Question 1163051
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Since one vessel is on a course east of north and the other is on a course west of north, simply adding the course values gives the difference in their courses.  Since the sum is 90 degrees, the horizontal distance is easily calculated by the Pythagorean theorem:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \sqrt{(3\,\times\,8.89)^2\ +\ (3\,\times\,13.3)^2}]


  But the distance radio waves would have to travel is dependent on the height above the waterline for the radio antenna for the boat and the actual depth of the radio antenna for the submarine.  Further, since radio waves are refracted when changing media from atmosphere to water, and the index of refraction for air is temperature dependent and the index of refraction for water is temperature and salinity dependent, there is insufficient information to answer your question as posed.


However, making the highly irrational assumption that the boat and submarine are point sources of radio frequency energy and discounting refraction, an additional application of Pythagoras with one leg of the triangle the results of the above calculation and the other leg measuring 30 meters, will provide the desired answer.
								
								
John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
My calculator said it, I believe it, that settles it
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