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Thread: Find the value R

  1. #1
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    Find the value R

    Given that the two equations:
    R sin⁡〖60〗=10 sin⁡〖θ〗+v
    R cos⁡〖60〗=10 cos⁡〖θ〗

    How do I get the value R ? Please show me how to do it. Thanks.
    Last edited by alexander9408; Sep 12th 2012 at 09:28 AM.
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    Re: Find the value R

    Is $\displaystyle \theta \hspace{2mm}or \hspace{2mm}v$ known.
    Last edited by kalyanram; Sep 12th 2012 at 09:34 AM.
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    Re: Find the value R

    Quote Originally Posted by alexander9408 View Post
    Given that the two equations:
    R sin⁡〖60〗=10 sin⁡〖θ〗+v
    R cos⁡〖60〗=10 cos⁡〖θ〗

    How do I get the value R ? Please show me how to do it. Thanks.
    I assume $\displaystyle v$ is known.

    $\displaystyle \left( \frac{\sqrt{3}}{2} R - v}\right)^2 + \frac{R^2}{4} = 100$

    This can be solved for $\displaystyle R=\frac{\sqrt{3}v \pm \sqrt{400-v^2}}{2}$
    Last edited by kalyanram; Sep 12th 2012 at 09:42 AM.
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    Re: Find the value R

    both are unknown
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    Member kalyanram's Avatar
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    Re: Find the value R

    Well then there are three unknowns $\displaystyle R, \theta, v$ and two equations. This cannot be solved for a unique solution.
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    Re: Find the value R

    well then, can you form this equation θ=60 - sin^(-1)⁡〖v/20〗using the two equations above?
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    Re: Find the value R

    Quote Originally Posted by alexander9408 View Post
    well then, can you form this equation θ=60 - sin^(-1)⁡〖v/20〗using the two equations above?
    Yes this can be done by eliminating $\displaystyle R$ from the above two equations?
    Spoiler:

    Dividing the given equations we have
    $\displaystyle \sqrt{3} = \frac{10 \sin \theta + v}{10 \cos \theta} \implies 10\sqrt{3} \cos \theta - 10 \sin \theta = v$
    $\displaystyle \implies 20 \left( \frac{\sqrt{3}}{2} \cos \theta - \frac{1}{2} \sin \theta \right) = v$
    $\displaystyle \implies \sin \left( 60^{o} - \theta \right) = \frac{v}{20}$
    $\displaystyle \implies \theta = 60^{o} - \sin^{-1} \left( \frac{v}{20} \right)$

    NOTE: Before division by $\displaystyle 10 \cos \theta$ you need to argue that you are not dividing by zero first.
    Last edited by kalyanram; Sep 12th 2012 at 11:36 AM.
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    Re: Find the value R

    Quote Originally Posted by kalyanram View Post

    NOTE: Before division by $\displaystyle 10 \cos \theta$ you need to argue that you are not dividing by zero first.
    [/SPOILER]
    What do you mean by this?
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    Member kalyanram's Avatar
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    Re: Find the value R

    Quote Originally Posted by alexander9408 View Post
    What do you mean by this?
    Have you considered the case when $\displaystyle \theta=\frac{\pi}{2}$.

    By the way the general solution is $\displaystyle \theta = n\pi + (-1)^n \left( \frac{\pi}{3} - \sin^{-1} \left( \frac{v}{20}\right) \right), n \in \mathbb{Z}$. (why?)
    Last edited by kalyanram; Sep 12th 2012 at 12:04 PM.
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    Re: Find the value R

    Quote Originally Posted by kalyanram View Post
    Have you considered the case when $\displaystyle \theta=\frac{\pi}{2}$.

    By the way the general solution is $\displaystyle \theta = n\pi + (-1)^n \left( \frac{\pi}{3} - \sin^{-1} \left( \frac{v}{20}\right) \right), n \in \mathbb{Z}$. (why?)
    Since the question have been given that the v<10 , when I substitute v=10 to the equation, then I will get θ = 30 , so the θ won't exceed 30.
    The value of sin is positive at first and second quadrant, and negative at third and fourth quadrant, is that so?
    Last edited by alexander9408; Sep 12th 2012 at 12:25 PM.
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  11. #11
    Member kalyanram's Avatar
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    Re: Find the value R

    Quote Originally Posted by alexander9408 View Post
    Since the question have been given that the v<10
    When $\displaystyle \theta = \frac{\pi}{2}$ we have $\displaystyle R=0$ and $\displaystyle v=-10$ which satisfies the condition $\displaystyle v<10$.

    Anyway the point I wanted to highlight was not that. I am probably confusing you more that I intend to so here is how I would like you to analyze.

    Step1: There are three unknown $\displaystyle R,\theta, v$, and two equations so if you want to relate any two you have to eliminate the other.
    Relating $\displaystyle R \& v$ by eliminating $\displaystyle \theta$ was shown above. Now here the discriminant of the quadratic $\displaystyle (400 - v^2)$ should be $\displaystyle > 0$ for the range of values specified on $\displaystyle v$ this gives us an additional criteria that $\displaystyle |v| \le 20$. So permissible values of $\displaystyle v$ are $\displaystyle -20 \le v < 10$.

    Step 2: Relating $\displaystyle \theta$ and $\displaystyle v$
    While eliminating any variable by trying to divide two expressions (like I did) you have to be careful as to the expression in the denominator is $\displaystyle \ne 0$. This is the case when $\displaystyle \theta = \frac{\pi}{2}$. So you can deal with it as a separate case and argue it out, or try to avoid division if possible. And in this case you can avoid division of the two expressions as shown below:
    From the first equation w
    We have $\displaystyle R = \frac{2}{\sqrt{3}} \left( 10 \sin \theta + v \right)$, $\displaystyle R = 20 \cos \theta$

    $\displaystyle 20 \cos \theta = \frac{2}{\sqrt{3}} \left( 10 \sin \theta + v \right) $ from here we again get

    $\displaystyle 10 \sqrt{3} \cos \theta - 10 \sin \theta = v$ which has been solved above. The point to be noted here is that eliminating $\displaystyle R$ as shown here we have avoided any possibility of division by $\displaystyle 0$.

    Step 3: General values of $\displaystyle \theta$.

    Quote Originally Posted by alexander9408 View Post
    when I substitute v=10 to the equation, then I will get θ = 30 , so the θ won't exceed 30. The value of sin is positive at first and second quadrant, and negative at third and fourth quadrant, is that so?
    As I have mentioned above when you obtain a solution for a variable it is all the more important to verify if the solution is valid. When solving for Step 2, we obtain $\displaystyle \theta = \frac{\pi}{3} - sin^{-1} \left( \frac{v}{20} \right)$ as the principle values of the angle. As stated by you we already know that $\displaystyle \sin (\pi - \theta) = \sin \theta$ and also $\displaystyle \sin (2n\pi + \theta) = \sin \theta, \forall n \in \mathbb{Z}$ so we need all possible solutions of $\displaystyle \theta$ we obtain the form $\displaystyle \theta = n\pi + (-1)^n \left( \frac{\pi}{3} - sin^{-1} \left( \frac{v}{20}\right)\right)$. Now we need to see if this solution is a valid solution for that we need $\displaystyle |v|< 20$(why?) which is also consistent with what we got in the solution for $\displaystyle R$.
    Last edited by kalyanram; Sep 12th 2012 at 10:17 PM.
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