Your ship leaves port on a bearing of N25E for 3.5 hours, at which time you change your course to N90E for 5.25 hours. If your constant speed is 20mph what bearing must you take to get back to port? How long will the return trip take?

Question

dumbcow · Answer

|dw:1392609386450:dw|
use Law of cosines to find length x

use Law of Sines to find angle a

anonymous · Answer

Thank you so much for the diagram and help. I really appreciate it! Would my x=27.58 or did I miss calculate? Would a = 41.8 degrees?

dumbcow · Answer

no i am getting different answers
$$x = \sqrt{70^2 + 105^2 -2(70)(105)\cos(115)}$$
$$\sin a = \frac{70 \sin(115)}{x}$$

anonymous · Answer

Okay. So x should be 63.44 and how would you get a?

dumbcow · Answer

no not 63.44 , look at diagram, "x" is the longest side so it must be greater than 105

anonymous · Answer

My bad I miss read x= 148.78 then. and a should be the 63.44. Trig is really not my best subject

dumbcow · Answer

yes on x
no on "a"

sin(115) = .9063
sin(a) = .9063*(70/148.78) = .42639

a = 25.24

anonymous · Answer

Okay. Thank you so much! I sincerely appreciate all of your help. So this means my bearing for back to port would be 25.24 degrees SE? and it would take be 148.78/20 to return.

dumbcow · Answer

SW not SE

not sure on bearing format
W 25.24 S ??

or
S 64.76 W

anonymous · Answer

Okay. Awesome! I will put down both. How did you come up with the 64.76? Thank you so much!

dumbcow · Answer

90 - 25 = 65

anonymous · Answer

Thank you!