## Mimi_x3 3 years ago Geometry Question.

1. Mimi_x3

2. Mimi_x3

3. lalaly

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

4. Mimi_x3

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

5. lalaly

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

6. 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?

7. lalaly

since*

8. Mimi_x3

Hm,okay; I will try from that..

9. Mimi_x3

thanks anyway

10. lalaly

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

11. lalaly

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

12. Mimi_x3

Nope, only the first one.

13. lalaly

ok:)

14. inkyvoyd

That's a stupid question.

15. inkyvoyd

* problem

16. razor99

may i come here

17. .Sam.

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

18. Mimi_x3

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

19. inkyvoyd

|dw:1337940741536:dw|

20. Mimi_x3

21. inkyvoyd

|dw:1337940823123:dw|

22. inkyvoyd

That's x.

23. inkyvoyd

I understand the problem now.

24. inkyvoyd

|dw:1337940877496:dw|

25. 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.

26. inkyvoyd

@Mimi_x3

27. inkyvoyd

|dw:1337941268446:dw|

28. 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

29. inkyvoyd

|dw:1337941693329:dw|

30. inkyvoyd

@lalaly , do you at least understand my explanation?

31. .Sam.

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

32. inkyvoyd

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

33. Mimi_x3

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

34. .Sam.

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

35. inkyvoyd
36. Mimi_x3

I'm lost..

37. 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)$

38. lalaly

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

39. lalaly

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

40. lalaly

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

41. lalaly

and TO=x+a

42. lalaly

did u get it @Mimi_x3

43. Mimi_x3

yeah; thank you!

44. lalaly

:D

45. inkyvoyd

@Mimi_x3 , rage

46. .Sam.

I was using angle JOB lol