Finding W and Theta in Simultaneous Equations

[camino]

Homework Statement

160 cos(theta) - W cos(69) = 0

160 sin(theta) - W cos(21) - W = 0

Can someone solve these equations for me to find W and theta?

 

[rock.freak667]

Homework Statement

160 cos(theta) - W cos(69) = 0

160cosθ=Wcos(69)

W=160cosθ/cos(69)


continue now

 

[camino]

W = W cos(69) / cos(69)?

 

[rock.freak667]

W = W cos(69) / cos(69)


?

no, put W=160cosθ/cos(69) into the second equation

 

[camino]

Sorry I am still not seeing it. Could you give me one more step?

 

[rock.freak667]

Sorry I am still not seeing it. Could you give me one more step?

In this equation

160 sinθ - W cos(21) - W = 0


everywhere you see 'W' put 160cosθ/cos(69) and try to simplify it.

 

[camino]

I do understand how to simplify one equation and substitute into the other, however I just can't seem to simplify these down. The theta with the sin and cos attached is what is really confusing me when trying to simplify. This is just a mess and I'm very confused.

 

[rock.freak667]

I do understand how to simplify one equation and substitute into the other, however I just can't seem to simplify these down. The theta with the sin and cos attached is what is really confusing me when trying to simplify. This is just a mess and I'm very confused.

just show where you've reached and we'll see if we can simplify it further

 

[camino]

160 sinθ - (160cosθ/cos69) cos(21) - (160cosθ/cos69) = 0

 

[rock.freak667]

160 sinθ - (160cosθ/cos69) cos(21) - (160cosθ/cos69) = 0

so you get


160 sin \theta - (\frac{160cos21}{cos69} - \frac{160}{cos69})cos \theta = 0


this is like Asinθ-Bcosθ=0, how do you solve this equation?

 

[camino]

Is there a trig identity that replaces either sin or cos? I'm thinking we want to narrow this down to only having either sin or cos in the equation.

 

[rock.freak667]

Is there a trig identity that replaces either sin or cos? I'm thinking we want to narrow this down to only having either sin or cos in the equation.

well yes but we can put it like this


Asinθ=Bcosθ

or sinθ/cosθ=B/A

do we know another way to express sinθ/cosθ ?

 

[camino]

I am really not sure.

 

[rock.freak667]

I am really not sure.

do you not know that tanθ=sinθ/cosθ ?

 

[camino]

Oh my it has been a long day.. Yes i see that now. So simplifying I would then do:

160 sinθ - (-29.6542) cosθ = 0

160 sinθ = 29.6542 cosθ

θ = tan^-1(160/29.6542)

θ = 79.5° ?

 

[rock.freak667]

Oh my it has been a long day.. Yes i see that now. So simplifying I would then do:

160 sinθ - (-29.6542) cosθ = 0

160 sinθ = 29.6542 cosθ

θ = tan^-1(160/29.6542)

θ = 79.5° ?

yes that should be correct. You are able to find W knowing θ=79.5° right?

 

[camino]

Yes:

160cos(79.5) - W cos(69) = 0

29.1577 = W cos(69)

W = 81.36

 

[camino]

Thank you so much for all your help!