Find the contraction of angles seen by an observer

[TSny]

You are getting very close. But you are making some careless errors regarding which length is the contracted length and the definition of gamma.

I also know that C'D = ϒCD, which is the contracted length is equal to the uncontracted length times gamma.
I know that gamma is equal to √(¾) is is √(3)/2 for this problem.

The relation C'D = ϒCD is correct. But C'D is the uncontracted length and CD is the contracted length.

Gamma is not equal to $\sqrt{1-v^2/c^2}$. Gamma is $\large \frac{1}{\sqrt{1-v^2/c^2}}$

Please try again.

 

[guyvsdcsniper]

You are getting very close. But you are making some careless errors regarding which length is the contracted length and the definition of gamma.The relation C'D = ϒCD is correct. But C'D is the uncontracted length and CD is the contracted length.

Gamma is not equal to $\sqrt{1-v^2/c^2}$. Gamma is $\large \frac{1}{\sqrt{1-v^2/c^2}}$

Please try again.

So it should be this? Which I then just repeat to β and then sum the two and subtract from 180?

 

[TSny]

Looks good.

 

[guyvsdcsniper]

Thank you all for your help. My apologies if I made it more difficult that necessary.