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Edexcel C2-January 2008,question 5 watch

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    In triangle ABC , AB=6cm, BC= 13cm and CA=9cm

    a) find the value of cos BAC as a fraction in its lowest terms.

    Calculations:

    a^2=b^2+c^2- 2bc Cos A

    13^2=9^2+6^2- 2(9)(6)cosA

    169=81+36 - 108 cos A

    Cos A = -52/108 = -13/27


    b) Show that the area of a triangle ABC is 4√35cm2


    Please can anybody show me the whole procedure how to do it?

    I know i`ve got to use a formula for area of triangle which is

    Area=1/2 bc sinA

    But I have no idea how to get the result and whole calculation in a surd form.

    Thanks. x
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    okey well you know that cosA = 52/108

    so draw a right angled triangle, and fill in the sides (you know that a/h = 52/108, so a = 52 and h = 108). from this you can find the opposite side: 16 root 35.

    then you can see that sinA = o/h = (16 root 35) / 108

    and simply put this back into the A = 0.5 b c sinA = 0.5 x 9 x 6 x (16 root 35)/108 = 4 root 35.

    Edit: just noticed i didn't use the cancelled down fraction, don't know why but yeah it works the same anyway.
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    (Original post by BrightGirl)
    okey well you know that cosA = 52/108

    so draw a right angled triangle, and fill in the sides (you know that a/h = 52/108, so a = 52 and h = 108). from this you can find the opposite side: 16 root 35.

    then you can see that sinA = o/h = (16 root 35) / 108

    and simply put this back into the A = 0.5 b c sinA = 0.5 x 9 x 6 x (16 root 35)/108 = 4 root 35.

    Edit: just noticed i didn't use the cancelled down fraction, don't know why but yeah it works the same anyway.

    Thank you so much, you have been so helpful.:yes:
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    (Original post by Janka3112)
    Thank you so much, you have been so helpful.:yes:
    s'ok
 
 
 
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