Question

Two ship leave port at the same time one travel at 5km/h on a bearing of 046 the other travel 9km/h on a bearing of 127 how far apart are the ship after 2 hours

Answer 1

You do not sail on a "bearing". You sail on a "heading". A bearing is a direction you look at another ship or a lighthouse or whatever. " That tanker bears south by Southwest for example. Math texts drive me nuts. Landlubbers.
Anyway:
first ship east (+x) distance = 10 cos 46 = 6.95 km
first ship north (+y) distance = 10 sin 46 = 7.19 km
180 - 127 = 53 degrees north of west
second ship east distance = -18 cos 53 = - 10.8 km
second ship north distance = 18 sin53 = 14.4 km

difference east-west = 6.95 + 10.8 =
difference north south = 14.4 - 7.9 =

you want the square root of the squares of those two numbers

Answer 2

All angles are measured CW from +y-axis.
AB = 5km/h * 2h = 10km[46o].
BA = 10km[46+180] = 10km[ 226o].
AC = 9km/h * 2h = 18km[127o].
BC = distance between the ships after 2h.

BC = BA+AC = 10[226] + 18[127]
BC = (10*sin226+18*sin127) + (10*cos226+18*cos127)i
BC = 7.2 - 17.8i
BC = sqrt(7.2^2 + 17.8^2) =

Answer 3

Two ship leave port at the same time one travel at 5km/h on a bearing of 046 the other travel 9km/h on a bearing of 127 how far apart are the ship after 2 hours

Answer 4

Help me with it with diagram

Answer 5

DIAGRAM PLS

Answer 6

Diagram pls

Answer 7

Thanks but can you help me with the diagram please

Answer 8

I needed answer for the question