Maths trigo question?

Question

At 0900 hour, ship A leaves port X at a speed of 50km/h on a bearing of 040 degrees and ship B leaves at 30km/h on a bearing of 120 degrees.
At 1200h, what is the
(i) distance between the two ships
(ii) bearing of A from B?

Answer 1

Let us consider the X-axis as the reference axis.
1200 hrs - 0900 hrs = 3 hrs
In 3 hrs ship A travels, 50*3 = 150 km at 40 deg with X-axis
In 3 hrs ship B travels, 30*3 = 90 km at 120 deg with X-axis

Thus we have the positions of A and B at 1200 hrs in the polar coordinates as (150, 40 deg) and (90,120 deg) respectively.

Draw this figure and join A and B.
Consider triangle OAB [O is origin].

(i)
Length of side AB is needed.
In this triangle we know AO = 150, BO = 90,
By cosine rule,
AB^2 = AO^2 + BO^2 - 2*AO*BO*cos( AB^2 = 150^2 + 90^2 - 2*150*90*cos(80)
Thus, AB = 160.97
Thus reqd. distance between the ships is 160.97 km.

(ii)
Angle between X-axis and line segment AB is needed.

By sine rule,
AO/sin( 150/sin( 150/sin( Thus,
Thus bearing of A from B is 66.6 - 60 = 6.66 deg

Answer 2

after 3 hrs, ship A has traveled 150 km, ship B has traveled 90km at an angle of 80 degress from the trajectory of the 1st.
From the law of cosines, the distance between the ships, x can be found from
x^2=150^2+90^2-2*90*50cos(80)=161 km

Fiddling with triangles a little (hard to explain without graph) I get ship A is bearing 6.6 degrees & 161 km relative to B