Find angles between two rope and the ceiling

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The discussion focuses on finding the angles between two ropes and the ceiling using trigonometric identities and vector properties. Participants emphasize the importance of resolving tension into vertical and horizontal components, noting that the horizontal components must balance while the vertical components support the weight. The challenge lies in separating the vectors into components without given angles, leading to confusion about using the weight as a shared vertical component. A suggestion is made to set up two equations based on vertical and horizontal considerations to solve for the unknown angles. The Law of Cosines is also mentioned as a potential method for finding the angles.
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***Excuse the pluralization error in the title***

Homework Statement



I attached a picture below...

Homework Equations



sin^2(theta)+cos^2(theta)=1

a^2+b^2=c^2

Vector properties

Trig identities

The Attempt at a Solution

I am familiar with trig identities and vector properties, but I can't get anywhere with this. I tried using the 697N weight as the y-component vector for either side, but that didn't work.

Homework Statement


Homework Equations


The Attempt at a Solution

 

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seanster1324 said:
***Excuse the pluralization error in the title***

Homework Statement



I attached a picture below...


Homework Equations



sin^2(theta)+cos^2(theta)=1

a^2+b^2=c^2

Vector properties

Trig identities

The Attempt at a Solution




I am familiar with trig identities and vector properties, but I can't get anywhere with this. I tried using the 697N weight as the y-component vector for either side, but that didn't work.

resolve each tension into vertical and horizontal components.
The horizontal components have to balance each other,
The vertical components together support the mass.
 
I understand that part. The part I don't understand is how to separate each vector into its components. I only have the magnitude with no given angles. Looking at the hint, I need to play around with some trig. But I can't see how to incorporate the identity it hints at.

And I can't use 697N as the shared vertical component, right? Because that would make this much easier, but it did not work for me.
 
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seanster1324 said:
I understand that part. The part I don't understand is how to separate each vector into its components. I only have the magnitude with no given angles. Looking at the hint, I need to play around with some trig. But I can't see how to incorporate the identity it hints at.

And I can't use 697N as the shared vertical component, right? Because that would make this much easier, but it did not work for me.

You have two unknowns, theta 1 and theta 2

Vertical considerations will give one equation involving them

Horizontal considerations will give a second equation connecting them

Two equations in two unknowns should mean a simultaneous equations solution is possible.

EDIT: you could also use the cosine rule to find the angles.
 
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Ah, I see what you mean. Thank you so much!
 
seanster1324 said:
Ah, I see what you mean. Thank you so much!

Note my edit about the cosine rule.
 
Witch cosine rule? The Law of Cosines?
 
seanster1324 said:
Witch cosine rule? The Law of Cosines?

yes: a2 = b2 + c2 - 2bc.cos(A)
 
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