Mimi_x3
Geometry Question.
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Mimi_x3
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Mimi_x3
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i don't know how to start..can someone please help..
lalaly
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i am still trying to do it but something is confusing me whats the x?
Mimi_x3
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I don't know; I'm confused as well, where did the x come from? lol
lalaly
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lol i did half of the work but got to an equatoin with x in it ...
Mimi_x3
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Well, I don't know how to do it; where the x come from?
Can you give me a hint how to start?
lalaly
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since*
Mimi_x3
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Hm,okay; I will try from that..
Mimi_x3
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thanks anyway
lalaly
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sorry:( .. im sure the other people viewing would have an idea
lalaly
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i can do part b xD do u need help with that?
Mimi_x3
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Nope, only the first one.
lalaly
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ok:)
inkyvoyd
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That's a stupid question.
inkyvoyd
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* problem
razor99
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may i come here
.Sam.
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where the 'x' come from in equation 'a'?
Mimi_x3
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i don't know that is why I'm lost; or probably there's a mistake :/
inkyvoyd
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|dw:1337940741536:dw|
Mimi_x3
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lol, please delete that inky..
inkyvoyd
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|dw:1337940823123:dw|
inkyvoyd
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That's x.
inkyvoyd
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I understand the problem now.
inkyvoyd
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|dw:1337940877496:dw|
inkyvoyd
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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.
inkyvoyd
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@Mimi_x3
inkyvoyd
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|dw:1337941268446:dw|
inkyvoyd
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|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
inkyvoyd
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|dw:1337941693329:dw|
inkyvoyd
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@lalaly , do you at least understand my explanation?
.Sam.
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Max I can get is
\[asin(\theta)-(x+a)sin(\phi)=0\]
assuming x was there
|dw:1337941920722:dw|
inkyvoyd
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@.Sam. , that's what I meant... I'm sure that that is x
Mimi_x3
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First, where did sine come from? I havent done geometry in a while..
.Sam.
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I've ended up with another solution
\[rcos(90+\phi-\theta)-(x+r)sin(\phi)=0\]
Mimi_x3
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I'm lost..
lalaly
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angle TBO= 180-(180-theta+phi)=theta-phi
\[\frac{(x+a)}{\sin(\theta-\phi)}=\frac{a}{\sin \phi}\]\[(x+a)\sin \phi=asin(\theta-\phi)\]
lalaly
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so\[asin(\theta-\phi)-(x+a)\sin \phi=0\]
lalaly
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i assumed that the part of TC that is outside the circle is x ... and it worked!! :D
lalaly
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and then i used the sin rule after finding TBO
O is the center
lalaly
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and TO=x+a
lalaly
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did u get it @Mimi_x3
Mimi_x3
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yeah; thank you!
lalaly
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:D
inkyvoyd
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@Mimi_x3 , rage
.Sam.
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I was using angle JOB lol