Two boats A and B left a port C at the same time on different route .B travelled on a bearing of 150 degree and A travelled 8km and B had travelled 10km ,the distance between the Two boats was found to be 12km .calculate the bearing of A's route from C.
Can I get the drawing for the above question pls
can i get a drawing
Can i get he drawing pls
12²=8²+10²-2*8²*10² =82.8 ,150_82.8=67.2
can i get the drawing please
Thanks
Pls the diagram
How will i drew the diagram
how did you please get 82.8 in this question
How did you get 82.8
Thanks
The diagram pls
pls can I get a drawing
The answer to tangle <A is -57
If add angle C and B it will equal to 233, minus it from 180° it
Try getting the angle of C then add all the angles together it will equal to 180°
The diagram please also I don't understand how you got the answer
B traveled on a HEADING of 150 (around SSE)
Draw position of B, 10 from origin at 60 degrees south of east (heading of 150)
With center at B, draw a circle of radius 12
A is 8 km from the origin (can draw another circle if you want) at the intersection of the 12 radius circle, two points, one north of east and one south of west.
we have triangles with lengths 8 10 and 12
find Angle ACB with law of cosines
12^2 = 8^2 + 10^2 - 2*8*10 cos ACB
144 = 164 -160 cos ACB
ACB = 82.8 deg
so CA is 82.8 - 60 = 22.8 degrees north of east
If you alternately figure A headed the other way (left)
then the total angle of A clockwise from east = 60 + 82.8 = 142.8 or 37.2 deg south of east