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Math Help - Cant do this question, someone please help!

  1. #1
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    Cant do this question, someone please help!

    The three points A, B, and C, with coordinates A(0,1), B(3,4), and C(1, 3) respectively, are joined to form a triangle:

    a) Show that COS ACB= -4/5

    b) Calculate the area of triangle ABC.



    Ive done this question and have come out with the answer a) 8.66 but apparently its 5. ??
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  2. #2
    MHF Contributor
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    Here is one way.

    Since the coordinates of the corners are given, then we can get the lengthjs of the 3 sides
    distance = sqrt[(x2 -x1)^2 +(y2 -y1)^2]

    Then we can get the cos (angle ACB) by using the Law of Cosines.
    (AB)^2 = (AC)^2 +(BC)^2 -2(AC)(BC)*cos(angle ACB)
    cos(angle ACB) = [(AC)^2 +(BC)^2 -(AB)^2] / [2(AC)(BC)] ---(1)

    Then we can get the area of triangle ABC by using the Heron's Formula.
    A = sqrt[s(s-a)(s-b)(s-c)]
    Or, by using the formula, A = (1/2)bc*sinA

    Those above need long computations. I don't need to show those computations here, so solve them yourself.

    I solved them on paper here and I got:
    AB = sqrt(18)
    BC = sqrt(5)
    AC = sqrt(5)

    Substituting those in (1), I got cos(angle ACB) = -8/10 = -4/5.

    Then using the formula A = (1/2)bc*sinA for the area of triangle A,
    where
    angle "A" = angle ACB
    "b" = AC = sqrt(5)
    "c" = BC = sqrt(5)

    cos(A) = -4/5
    Square both sides,
    cos^2(A) = 16/25
    So,
    1 -sin^2(A) = 16/25
    sin^2(A) = 1 -16/25 = 9/25
    sinA = sqrt(9/25) = 3/5

    So, area of triangle ABC
    = (1/2)*sqrt(5)*sqrt(5)*(3/5)
    = (1/2)(5)(3/5)
    = 3/2 or 1.5 sq. units ----------answer.
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  3. #3
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    Sep 2005
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    Unhappy eep my bad

    haha i just went to write it down and do the working myself then i realised i have actually done this. i typed out the wrong question godammit


    the question i meant to type was:




    In triangle ABC, AB= 10cm, BC= a√3 cm, AC=5√13 cm and angle ABC=150 degrees.
    Calculate:

    a) the value of a

    b) the exact area of triangle ABC







    this is the question that i got 8.66 for, but the textbook answer says it is 5.

    howww?
    thankss

    xxx
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  4. #4
    MHF Contributor
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    I don't know, but this looked like a trap. Not so good. But I'd give you the benefit of the doubt. Let us say you really made a mistake.

    --------------
    Same thing:
    >>>Law of Cosines.
    >>> Area = (1/2)ab*sinA

    [5sqrt(13)]^2 = (10)^2 +[a*sqrt(3)]^2 -2(10)[a*sqrt(3)]*cos(150deg)
    25(13) = 100 +3a^2 +30a
    325 = 3a^2 +30a +100
    0 = 3a^2 +30a +100 -325
    0 = 3a^2 +30a -225
    Divide both sides by 3,
    0 = a^2 +10a -75
    a^2 +10a -75 = 0
    (a+15)(a-5) = 0
    a = -15 or 5 ------------***
    Since there are no negative dimensions, reject a = -15.
    Therefore, a = 5 ----------answer.

    -------------------
    Area = (1/2)(10)[a*sqrt(3)]*sin(150deg)
    Area = (1/2)(10)[5sqrt(3)]*sin(150deg)
    Area = (1/2)(10)[5sqrt(3)](0.5)
    Area = (12.5)*sqrt(3) sq.units -------answer.
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