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Math Help - Law of Sines

  1. #1
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    Law of Sines

    There is a trig application problem I have been working on over and over and arriving at the same answer, even when approaching the problem different ways. My answer is \approx 126 but the answer in the back of my book is \approx 130. Could you guys try to solve it?

    Distance Across a Canyon
    To find the distance across a canyon, a surveying team locates points A and B on one side of the canyon and point C on the other side of the canyon. The distance between point A and B is 85 yards. The measure of angle CAB is 68 degrees and the measure of angle CBA is 75 degrees. Find the distance across the canyon.

    Thanks
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  2. #2
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    Quote Originally Posted by c_323_h
    There is a trig application problem I have been working on over and over and arriving at the same answer, even when approaching the problem different ways. My answer is \approx 126 but the answer in the back of my book is \approx 130. Could you guys try to solve it?

    Distance Across a Canyon
    To find the distance across a canyon, a surveying team locates points A and B on one side of the canyon and point C on the other side of the canyon. The distance between point A and B is 85 yards. The measure of angle CAB is 68 degrees and the measure of angle CBA is 75 degrees. Find the distance across the canyon.

    Thanks
    If the problem is correct as posted, the the answer given by the book is wrong. Books, like people, are not perfect.

    Although the title of your posting here is Law of Sines, the solution of the problem will not even use any sine function. The Law of Sines will give only the other two sides of the triangle, which are not the distance across the canyon. This distance accross the canyon is the distance from corner C to side AB, such that this said distance is perpendicular to AB. Or, this distance is the altitude of the triangle if AB were the base.

    Let d = this distance, in yards.
    And point D be the intersection of d and AB.

    In right triangle ADC,
    hypotenuse = AC
    one leg = d
    the other leg is unknown, = say, x.
    angle opposite d is 68deg

    tan(68deg) = d/x
    x = d/tan(68deg) --------(i)

    In right triangle BDC,
    hypotenuse = BC
    one leg = d also
    the other leg is (85-x).
    angle opposite d is 75deg

    tan(75deg) = d/(85-x)
    85-x = d/tan(75deg) --------(ii)

    (i) plus (ii),
    85 = d/tan(68deg) +d/tan(75deg)
    85 = d[1/tan(68deg) +1/tan(75deg)]
    d = 85/[1/tan(68deg) +1/tan(75deg)]
    d = 85/[0.671975418]
    d = 126.49 yards ----------------answer.
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  3. #3
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    i actually used the law of sines and got the same answer as you, approximately 126. but the answer in the back of my trig textbook says 130. i guess the book must be wrong. thanks
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  4. #4
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    Quote Originally Posted by c_323_h
    i actually used the law of sines and got the same answer as you, approximately 126. but the answer in the back of my trig textbook says 130. i guess the book must be wrong. thanks
    It could merely be rounding...all of the "data" in the problem are given to two significant digits, so the answer should be given that way as well. I don't usually see significant digits used in a Math problem, but it would explain it.

    -Dan
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  5. #5
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    Hi:

    Let h denote the height of triangle ABC extending from point C to side AB at P (i.e., h is the perpendicular distance across the canyon). Note that sin(75)=h/(BC). Therefore, h=(BC)sin75 (eq#1). Now, by L.of.Sines, BC/sin(68) = 85/sin(37). Thus BC=85sin(68)/sin(37) (eq#2). Substitutng the right member of eq#2 into eq#1 for BC gives,

    h = 85*sin68*sin75 / sin(37) = 126yd, 1ft, 5_ 3/4 in.

    Regards,

    Rich B.
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