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August 29, 2014

August 29, 2014

Posted by **Sonia** on Wednesday, July 21, 2010 at 7:07pm.

- Trig -
**bobpursley**, Wednesday, July 21, 2010 at 7:16pmdistance 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+15^2-2*20*15Cos130

that is it, solve for c.

- Trig -
**Sonia**, Wednesday, July 21, 2010 at 7:24pmOk thank u:)

- Trig -
**Sonia**, Wednesday, July 21, 2010 at 7:41pmUm the answer that I got is 24.72km can u please help me get the answer above?

- Trig -
**Reiny**, Wednesday, July 21, 2010 at 9:11pmin 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**, Wednesday, July 21, 2010 at 9:36pmOk 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**, Thursday, July 22, 2010 at 11:09pmUsing vector analysis, i calculated 31.79 km. So my cal. agree with bobpursley.

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