Question

I'm trying to find an angle to intercept two objects. Object B is 15 degrees East of North at a distance of 20km from object A. Object B is travelling 40 degrees East of North at a speed of 30km/h. Object A is travelling at 100km/h. At what angle should then object A travel to intercept with object B? I'm trying to treat them as vectors, but am unsure if it works due to the interception point being relative to the time taken to travel?

Answer 1

Is the following possible?
\[\frac{ \sin(x°) }{ 30 \times t } = \frac{ \sin(40°-15°) }{ 100 \times t } \rightarrow x° = \sin^{-1} (\frac{ 30 \times t \times \sin(25°) }{ 100 \times t }) = 7.284°\]
Where I then add 15°+7.284°, so the angle to intercept with object B would be 22.284°? Would this be correct?

Answer 2

now, first off, B 15 degree east of north compare to A, right? so draw it out

Answer 3

agree?