Solving Trig Equations: Help Needed!

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Discussion Overview

The discussion revolves around solving a system of trigonometric equations related to the equilibrium of a particle. Participants explore methods to isolate variables and derive solutions for the force \( W \) and angle \( \theta \) while addressing potential multiple solutions.

Discussion Character

  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant presents a system of equations and expresses difficulty in solving them, leading to a derived equation involving tangent.
  • Another participant suggests isolating \( \cos(\theta) \) and \( \sin(\theta) \), squaring them, and adding to eliminate \( \theta \) to solve for \( W \).
  • A participant reports deriving a quadratic equation from the suggested method, yielding two values for \( W \) and corresponding angles \( \theta \).
  • There is a discussion about the validity of both sets of solutions found and whether both should be considered, given that they satisfy the original equations.
  • Participants question why one solution might be preferred over the other, considering the context of the equilibrium problem.

Areas of Agreement / Disagreement

Participants generally agree that both sets of solutions are mathematically valid; however, there is no consensus on why one solution is preferred in the context of the original problem.

Contextual Notes

The discussion highlights the potential for multiple solutions in trigonometric equations and the importance of context in determining which solution to use. There are unresolved questions regarding the criteria for selecting between the solutions.

Drain Brain
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Can you help me how to solve this system of trig eqns

$W\cos(30^{\circ})-275\cos(\theta)=0$
$W\sin(30^{\circ})+275\sin(\theta)=300$

I have tried to divide the first eqn by 2nd and I get

$\tan(30^{\circ})=\frac{275}{300\cos(\theta)}-\tan(\theta)$ I'm stuck here! Kindly help me please!
 
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Drain Brain said:
Can you help me how to solve this system of trig eqns

$W\cos(30^{\circ})-275\cos(\theta)=0$
$W\sin(30^{\circ})+275\sin(\theta)=300$

I have tried to divide the first eqn by 2nd and I get

$\tan(30^{\circ})=\frac{275}{300\cos(\theta)}-\tan(\theta)$ I'm stuck here! Kindly help me please!

Hey Drain Brain!

Suppose we isolate $\cos(\theta)$ and $\sin(\theta)$ in each equation, squared them, and add them.
Then we would be rid of $\theta$ and can solve for $W$.

Afterwards, we can substitute $W$ back in the first equation and solve for $\theta$.
Finally, we should check if the solutions found are actually solutions, since we may have introduced new solutions.
 
HI I LIKE YOU!:o

I just did what said and came up with a quadratic equation

$W^2-600W\sin(30)+300^2-275^2=0$

the two solutions are

$W=240.1$ and $W=59.86$

$\theta = 40.9$ and $\theta = 79.13$

which of them should I choose?
 
Drain Brain said:
HI I LIKE YOU!:o

I just did what said and came up with a quadratic equation

$W^2-600W\sin(30)+300^2-275^2=0$

the two solutions are

$W=240.1$ and $W=59.86$

Good! ;)
$\theta = 40.9$ and $\theta = 79.13$

which of them should I choose?

Actually, you should have 2 solutions for $\theta$ for each value of $W$...

What happens if we substitute them in the second equation?
 
I like Serena said:
Good! ;)

Actually, you should have 2 solutions for $\theta$ for each value of $W$...

What happens if we substitute them in the second equation?

$240\sin(30^{\circ})+275\sin(40.9)=300$-->>$300=300$ $59.86\sin(30^{\circ})+275\sin(79.13)=300$---->>$300=300$

Does this mean that the values of W and $\theta$ that I get are valid solutions?

Actually the system of equations that I posted above came from a problem about equilibrium of a particle.
I was asked to find force W and angle $\theta$ to satisfy equilibrium conditions. The answer to this problem was 240 lb for W, and $\theta=40.9$. If both sets of solution are valid, why the other solution was not chosen?
 
Drain Brain said:
$240\sin(30^{\circ})+275\sin(40.9)=300$-->>$300=300$ $59.86\sin(30^{\circ})+275\sin(79.13)=300$---->>$300=300$

Does this mean that the values of W and $\theta$ that I get are valid solutions?

Actually the system of equations that I posted above came from a problem about equilibrium of a particle.
I was asked to find force W and angle $\theta$ to satisfy equilibrium conditions. The answer to this problem was 240 lb for W, and $\theta=40.9$. If both sets of solution are valid, why the other solution was not chosen?

Yes, they are both solutions.
It depends on the actual problem statement what to do with them.
Perhaps just a single solution was requested, perhaps there is another reason to discard one of them, or perhaps the given answer is simply incomplete.
 

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