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17 March, 04:29

In a triangle ABC, AB = 7, AC = 8, BC = 3, what is cos (A) ?

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  1. 17 March, 06:50
    0
    Use the cosine rule.

    c=AB=7

    b=AC=8

    a=BC=3

    a^2=b^2+c^2-2bc (cos (A))

    =>

    cos (A) = (b^2+c^2-a^2) / (2bc)

    = (7^2+8^2-3^2) / (2*7*8)

    =13/14

    =0.9286
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