Asked by Ade
A ship P sails at an average speed of 50 km/h on a bearing of N35°E from a port R. At the same time another ship Q sails at an average speed of 40 km/h on a bearing of S25°W. How far are they after 4 hours?
Answers
Answered by
oobleck
After 4 hours,
P has gone 200 km
Q has gone 160 km
The angle between their headings is 170°
Now just use the law of cosines to find the distance PQ
P has gone 200 km
Q has gone 160 km
The angle between their headings is 170°
Now just use the law of cosines to find the distance PQ
Answered by
henry2,
All angles are measured CW from +y-axis.
RP = 50[35o] * 4 = 200km[35o].
PR = 200[35+180] = 200km[215o].
RQ = 40[205o] * 4 = 160km[205o].
PQ = PR+RQ = 200[215o] + 160[205o]
PQ = (200*sin215+160*sin205) + (200*cos215+160*cos205)I
PQ = -182. - 309i. = 359km[30.5o].
RP = 50[35o] * 4 = 200km[35o].
PR = 200[35+180] = 200km[215o].
RQ = 40[205o] * 4 = 160km[205o].
PQ = PR+RQ = 200[215o] + 160[205o]
PQ = (200*sin215+160*sin205) + (200*cos215+160*cos205)I
PQ = -182. - 309i. = 359km[30.5o].
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