looks like a rather straight-forward case of the cosine law.
distance^2 = 42^2 + 28^2 - 2(42)(28)cos60
= 1372
distance = 37.04 nm
or
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
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
PLEASE HELP!
6 answers
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 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
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
= 1519.62
distance = √1519.62 = appr 38.98 nm
which is choice B
The next question you posted is a repeat of your first
The third:
same setup;
dist^2 = 36^2+ 12^2 - 2(36)(12)cos150
= 2188.25
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.
again by the cosine law,
dist^2 = 30^2 + 10^2 - 2(30)(10)cos150
= 1519.62
distance = √1519.62 = appr 38.98 nm
which is choice B
The next question you posted is a repeat of your first
The third:
same setup;
dist^2 = 36^2+ 12^2 - 2(36)(12)cos150
= 2188.25
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
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