Author Topic: Help for Circular Functions question!!! (Read 1425 times) Tweet Share

Electroflame1395 · « **on:** April 18, 2013, 09:26:18 pm »

Hey guys,
I really need help for this question.
It is enclose in the attachment and is question 1. I am pretty sure the question is from the Essentials textbook.
If anyone can help me out it would be great.
Thanks

Jeggz · « **Reply #1 on:** April 19, 2013, 10:58:09 am »

I'll give part (a) a shot - but I'm really not sure if this is correct and I'm sorry if this doesn't make any sense!

We are trying to find $\theta$ . From the diagram we can see that the find the angle $\theta$ we need to subtract that larger angle (let's call this $\alpha$ ) from the smaller angle inside (let's call this $\beta$ )

$tan(\alpha) = \frac{a+b}{x}$ (using basic pythag)
$tan(\beta)=\frac{b}{x}$ (using basic pythag)

And as mentioned above $tan(\theta) =$ $tan(\alpha)-$ $tan(\beta)$
We know what $tan(\alpha)$ and $tan(\beta)$ hence we can replace it into the equation.

$tan(\theta) =$ $\frac{a+b}{x}- \frac{b}{x}$

$(\theta) = tan^-^1(\frac{a+b}{x}$ ) $- tan^-^1(\frac{b}{x}$ )

Alwin · « **Reply #2 on:** April 19, 2013, 03:55:02 pm »

Hiii,

I hope this will help, attached is an image of how I'd tackle that problem. Apologies in advance if I have made some silly errors or mistakes

If you have any more difficulties, feel free to let me know

btw, i can't make the file much smaller, so its split in two.

Alwin · « **Reply #3 on:** April 19, 2013, 03:57:01 pm »

Part two, since the attachments were being buggy:

Electroflame1395 · « **Reply #4 on:** April 21, 2013, 06:15:26 pm »

Thanks guys you have been a great help. Sorry about the late reply as i have been so busy.

silverpixeli · « **Reply #5 on:** April 23, 2013, 10:16:10 pm »

I've already been beaten and the previous explanations were fantastic, but I'd like to add to Alwin's explanation of part c

we already worked out;
$tan(\theta)=\frac{ax}{x^2 + ab + b^2}$
so yeah, if we let $\theta=\frac{\pi}{4}$ then that becomes $1=\frac{ax}{x^2 + ab + b^2}$
multiply both sides by the denominator,

$x^2 + ab + b^2=ax$

Alwin, you rearranged this to get $x^2-ax=-ba-b^2$ which is correct, but it's not x in terms of a and b which is what the question is asking

if you rearrange it as a quadratic = 0, you can find the roots

$x^2 -ax + ab + b^2=0$

$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$ (quadratic formula)

$x=\frac{-(-a)\pm\sqrt{(-a)^2-4(1)(ab+b^2)}}{2(1)}$

$x=\frac{a\pm\sqrt{a^2-4ab-4b^2}}{2}$ which I think is as simple as it can get. I'd be interested to know what the solutions said for this one.

Alwin · « **Reply #6 on:** April 23, 2013, 10:27:18 pm »

Quote from: silverpixeli on April 23, 2013, 10:16:10 pm

$x=\frac{a\pm\sqrt{a^2-4ab-4b^2}}{2}$ which I think is as simple as it can get. I'd be interested to know what the solutions said for this one.

haha don't worry! I know you're right, I can recall doing that problem in spesh class and had to keep on hiding my working from the teacher LOL. (yes, I do get a tad bored in class haha)

Thnaks for the correction silverpixeli!

Search the forums now!

We have moved!

Author Topic: Help for Circular Functions question!!! (Read 1425 times) Tweet Share

Electroflame1395

Help for Circular Functions question!!!

Jeggz

Re: Help for Circular Functions question!!!

Alwin

Re: Help for Circular Functions question!!!

Alwin

Re: Help for Circular Functions question!!!

Electroflame1395

Re: Help for Circular Functions question!!!

silverpixeli

Re: Help for Circular Functions question!!!

Alwin

Re: Help for Circular Functions question!!!

Recent Posts

User:
Password: