Asked by Doris

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

Answers

Answered by Damon
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

Answered by henry2,
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) =
Answered by Anonymous
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
Answered by Jonas Dezebel
Help me with it with diagram
Answered by Anonymous
DIAGRAM PLS
Answered by Brainstorm
Diagram pls
Answered by Felix Gideon
Thanks but can you help me with the diagram please
Answered by Chisom
I needed answer for the question
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