## Doppler effect, offset from line of effect

sorry lots of reading but i wanted to explain what I've done (tried to do)
1. The problem statement, all variables and given/known data
An ambulance has a loud siren, which emits a pure note at 5100Hz. A man 20 m from a straight road along which the ambulance is travelling, and records the frequency of the sound he hears from the siren. There is no wind. When the ambulance is approaching the man and a long way from the man what frequency is recorded? what frequency is heard after the ambulance pass the observer and recedes into the distance? the ambulance speed is 20 m/s.

Bonus marks if you find the frequency by the man at t=-1s and t=2s where t=0 is when the ambulance is closest to the man

2. Relevant equations
Speed of sound 343 m/s
f'= $$\frac{v}{v-v(s)}$$ f

3. The attempt at a solution
So far off into each distance is just straight forward Doppler effect
f'= $$\frac{v}{v-v(s)}$$ f

getting when travelling towards the man f'=5415Hz which seems correct since its travelling towards the detector yes?
and when travelling away from man f'=4819Hz which also seems to be correct since its travelling away the detectors yes?

here's when I get into a bit of trouble if he is off the road (I'm taking it as he is on a side road perpendicular to the main road) I have found two sources saying take the cos angle and one saying take sin angle (so I going to use cos angle for now, but is this right?) and I'm not sure how to multiply it into the equation this is what I tried,

this was a guess am I doing it right of do I do it a different way (my main problem)
f'= $$\frac{v}{v-sin (\theta)v(s)}$$ f

after I get the equation right would I have to do a couple triangles to find the angle created when the ambulance is 20 meters away (1 second) and 40 meters away (2 seconds)
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 The only difference, when the ambulance is close, is that it's not coming directly at him. Just find the speed along the direction from him to the ambulance, and plug that into your equation and you're good.
 Ahh good idea, so your saying find the an equivalent velocity (or proportion) of the ambulance in the direction straight towards the observer that makes sense. thanks