1. ## relative velocity

A ship, A is travelling due east at 24 km/h. A second ship, B, travelling at 33 km/h is first sighted 20 km due south of A. From A the ship B appears to be moving north east.
Find
(i) the direction in which B is actually moving

there are a few ways to do this. the easiest is to set up a triangle.

so using the sin rule. 33/sin45 = 24/sinx when i solve for x i get 30.1 degrees. the answer at the back of the book is 14.1 degrees. i thought i was doing everything right

2. ## Re: relative velocity

$v_B = v_A + r$

$\dfrac{\sin(45^\circ-\theta)}{24} = \dfrac{\sin(135^\circ)}{33}$

$\sin(45^\circ-\theta) = \dfrac{24 \sin(135^\circ)}{33}$

$45^\circ-\theta = \arcsin\bigg[\dfrac{24 \sin(135^\circ)}{33}\bigg]$

$\theta = 45^\circ - \arcsin\bigg[\dfrac{24 \sin(135^\circ)}{33}\bigg] \approx 14.1^\circ$

3. ## Re: relative velocity

i dont understand your diagram. is ship B not sighted south of ship A, so it would be heading towards it.

4. ## Re: relative velocity

the sketch shows velocity vectors, not position vectors.

5. ## Re: relative velocity

ok and i solved for the wrong angle in the OP thats why i got it wrong