Results 1 to 3 of 3

Math Help - Express the length x in terms of the trigonometric ratios of θ.

  1. #1
    Newbie
    Joined
    Oct 2011
    Posts
    17

    Express the length x in terms of the trigonometric ratios of θ.



    Don't know where to start.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    11,409
    Thanks
    1294

    Re: Express the length x in terms of the trigonometric ratios of θ.

    Quote Originally Posted by Remriel View Post


    Don't know where to start.
    To start with, the hypotenuse for the triangle with \displaystyle \begin{align*} x \end{align*} in it is the same as the side opposite angle \displaystyle \begin{align*} \theta \end{align*} in the upper triangle. Call this side \displaystyle \begin{align*} y \end{align*} and we have

    \displaystyle \begin{align*} \tan{\theta} &= \frac{y}{10} \\ y &= 10\tan{\theta} \end{align*}


    As for the third side of the triangle with \displaystyle \begin{align*} x \end{align*} in it, call this side \displaystyle \begin{align*} z \end{align*}. If you complete the rectangle, you'll have another right angle triangle with that side opposite \displaystyle \begin{align*} \theta \end{align*} and with a hypotenuse of \displaystyle \begin{align*} 10 \end{align*}. That means

    \displaystyle \begin{align*} \sin{\theta} &= \frac{z}{10} \\ z &= 10\sin{\theta} \end{align*}


    So finally, in the triangle with \displaystyle \begin{align*} x \end{align*} in it, we have by Pythagoras

    \displaystyle \begin{align*} x^2 + \left(10\sin{\theta}\right)^2 &= \left(10\tan{\theta}\right)^2 \\ x^2 + 100\sin^2{\theta} &= 100\tan^2{\theta} \\ x^2 &= 100\tan^2{\theta} - 100\sin^2{\theta} \\ x^2 &= 100\left(\tan^2{\theta} - \sin^2{\theta}\right) \\ x^2 &= 100\left(\frac{\sin^2{\theta}}{\cos^2{\theta}} - \sin^2{\theta}\right) \\ x^2 &= 100\left( \frac{\sin^2{\theta}}{\cos^2{\theta}} - \frac{\sin^2{\theta}\cos^2{\theta}}{\cos^2{\theta}  } \right) \\ x^2 &= 100\left( \frac{\sin^2{\theta} - \sin^2{\theta}\cos^2{\theta}}{\cos^2{\theta}} \right) \\ x^2 &= 100\left[ \frac{\sin^2{\theta}\left( 1 - \cos^2{\theta} \right)}{\cos^2{\theta}} \right] \\ x^2 &= 100\left( \frac{\sin^2{\theta}\sin^2{\theta}}{\cos^2{\theta}  } \right) \\ x^2 &= 100\left( \frac{\sin^4{\theta}}{\cos^2{\theta}} \right) \\ x &= 10\left( \frac{\sin^2{\theta}}{\cos{\theta}} \right) \\ x &= 10\sin{\theta}\tan{\theta} \end{align*}


    I think that's the simplest you can get it
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member
    earboth's Avatar
    Joined
    Jan 2006
    From
    Germany
    Posts
    5,829
    Thanks
    123

    Re: Express the length x in terms of the trigonometric ratios of θ.

    Quote Originally Posted by Remriel View Post

    Don't know where to start.
    1. I've completed the quadrilateral to a rectangle (see attachment).

    2. The side of the rectangle in orange is:

    \displaystyle{s = 10 \cdot \cos(90^\circ - \theta) = 10 \cdot \sin(\theta)}

    3. Then

    \displaystyle{x = s \cdot \tan(\theta) = 10 \cdot \sin(\theta) \cdot \tan(\theta)}

    which is the solution Prove It has posted.
    Attached Thumbnails Attached Thumbnails Express the length x in terms of the trigonometric ratios of θ.-unregelm_4eck.png  
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Express cos^2 (2x) in terms of cos 4x
    Posted in the Trigonometry Forum
    Replies: 3
    Last Post: May 13th 2012, 06:00 AM
  2. [SOLVED] Given that y = (x-a)/(x+b), express in terms of a, b and y
    Posted in the Algebra Forum
    Replies: 4
    Last Post: September 19th 2011, 04:43 AM
  3. Express log in Terms of x and y
    Posted in the Pre-Calculus Forum
    Replies: 3
    Last Post: April 20th 2009, 05:58 AM
  4. express y in terms of x
    Posted in the Algebra Forum
    Replies: 3
    Last Post: January 29th 2009, 09:02 PM
  5. express sin x/2 in terms of cos x
    Posted in the Trigonometry Forum
    Replies: 3
    Last Post: November 15th 2007, 07:23 PM

Search Tags


/mathhelpforum @mathhelpforum