looks like a rather straight-forward case of the cosine law.
distance^2 = 42^2 + 28^2 - 2(42)(28)cos60
distance = 37.04 nm
vector of first leg = (42cos270, 42sin270)=(0, -42)
vector for 2nd leg = (28cos30,28sin30) = (24.249, 14)
adding them to get (24.249 , -28)
magnitude = √(24.249^2 + (-28)^2) = √1372 = 37.04 nm
None of your answers match this.
My answer was obtained in 2 totally different methods
Wel, I'm not sure, I have two other questions with practically the same things, please take a look!
If a ship leaves port at 9:00 a.m. and sails due east for 3 hours at 10 knots, then turns N 60° E for another hour, how far from port is the ship?
A. 35 nm
B. 39 nm
C. 43 nm
D. 47 nm
If a ship leaves port at 9:00 a.m. and sails due south for 3 hours at 14 knots, then turns N 60° E for another 2 hours, how far from port is the ship?
A. 14 nm
B. 17.75 nm
C. 21.5 nm
D. 22.6 nm
Thank you; you'tr a life saver!
If a ship leaves port at 9:00 a.m. and sails due north for 3 hours at 12 knots, then turns N 30° E for another hour, how far from port is the ship?
A. 45 nm
B. 47 nm
C. 51 nm
D. 53 nm
This time I get a triangle with sides 30 and 10 and the contained angle between them is 150° , (90+60)
again by the cosine law,
dist^2 = 30^2 + 10^2 - 2(30)(10)cos150
distance = √1519.62 = appr 38.98 nm
which is choice B
The next question you posted is a repeat of your first
dist^2 = 36^2+ 12^2 - 2(36)(12)cos150
dist = 46.78 which looks like it is B.
Something fishy about your first one, since I followed exactly the same steps
check for data or answers.
Thank you very much; but these are the only answers! I don't know why, but I'll try to check with my teacher. Again, thanks for the help and showing your work, you're awesome!
Just tried you first question again, using cos30° in my cosine law
and got 22.6 which is D
BUT, according to your wording of N 60° E, the angle between the two paths would be 60, and not 30
So to get your answer it should have been N 30° E