = 30 degrees...

Setting lengths AB and AC to an arbitrary value of 1:

Bitches coming and not sleeping and solving before me.

I'm going out but I'll do it while I'm out so just sit tight buddy.

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The middle of the triangles is defined by the fact that all internal angles of a triangle make 180 degrees, using that you can find 70 for the top angle. The angles that make up a line are 180 degrees as well, allowing you to find the 110s, which lets you get the other 70 because of the total being 360.

Then you can find the two missing angles at the bottom because the triangle is isosceles, meaning that the two angles must equal 180-20=160. Each angle will then be 80 degrees, so solve for the missing piece.

This leaves you with only a few missing sections, so I labeled them with variables. The equations for finding these variables are:

Theta + a = 140 (since the angle of the side will be 180 total)
a + b = 160 (since the top triangle's internal angles make 180 total)
b + c = 130 (since the angle of the side will be 180 total)
c + Theta = 110 (since the center triangle will have 180 total)

From there, you need to locate the number that will make all 4 true, which is pretty easy since Theta isn't very big to begin with. If somebody has a faster way of finding the angle through those 4 equations, I'd be glad, because I used guess and check, though I suppose some insane substitution could work.[/COLOR]

My math teacher gave us a geometry problem as a sort of fun extracurricular thing and I found it interesting so maybe the mathy people here will enjoy it.

Triangle ABC is isoceles, angle BAC = 20 degrees
Angle EBC is 50 degrees.
Angle DCB is 60 degrees.
Find theta.

Here's a crappy diagram done in paint:

 Spoiler

 Spoiler

Angle DBE is 30
Angle DCE is 20
*makes point at intersection of DC and BE, names it F*
Angle BFC is 70
Angle EFC is 110
Angle ADC is 140
Angle AEB is 150
ADE+theta=ADC=140
20+ADE+AED=180
AED+ADE=160
ADE=160-AED
ADE=140-theta
160-AED=140-theta
20=-theta+AED
F'uck, I'm lost.
Don't feel like using trig

Omg it's isoleces? Are you serious.... This question is so easy then lol.

Omg it's isoleces? Are you serious.... This question is so easy then lol.

If it's so easy then solve it. I mean, it's just a missing angle problem, how hard could it be?

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 picture

Theta + a = 140 (since the angle of the side will be 180 total)
a + b = 160 (since the top triangle's internal angles make 180 total)
b + c = 130 (since the angle of the side will be 180 total)
c + Theta = 110 (since the center triangle will have 180 total)

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I would love for someone to prove me wrong, but you can't solve it with just those equations; I've tried it on my first attempt.
With those only those equations, the solution a = 85, b = 75, c = 55, theta = 55 is possible, but looking at the picture, clearly, c =/= theta.

EDIT: Oh, I didn't see that you used guess and check, but I'm just gonna point it out there that using substitution on those 4 equations will always get you 0 for whatever variable you decide to solve for first. The incorrect solution above was obtained from plugging the same equations into a matrix in my calculator. The correct answer of 30 is explained in my earlier post.

I would love for someone to prove me wrong, but you can't solve it with just those equations; I've tried it on my first attempt.
With those only those equations, the solution a = 85, b = 75, c = 55, theta = 55 is possible, but looking at the picture, clearly, c =/= theta.

EDIT: Oh, I didn't see that you used guess and check, but I'm just gonna point it out there that using substitution on those 4 equations will always get you 0 for whatever variable you decide to solve for first. The incorrect solution above was obtained from plugging the same equations into a matrix in my calculator. The correct answer of 30 is explained in my earlier post.

We could also include a rule for the angles to say that a specific one must be bigger than another or that some can't be the same, but yeah, I can't really think how you can solve it beyond just educated guessing.

Yeah this rethik guy got it so meh. Honestly I would have drawn a picture after solving a few of the angles with my protractor and ruler and it have been easier than any math lol.

Well obviously finding a missing angle is trivial if you draw the thing and measure it. The fun is in SOLVING THE PROBLEM. Not "so easy" after all then?

You're so hostile toward me lol. Fun is one thing, but I'm looking at efficiency. When I make Rubik's Cube threads asking for help, it's mostly about me getting the answer for credit in my class and/or for me to understand the/a way to do the problem that works. It has nothing to do with fun.

I looked at the problem with a pen and paper type approach and it is very complicated, you need to label sides, and use the law of sin/cosine to find other angles/sides. It's not simple to solve at all that way.

Which is why I took the liberty of drawing the triangle out (on a ghetto piece of loose leaf without being that accurate *twiddles thumbs...*) and I got the angle to measure 32 degrees, I'm sure I'm off by some degrees so I'd stick with Rethik's response of 30 degrees.