Posted by Mei on Thursday, December 1, 2011 at 10:32pm.
just start drawing your lines, and figure out how far each leg of the trip takes the ship. Start at (0,0)
go 10.4 mi at N34°E puts you at
(10.4*sin34°,10.4*cos34°) = (5.82,8.62)
Now go E for 4.6, and that adds (4.6,0) to the displacement. Now you are at
(10.42,8.62)
The new bearing θ is given by
tanθ = 10.42/8.62 = 1.2088
θ = 50.4°
distance is sqrt(182.88) = 13.52
or, 13.52mi at N50.4°E
Related Questions
TRIG HONORS CH: 5.8 - a ship leaves port with a bearing of S 42deg W after ...
TRIGONOMETRY - a ship leaves port with a bearing of S 42deg W after traveling 7 ...
trig - a ship leaves port with a bearing of S 42deg W after traveling 7 miles, ...
Trig - A ship leaves port with a bearing of S 40 W. After traveling 7 miles, the...
Trig - A ship leaves port with a bearing of S 40 W. After traveling 7 miles, the...
Trig - A ship leaves port with a bearing of S 40 W. After traveling 7 miles, the...
Trig - A ship leaves port with a bearing of S 40 W. After traveling 7 miles, the...
trig - Q1: Prove cos^2t+4cost+4/cost+2=2sect+1/sect Q2: A ship leaves port with ...
math- precalculus - I've attempted this problem a few times but I can't ...
trig - A freighter leaves on a course of 120 degrees and travels 515 miles. The...
For Further Reading