Asked by Bethany
Two ships left the same port at the same time. One travelled at 16kn/h on a course of 275 degrees while the other at 20 km/h of 225 degrees. How far apart are the ships after a. 2 hours? b. 4 hours?
*note: this is a trigonometry question using the cosine law.
Thanks!
*note: this is a trigonometry question using the cosine law.
Thanks!
Answers
Answered by
Damon
275 - 225 = 50 degrees difference in heading
In two hours
Ship A went 16 * 2 = 32
Sbip B went 20 * 2 = 40
c^2 = a^2 + b^2 - 2 a b cos C
c^2 = (32)^2 + (40)^2 -2(32)(40)cos 50
c^2 = 1024 + 1600 - 1646.5
c^2 = 978.4
c = 31.3
For pert B, similar triangle, double time, double distance
= 62.6
In two hours
Ship A went 16 * 2 = 32
Sbip B went 20 * 2 = 40
c^2 = a^2 + b^2 - 2 a b cos C
c^2 = (32)^2 + (40)^2 -2(32)(40)cos 50
c^2 = 1024 + 1600 - 1646.5
c^2 = 978.4
c = 31.3
For pert B, similar triangle, double time, double distance
= 62.6
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