# Thread: Find the value R

1. ## 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.

2. ## Re: Find the value R

Is $\theta \hspace{2mm}or \hspace{2mm}v$ known.

3. ## Re: Find the value R

Originally Posted by alexander9408
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 $v$ is known.

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

This can be solved for $R=\frac{\sqrt{3}v \pm \sqrt{400-v^2}}{2}$

4. ## Re: Find the value R

both are unknown

5. ## Re: Find the value R

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

6. ## Re: Find the value R

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

7. ## Re: Find the value R

Originally Posted by alexander9408
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 $R$ from the above two equations?
Spoiler:

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

NOTE: Before division by $10 \cos \theta$ you need to argue that you are not dividing by zero first.

8. ## Re: Find the value R

Originally Posted by kalyanram

NOTE: Before division by $10 \cos \theta$ you need to argue that you are not dividing by zero first.
[/SPOILER]
What do you mean by this?

9. ## Re: Find the value R

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

By the way the general solution is $\theta = n\pi + (-1)^n \left( \frac{\pi}{3} - \sin^{-1} \left( \frac{v}{20}\right) \right), n \in \mathbb{Z}$. (why?)

10. ## Re: Find the value R

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

By the way the general solution is $\theta = n\pi + (-1)^n \left( \frac{\pi}{3} - \sin^{-1} \left( \frac{v}{20}\right) \right), n \in \mathbb{Z}$. (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?

11. ## Re: Find the value R

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

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

Step 2: Relating $\theta$ and $v$
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 $\ne 0$. This is the case when $\theta = \frac{\pi}{2}$. 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 $R = \frac{2}{\sqrt{3}} \left( 10 \sin \theta + v \right)$, $R = 20 \cos \theta$

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

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

Step 3: General values of $\theta$.

Originally Posted by alexander9408
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 $\theta = \frac{\pi}{3} - sin^{-1} \left( \frac{v}{20} \right)$ as the principle values of the angle. As stated by you we already know that $\sin (\pi - \theta) = \sin \theta$ and also $\sin (2n\pi + \theta) = \sin \theta, \forall n \in \mathbb{Z}$ so we need all possible solutions of $\theta$ we obtain the form $\theta = n\pi + (-1)^n \left( \frac{\pi}{3} - sin^{-1} \left( \frac{v}{20}\right)\right)$. Now we need to see if this solution is a valid solution for that we need $|v|< 20$(why?) which is also consistent with what we got in the solution for $R$.