Rocket A moves with speed .75c in a northerly direction relative to an origin. Rocket B moves west (relative to that origin) with speed .4c. As B moves, it emits radiation with a wavelength 100nm normal to its line of motion, in the northerly direction. What is the wavelength of the radiation as seen by A?
Doppler shift equations
Velocity addition formulas
The Attempt at a Solution
So far I have two different approaches to a solution, the first approach finds the wavelength of the radiation as seen by a person at the origin (which is redshifted) because the source is moving away at .4c. Then use that wavelength to find the wavelength as seen by a person moving away at .75c (to represent the rocket A). I'm not so sure about this solution...
The second idea I had was to find the relative velocity of B as seen by A (since it would be a right triangle) and using those to find the angle between the light source and the observer. However, I am unsure how to find the separate components of the relative velocity. Do I keep the .75c speed in one direction and uy = uy' /(gamma(1-ux*v/c^2) to find the other direction, if that is so I have no idea what the relative velocity would be... This method is really tripping me up and I think it is the correct way. Could someone help me get on the right track?
All incite is appreciated