Wednesday

October 1, 2014

October 1, 2014

Posted by **John** on Wednesday, February 5, 2014 at 2:56pm.

A. 14 nm

B. 17.75 nm

C. 21.5 nm

D. 22.6 nm

PLEASE HELP!

- Trigonometry -
**Reiny**, Wednesday, February 5, 2014 at 3:28pmlooks 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

- Trigonometry -
**John**, Wednesday, February 5, 2014 at 3:35pmWel, 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!

- Trigonometry -
**John**, Wednesday, February 5, 2014 at 3:36pmIf 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

- Trigonometry -
**Reiny**, Wednesday, February 5, 2014 at 4:06pmThis 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.

- Trigonometry -
**John**, Wednesday, February 5, 2014 at 4:08pmThank 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!

- Trigonometry -
**Reiny**, Wednesday, February 5, 2014 at 4:32pmJust 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

**Answer this Question**

**Related Questions**

Trigonometry - Navigation A ship leaves port at noon and has a bearing of S 29° ...

math- precalculus - I've attempted this problem a few times but I can't get the ...

geometry - A ship leaves port and heads due east at a rate of 32 miles per hour...

Math Analysis - Can someone help me with this problem, please? Thanks!! "A ship ...

physics - A viking sails from his home port and travels east at 15km/hr.after 10...

pre-calc - a ship leaves port and sails for 2 hours north east and than 3 hours ...

I NEED URGENT HELP!!! - A ship leaves port at 1:00 P.M. and sails in the ...

Math - "A ship leaves port on a bearing of 34.0 degrees and travels 10.4 mi. ...

math - at 12 noon ship A is 65 km due north of a second ship B. Ship A sails ...

Calculus - at 12 noon ship A is 65 km due north of a second ship B. Ship A sails...