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Posted by on Saturday, July 23, 2011 at 3:59am.

Trigonometry problem:

The sides AB and AC of triangle ABC are of lengths 4 and 7 respectively. M is the midpoint of BC. AM is of length 3.5, Find the length of BC.

  • Math - , Saturday, July 23, 2011 at 7:12am

    Complete the parallelogram ABPC , where BP=7 and PC=4
    In a ||gram , diagonals bisect each other, so M is the midpoint of both diagonals and AP = 2(3.5) = 7

    (looks like we have isosceles triangles, but that doesn't matter)

    let angle PAC =Ø
    let MC = a , thus BC = 2a for later

    In triangle APC, by cosine law
    4^2 = 7^2 + 7^2 - 2(7)(7)cosØ
    cosØ = 82/98 = 41/49

    in triangle AMC
    a^2 = 3.5^2 + 7^2 - 2(3.5)(7)cosØ
    = 61.25 - 49(41/49) = 20.25
    a=√20.25 = 4.5

    So BC = 2a = 2(4.5) = 9

    better check my arithmetic, I have made some silly errors lately.

  • Math - , Saturday, July 23, 2011 at 9:22am

    Thank You!

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