Radio direction finders asre set up at points A and B, 8.68 mi apart on an east-west line. From A it is found that the bearing of a signal from a transmitter is N 54.3degE, while from B it is N 35.7degW. Find the distance of the transmitter from B, to the nearest hundredth of a mile.
i got 7.05mi

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  1. If you draw this, you have a triangle with ASA. With those two angles, you can get the third angleC.

    Use the law of sines:
    solve for side a.

    I agree with your answer.

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  2. My diagram has triangle ABC with the interior angle at B equal to 54.3 and the interior angle at A equal to 35.7

    (the bearings as given would not be inside the triangle as given)
    BTW, angle C = 90

    Your answer is what I got for the distance to A,
    the distance to B would be 5.064

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  3. k thnks

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