Equation for Solving v: Steps, Tips & Results | Did I Solve it Right?

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The equation presented involves complex variables including mass (m), gravitational force (g), and coefficients related to drag (C_a) and friction (μ). The user attempted to solve for v and derived a formula, but expressed uncertainty about its accuracy due to a lack of understanding in trigonometry. A response clarified that solving for v does not necessitate knowledge of trigonometry; instead, it involves rearranging the equation to isolate v². The recommended approach is to group all terms involving v² on one side and then solve for v by dividing and taking the square root. This discussion highlights the importance of algebraic manipulation over trigonometric knowledge in solving the equation.
ektrules
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Here's the equation.
\frac{mv^{2}}{r}=\frac{mg+C_{a}v^{2}cos\vartheta}{cos\vartheta-\mu sin\vartheta}\left(\mu cos\vartheta+sin\theta\right)-C_{a}v^{2}sin\vartheta
After trying to solve for v, I got this:
v=\sqrt{\frac{mg\mu cos\vartheta+mg sin\vartheta}{-2\mu C_{a}cos\vartheta-2\mu C_{a}sin\vartheta+\frac{mcos\vartheta-\mu msin\vartheta}{r}}}
I'm not sure if this is correct though. I never took trig, so all the sin's and cos's confuse me, and I'm not exactly sure how to perform algebra on them.
 
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I haven't checked your result. However, solving for v doesn't require knowing trig. Just put all the v2 on one side of the equation, and put all the other terms on the other side. Then divide by the coefficent of v2 and take the square root.
 
Here is a little puzzle from the book 100 Geometric Games by Pierre Berloquin. The side of a small square is one meter long and the side of a larger square one and a half meters long. One vertex of the large square is at the center of the small square. The side of the large square cuts two sides of the small square into one- third parts and two-thirds parts. What is the area where the squares overlap?

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