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Geometry Question.

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i don't know how to start..can someone please help..
i am still trying to do it but something is confusing me whats the x?

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Other answers:

I don't know; I'm confused as well, where did the x come from? lol
lol i did half of the work but got to an equatoin with x in it ...
Well, I don't know how to do it; where the x come from? Can you give me a hint how to start?
Hm,okay; I will try from that..
thanks anyway
sorry:( .. im sure the other people viewing would have an idea
i can do part b xD do u need help with that?
Nope, only the first one.
That's a stupid question.
* problem
may i come here
where the 'x' come from in equation 'a'?
i don't know that is why I'm lost; or probably there's a mistake :/
lol, please delete that inky..
That's x.
I understand the problem now.
As to how I figured it out, I began by constructing that perpendicular line segment, and then I assumed the statement was true. I found that sin(theta-phi)=sin(angle b), and a sin(b)= that line segment I constructed. Then, sin(phi)=opp/hyp=contructed line segment/(a+something), or (a+something)sin(phi) =a sin(theta-phi) and the only value for something that makes sense is that segment.
|dw:1337941445676:dw| a sin(theta-phi)=constructed line segment -> let's call that segment length y Then, (a+x) sin(phi)=opp/hyp ->opp is segment y, and hyp would be x+a - the only definition of x that works is in my drawing. Thus, (a+x)sin(phi)=(a+x)*(opp/hyp)=(a+x)(y)/(x+a)=y and a sin(theta-phi) is also y, and so we can substitute (y)-(y)=0
@lalaly , do you at least understand my explanation?
Max I can get is \[asin(\theta)-(x+a)sin(\phi)=0\] assuming x was there |dw:1337941920722:dw|
@.Sam. , that's what I meant... I'm sure that that is x
First, where did sine come from? I havent done geometry in a while..
I've ended up with another solution \[rcos(90+\phi-\theta)-(x+r)sin(\phi)=0\]
I'm lost..
angle TBO= 180-(180-theta+phi)=theta-phi \[\frac{(x+a)}{\sin(\theta-\phi)}=\frac{a}{\sin \phi}\]\[(x+a)\sin \phi=asin(\theta-\phi)\]
so\[asin(\theta-\phi)-(x+a)\sin \phi=0\]
i assumed that the part of TC that is outside the circle is x ... and it worked!! :D
and then i used the sin rule after finding TBO O is the center
and TO=x+a
did u get it @Mimi_x3
yeah; thank you!
@Mimi_x3 , rage
I was using angle JOB lol

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