Solving Trig Equations: Help Needed!

[Drain Brain]

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!

 

[I like Serena]

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.

 

[Drain Brain]

HI I LIKE YOU!

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?

 

[I like Serena]

HI I LIKE YOU!

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?

 

[Drain Brain]

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?

 

[I like Serena]

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

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.