Tribot signup question.

[FerretBuster]

Before everyone starts flaming me - I am already signed up at the Tribot website, I signed up using the question about the original creator of scar.

However here in the UK I have not been taught how to solve a mathematical problem such as the one on the Tribot site. I am no idiot either, the college I attend is in the top 5 in the entire country - I am currently studying maths at AS level there.

If someone else is in the UK and has been taught how to solve this type of equation, could you give me the topic that it is under as I would love to actually figure out how to solve it myself.

However if someone could just walk me through how to solve it that would be great.

here is the equation for reference:

Cos((11(Pi))/6) = sqrt(x)/(y/24). Express your answer in the form of "x,y", without quotations.

Thanks.

 

[mistatic]

Firstly what is tribot? And I assume it is cos x 11 x pi which I think is 3.14? I am just taking a guess here though as I have never done it.

 

[weeps]

Nvm

Do the y your self now (a normal equcation)

 

[TRiLeZ]

Firstly what is tribot? And I assume it is cos x 11 x pi which I think is 3.14? I am just taking a guess here though as I have never done it.

To solve this question, you would first have to find the exact value of (cos(pi)/6). You could easily do that with a unit circle. After that, it should be pretty straight-forward.

 

[FerretBuster]

Just to confirm, should the answer be in terms of x and y? or should it be x=answer and y=answer. If it is the latter I would have absolutely no idea how to go about it, I have never had to solve an equation with two unknowns.

Also TRiLeZ, I had sent you a pm with this question on your website, however you have not responded as of yet. So i'll just ask it here. Did you self-teach yourself java, and if you did what resources did you use to learn it? (I am currently using the java docs)

Thanks.

 

[sonic_fanrs]

The unit circle values' coordinates (x, f(x)) are (cos, sin), so cos 11pi/6 = (sqrt(3)/(2))

and then

And I guess just simplify that 'til the form accepts it. I'm not sure what they mean by "x,y". It's a pretty shitty question, because if you type your answer with wrong spacing or in a different yet acceptable notation, it won't be accepted.

 

[FerretBuster]

It is a pretty annoying question, I would be happy if I could actually do it. But the fact is I have never came across unit circles and I am an A*/A grade student. I think the question should maybe be changed to something that we know is covered by a large array of countries syllabus'.

Looking over unit circles now though! I will not be beaten!

 

[FerretBuster]

I can't do it, I just don't understand how you are meant to use unit circles. I have never used one before in my life...

 

[sonic_fanrs]

I can't do it, I just don't understand how you are meant to use unit circles. I have never used one before in my life...

You don't really need to know how the unit circle works, just know that the Cosine of 11pi/6 = sqrt(3)/2.

You can check that on a calculator:

cos(11pi/6) = .8660254038
sqrt(3)/2 = .8660254038

And it pisses me off so much that I can't figure out what the form wants. I can solve for X or Y just fine, but nothing I've entered satisfies the form.

 

[FerretBuster]

Where did you get the 3 and the 2 from? (in the sqrt(3)/2. bit )

 

[poopy2177]

R=Radius
Cos=X/R
Sin=Y/R
Tan=Y/X

Basic college level Trig
sqrt(3)/2 is The square root of 3 over 2

The radias of the unit circle is always 1

now look at 11pi/6 rads on the unit circle

the question is asking for cos so think of X/R

the X value of 11pi/6 is (sqrt(3)/2)

helpful link: http://www.wolframalpha.com/

oh btw 0 with line through middle is theta which means Angle value pretty much

hope this helps

edit: looked at question really easy to solve but nothing I enter is right so I just gave up and with it does it want y=sqrt(3)sqrt(x) or like sqrt(3)/2=24sqrt(x)/y Idk neither work lol

 

[Jimmy]

I seriously doubt that somebody in a college-level math class can't do basic trigonometry.

"3,48"?

 

[sonic_fanrs]

Where did you get the 3 and the 2 from? (in the sqrt(3)/2. bit )

The first part of the equation is cos(11pi/6). The unit circle tells us that it is equal to sqrt(3)/2. If you check each value on a calculator, you'll get the same decimal. There isn't any algebra involved (in this step); both are equal.

I seriously doubt that somebody in a college-level math class can't do basic trigonometry.

"3,48"?

worked. And also, made it through Calc 2. Couldn't figure it out. I suck at Algebra though. Calc 3 this summer.

 

[FerretBuster]

Ok Jimmy that made a lot more sense now - I got to the second step (that is how I would logically do it), but I didn't realise that you could then just say the top of a fraction = the top, and the bottom = the bottom.

And yes I am studying at college, but we have never done those unit circle things (let alone anything with two unknowns.)

 

[TRiLeZ]

I seriously doubt that somebody in a college-level math class can't do basic trigonometry.

"3,48"?

Finally someone who could solve that question! I made that question when I was 16 and didn't think it was too hard...

Did you self-teach yourself java, and if you did what resources did you use to learn it? (I am currently using the java docs)

Thanks.

Yes, I self-taught myself java. I started with modifying private servers/clients, then eventually made my own. After I finished with private servers, I moved on to making my own BCEL updater for the client, then I made TRiBot after that. I learned by jumping right into reading/writing code. I've never even read a Java tutorial.

 

Lol belongs more in homework help don't you think?