MHB Solving Trig Equations: Help Needed!

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The discussion revolves around solving a system of trigonometric equations related to the equilibrium of a particle. The user initially struggles with the equations but successfully derives a quadratic equation for W and finds two potential solutions for W and corresponding angles θ. Both sets of solutions are validated against the original equations, confirming their correctness. However, the choice of which solution to use depends on the specific requirements of the problem, as both solutions are mathematically valid. Ultimately, the user concludes that the preferred solution is W = 240 lb and θ = 40.9°.
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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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