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4 February, 08:35

Solve triangle ABC, when A = 6, B=10 and c=12

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  1. 4 February, 09:05
    0
    Use cosine rule,

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

    = (10^2+12^2-6^2) / (2*10*12)

    =13/15

    A=29.926 degrees ... (A)

    cos (B) = (c^2+a^2-b^2) / (2ca)

    = (12^2+6^2-10^2) / (2*12*6)

    =5/9

    B=56.251 degrees ... (B)

    cos (C) = (a^2+b^2-c^2) / (2ab)

    = (6^2+10^2-12^2) / (2*6*10)

    =-1/15

    C=93.823 degrees ... (C)

    Check:29.926+56.251+93.823=180.0 degrees ... ok
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