Question 1196965
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Lookouts on two naval vessels, about 2 miles apart, spot a disabled boat off the port side. 
One lookout spots the boat at an angle of 42° with the line connecting the naval vessels. 
The other lookout spots the boat at an angle of 28° with the same line. 

(a) Which naval vessel is closer to the boat?

(b) How much closer it? 
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<pre>
You have a triangle ABC, where A is the point of the first lookout and B is the point 
of the second lookout. The point C is where the disable boat is located.


Answer to (a) is easy: in any triangle, the longer the side is, the greater is the opposite angle.

In our case, angle at A is 42°, while angle at B is 28°.

Since angle A is greater than angle B, it means that side BC is longer than side AB.

Thus, vessel A is closer to the disable boat than vessel B.    It is the <U>ANSWER to question (a)</U>



To solve (b), use the sine law

    {{{a/sin(A)}}} = {{{b/sin(B)}}} = {{{c/sin(C)}}}.


We know c = 2 kilometers, C = 180° - 42° - 28 = 110°. 

So,  {{{c/sin(C)}}} = {{{2/sin(110^o)}}} = {{{2/0.93969}}} = 2.12836  (rounded).


Therefore,  a = 2.12836*sin(A) = 2.12836*sin(42°) = 2.12836*0.66913 = 1.424 kilometers,

            b = 2.12836*sin(B) = 2.12836*sin(28°) = 2.12836*0.46947 = 0.999 kilometers.


Thus  A  is closer than  B  by  a - b = 1.424 - 0.999 = 0.425 kilometers = 425 meters, approximately.
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Solved.