Geometry Question.

- Mimi_x3

Geometry Question.

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- Mimi_x3

##### 1 Attachment

- Mimi_x3

i don't know how to start..can someone please help..

- lalaly

i am still trying to do it but something is confusing me whats the x?

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## More answers

- Mimi_x3

I don't know; I'm confused as well, where did the x come from? lol

- lalaly

lol i did half of the work but got to an equatoin with x in it ...

- Mimi_x3

Well, I don't know how to do it; where the x come from?
Can you give me a hint how to start?

- lalaly

since*

- Mimi_x3

Hm,okay; I will try from that..

- Mimi_x3

thanks anyway

- lalaly

sorry:( .. im sure the other people viewing would have an idea

- lalaly

i can do part b xD do u need help with that?

- Mimi_x3

Nope, only the first one.

- lalaly

ok:)

- inkyvoyd

That's a stupid question.

- inkyvoyd

* problem

- razor99

may i come here

- .Sam.

where the 'x' come from in equation 'a'?

- Mimi_x3

i don't know that is why I'm lost; or probably there's a mistake :/

- inkyvoyd

|dw:1337940741536:dw|

- Mimi_x3

lol, please delete that inky..

- inkyvoyd

|dw:1337940823123:dw|

- inkyvoyd

That's x.

- inkyvoyd

I understand the problem now.

- inkyvoyd

|dw:1337940877496:dw|

- inkyvoyd

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

@Mimi_x3

- inkyvoyd

|dw:1337941268446:dw|

- inkyvoyd

|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

|dw:1337941693329:dw|

- inkyvoyd

@lalaly , do you at least understand my explanation?

- .Sam.

Max I can get is
\[asin(\theta)-(x+a)sin(\phi)=0\]
assuming x was there
|dw:1337941920722:dw|

- inkyvoyd

@.Sam. , that's what I meant... I'm sure that that is x

- Mimi_x3

First, where did sine come from? I havent done geometry in a while..

- .Sam.

I've ended up with another solution
\[rcos(90+\phi-\theta)-(x+r)sin(\phi)=0\]

- inkyvoyd

http://www.twiddla.com/847714

- Mimi_x3

I'm lost..

- lalaly

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

so\[asin(\theta-\phi)-(x+a)\sin \phi=0\]

- lalaly

i assumed that the part of TC that is outside the circle is x ... and it worked!! :D

- lalaly

and then i used the sin rule after finding TBO
O is the center

- lalaly

and TO=x+a

- lalaly

did u get it @Mimi_x3

- Mimi_x3

yeah; thank you!

- lalaly

:D

- inkyvoyd

@Mimi_x3 , rage

- .Sam.

I was using angle JOB lol

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