How does sin(65)/sin(37) become sin(65)/tan(37)?

[foreverlost]

Homework Statement

This is collision in 2D problem. I already did most of the work except I'm stuck on the trigonometry and algebra for solving for the final velocity of object a.

I will put the problem here just in case someone wants to show me a better way for getting to the answer.

Collision between two pucks. Puck "a" has mass = 0.025kg Velocity of "a" = +5.5m/s

along x-axis "a" makes collision with puck "b" which has mass = 0.050kg and "b" starts at rest. Collision is not head on. So, after collision puck "a" flies apart from "b" at angle 65° and puck "b" flies off at angle 37°

Homework Equations

I am trying to solve for Vfa;

Ma*Voa = Ma*Vfa(cos65) + [Ma*Vfa(sin65)/(sin37)](cos37)

Ma = X
Voa = Y
Vfa = Z

XY = XZ(cos65) + [ XZ(sin65)/(sin37)](cos37) solve for Z

Hopefully that's better.

The solutions in textbook managed to solve for Vfa or Z like this:

Z = Y/[cos65 + (sin65/tan37)]

My question how to get Z by itself and where did (sin65/tan37) come from?

 

[Doc Al]

My question how to get Z by itself

Factor out the Z from the right hand side.

and where did (sin65/tan37) come from?

What does tanθ equal? 1/tanθ?

 

[foreverlost]

Factor out the Z from the right hand side.

Okay What confused me was that I didn't know how to get rid of Z from inside the parenthesis.

What does tanθ equal? 1/tanθ?

I'm not sure could I just plug in any number to θ and figure it out? I just get different numbers.

 

[foreverlost]

Factor out the Z from the right hand side.

What does tanθ equal? 1/tanθ?

tanθ = sin/cos

1/tanθ = cotθ = -sinθ/cosθ

 

[Doc Al]

tanθ = sin/cos

Right.

So 1/tanθ = 1/(sin/cos) = cosθ/sinθ

1/tanθ = cotθ = [STRIKE]-sinθ/cosθ[/STRIKE]

 

[foreverlost]

Right.

So 1/tanθ = 1/(sin/cos) = cosθ/sinθ

Thanks Doc Al. I always did have trouble with trigonometric identities. Fortunately this exact problem was on the exam, and I practically remembered every step without much thought lol.