Results 1 to 4 of 4

Math Help - rectangular section PQRS

  1. #1
    Junior Member
    Joined
    May 2007
    Posts
    36

    rectangular section PQRS

    hi to all i would like some help on the that math project
    Attached Thumbnails Attached Thumbnails rectangular section PQRS-scan0003_001.jpg  
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Member
    Joined
    Feb 2008
    From
    Westwood, Los Angeles, CA
    Posts
    176
    Thanks
    1
    Note that angle QAB has measure 90 - x.

    Also, cos(QAB) = QB/AB = QB/18, so QB = 18(cos(90-x)).

    Similarly, cos(CBR) = BR/27, so BR = 27(cos(x)).

    Note that QB + BR = 30 (the bottom part of your diagram).

    So, 18(cos(90-x)) + 27(cos(x)) = 30.

    Use the fact that the graph of cosine is the same as the graph of sine (just shifted) to convert cos(90-x) to sin(x).

    Finally, divide everything by 3.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Moo
    Moo is offline
    A Cute Angle Moo's Avatar
    Joined
    Mar 2008
    From
    P(I'm here)=1/3, P(I'm there)=t+1/3
    Posts
    5,618
    Thanks
    6
    Hello,

    Note that angle QAB has measure 90 - x.
    I'd rather say that it's angle QBA whose measure is 90-x.

    Angle QAB has measure x
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,548
    Thanks
    541
    Hello, carlasader!

    a) Prove that: . 9\cos x + 6\sin x \:=\:10
    We have: . AB = DC = 18
    Note that: . \angle BAQ = x

    In right triangle AQB\!:\;\;\sin x \,=\,\frac{QB}{18}\quad\Rightarrow\quad QB \,=\,18\sin x


    We have . BC = AD = 27
    In right triangle CRB\!:\;\;\cos x \:=\:\frac{BR}{27}\quad\Rightarrow\quad BR \,=\,27\cos x


    Since QR \:=\:BR + QB \:=\:30

    . . we have: . 27\cos x + 18\sin x \:=\:30 \quad\Rightarrow\quad\boxed{ 9\cos x + 6\sin x \:=\:10}



    b) Express 9\cos x + 6\sin x in the form: R\cos(x - a)
    . . where R > 0\text{ and }0^o < a < 90^o,
    . . giving the values of R\text{ and }a to 2 decimal places.
    This takes a bit of work . . .
    If you've never seen this done, no wonder you are baffled.


    We have: . 9\cos x + 6\sin x

    . . Find: . \sqrt{9^2+6^2} \;=\;\sqrt{117} \:=\:3\sqrt{13}

    Multiply and divide by 3\sqrt{13}\!:\;\;3\sqrt{13}\,\left[\frac{9}{3\sqrt{13}}\cos x + \frac{6}{3\sqrt{13}}\sin x\right]

    . . and we have: . 3\sqrt{13}\,\left(\frac{3}{\sqrt{13}}\cos x + \frac{2}{\sqrt{13}}\sin x\right)


    Let \cos a \,=\,\frac{3}{\sqrt{13}}\,\text{ and }\,\sin a \,=\,\frac{2}{\sqrt{13}}

    . . and we have: . 3\sqrt{13}\,\left(\cos a\cos x + \sin a\sin x\right) \;=\;\boxed{3\sqrt{13}\,\cos(x-a)}

    Hence: . R\:=\:3\sqrt{13} \quad\Rightarrow\quad\boxed{R \:\approx\:10.82}

    And: . a \,=\,\cos^{-1}\left(\frac{3}{\sqrt{13}}\right) \quad\Rightarrow\quad\boxed{a \:\approx\:33.69^o}




    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Please help me in section 6.5 : Work
    Posted in the Calculus Forum
    Replies: 2
    Last Post: May 31st 2010, 09:09 PM
  2. Section
    Posted in the Geometry Forum
    Replies: 2
    Last Post: March 22nd 2010, 10:10 PM
  3. SAT Math Section
    Posted in the Math Topics Forum
    Replies: 2
    Last Post: May 26th 2009, 04:07 PM

Search Tags


/mathhelpforum @mathhelpforum