Posted by Sonia on .
A sail boat on lake Huron leaves Southampton and sails 20degrees west of north of 20km. At the same time, a fishing boat leaves Southampton and sails 30degrees west of south for 15 km. At this point, hot far apart are the boats, to the nearest metre? ( the answer is 4139m, but I need to know the method to understand how to get that answer)

Trig 
bobpursley,
distance is the difference in positions.
But that involves vectors, so here, I think your teacher wants you to use trig.
draw the figure.
Notice the angles first: the interior angle between the paths is 130degrees
Next, the law of cosines
c^2=20^2+15^22*20*15Cos130
that is it, solve for c. 
Trig 
Sonia,
Ok thank u:)

Trig 
Sonia,
Um the answer that I got is 24.72km can u please help me get the answer above?

Trig 
Reiny,
in my previous reply to your question,
I made the silly mistake of adding 180(20+15_ instead of 180(20+30) to get the angle between their paths.
http://www.jiskha.com/display.cgi?id=1279674199
I agree with bobpursley's equation, and the result of that is
c^2 = 400 + 225  (385.6726)
c^2 = 1010.672566
c = 31.8 km
Your answer of 4139 m or 4.139 km makes no sense,
I noticed you changed it from 4139km to 4139 m but it is still way off 
Trig 
Sonia,
Ok but do think u can help me with this question:
a sailor out in a lake sees two likght houses 11km apart along the shore and gets bearings of 285degrees from his present position for light house A and 237degrees for light house B. From light house B, light house A has a bearing of 45degrees. How far to the nearest kilometre, is the sailor from each light house? What is the shortest distance, the nearest kilometre, from the sailor to the shore? ( the answers are 3km, 13km, 3km) 
Trig 
Henry,
Using vector analysis, i calculated 31.79 km. So my cal. agree with bobpursley.

Trig 
Henry,
20km[110o]  15km[240o] =
6.84+18.8i (7.513i) = 0.66 + 31.8i = 31.81 km[88.8o] N. of E.