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.

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- Sep 12th 2012, 09:25 AMalexander9408Find 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. - Sep 12th 2012, 09:32 AMkalyanramRe: Find the value R
Is $\displaystyle \theta \hspace{2mm}or \hspace{2mm}v$ known.

- Sep 12th 2012, 09:39 AMkalyanramRe: Find the value R
- Sep 12th 2012, 09:41 AMalexander9408Re: Find the value R
both are unknown

- Sep 12th 2012, 11:09 AMkalyanramRe: 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.

- Sep 12th 2012, 11:17 AMalexander9408Re: Find the value R
well then, can you form this equation θ°=60° - sin^(-1)〖v/20〗using the two equations above?

- Sep 12th 2012, 11:30 AMkalyanramRe: Find the value R
- Sep 12th 2012, 11:49 AMalexander9408Re: Find the value R
- Sep 12th 2012, 11:56 AMkalyanramRe: Find the value R
- Sep 12th 2012, 12:11 PMalexander9408Re: Find the value R
- Sep 12th 2012, 10:10 PMkalyanramRe: Find the value R
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 $\displaystyle \theta$.

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 $\displaystyle R$.