Results 1 to 4 of 4

Thread: Circle

  1. #1
    Junior Member Singular's Avatar
    Joined
    Dec 2006
    From
    Solo, Java
    Posts
    57

    Circle

    O is the center of the circle; AB is the diameter
    P is on OB.
    the angle OQP = angle ORP = 10 degree
    The angle AOQ = 40 degree

    Find angle ROB

    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member
    earboth's Avatar
    Joined
    Jan 2006
    From
    Germany
    Posts
    5,854
    Thanks
    137
    Quote Originally Posted by Singular View Post
    O is the center of the circle; AB is the diameter
    P is on OB.
    the angle OQP = angle ORP = 10 degree
    The angle AOQ = 40 degree

    Find angle ROB
    Hello,

    maybe this comes too late, but nevertheless..

    I only can offer you a more or less precise drawing with a result.

    The thick black lines describe the situation of your problem. The thin blue lines are necessary to place Point R on the circle line. The red value is the result you are looking for.

    Obviously the center of the blue circle is placed on the black circle too. Maybe this property is of some use for you.

    EB

    PS.: I corrected my drawing because I measured the wrong angle.
    Attached Thumbnails Attached Thumbnails Circle-zweiwinkl_winkl2.gif  
    Last edited by earboth; Mar 27th 2007 at 06:26 AM.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member
    earboth's Avatar
    Joined
    Jan 2006
    From
    Germany
    Posts
    5,854
    Thanks
    137
    Quote Originally Posted by Singular View Post
    O is the center of the circle; AB is the diameter
    P is on OB.
    the angle OQP = angle ORP = 10 degree
    The angle AOQ = 40 degree

    Find angle ROB
    Hello,

    it's me again.
    I've modified your sketch a little bit (see attachment):

    OP is the base of the 2 triangles, therefore I called this line b.
    The angle <(QPA) = α (alpha) and the angle <(RPA) = β (beta).
    OQ = OR = r

    You know that α + 10 = 40 thus α = 30

    Now use Sine rule to calculate b:

    b/r = sin(10)/sin(30). Because sin(30) = 0.5 you get:

    b = 2rsin(10)

    Now use Sine rule to calculate β:

    sin(β)/sin(10) = r / b Solve for sin(β) and plug in the term for b:

    sin(β) = (r sin(10))/(2r sin(10)) = 1/2

    Therefore β = 30 or β = 150. β must be greater than 90 therefore β = 150. The angle <(ROB) = 180 - 150 - 10 = 20

    EB
    Attached Thumbnails Attached Thumbnails Circle-zweiwinkel_skizze.gif  
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Junior Member Singular's Avatar
    Joined
    Dec 2006
    From
    Solo, Java
    Posts
    57
    Thanks earboth
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Circle, tangent line, and a point not on the circle
    Posted in the Pre-Calculus Forum
    Replies: 5
    Last Post: Mar 31st 2011, 02:40 PM
  2. Replies: 6
    Last Post: Jul 8th 2010, 06:39 PM
  3. Replies: 7
    Last Post: Mar 15th 2010, 05:10 PM
  4. Replies: 2
    Last Post: Feb 6th 2010, 09:31 AM
  5. Replies: 0
    Last Post: Oct 12th 2008, 05:31 PM

Search Tags


/mathhelpforum @mathhelpforum