Asked by Idris

Two ships A and B leaves the some harbour at the same time in directions 152° and 142° respectively calculate the distance between the two ships when A has travelled 10km and B has travelled 12km. What is the bearing of A from B at this time
Answered by bobpursley
sound much as the law of cosines and sines on this one.
sketch it. first you have two sides and angle between to get opposite side. Now use the law of sines to get the last angle, Notice you have to convert that to a bearing relative ot North.
Answered by henry2,
All angles are measured CW from +y-axis.
Given: HA = 10km[152o], HB = 12km[142o].
BH = 12km[142+180] = 12km[322o].

BA = BH + HA = 12[322] + 10[152]
BA = (12*sin322+10*sin152) + (12*cos322+10*cos152)i
BA = -2.69 + 0.627i = 2.76km[-76.9o] = 2.76km[283o] CW.

Bearing of A from B = 283 degrees.
Answered by Dicta
The sketch of the journey was not drawn
Answered by Akwem
Why formulary was used here
I'm not sure it's cosine
Answered by Daniel chang
The answer is 2.765
There are no AI answers yet. The ability to request AI answers is coming soon!