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Math Help - Trig. Problem

  1. #1
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    Exclamation Trig. Problem

    The triangle ABC has sides of length, a,b and c, as shown in the diagram. The point D lies on AB and CD is perpendicular to AB.
    a) Show that asinB = bsinA.
    b) Show that c = acosB + bcosA.
    c) Given that c^2 = 4ab cosA cos B, show that a=b.

    Could someone please show me how to do (c) of the question?

    I've tried equating (b) squared to (c) but I end up with asinB = b sinA. How do I use this to prove a=b?
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  2. #2
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    Quote Originally Posted by xwrathbringerx View Post
    The triangle ABC has sides of length, a,b and c, as shown in the diagram. The point D lies on AB and CD is perpendicular to AB.
    a) Show that asinB = bsinA.
    b) Show that c = acosB + bcosA.
    c) Given that c^2 = 4ab cosA cos B, show that a=b.

    Could someone please show me how to do (c) of the question?

    I've tried equating (b) squared to (c) but I end up with asinB = b sinA. How do I use this to prove a=b?
    (a\cos{B}+b\cos{A})^2 = a^2\cos^2{B} + 2ab\cos{A}\cos{B} + b^2\cos^2{A}


    a^2\cos^2{B} + 2ab\cos{A}\cos{B} + b^2\cos^2{A} = 4ab\cos{A}\cos{B}

    a^2\cos^2{B} - 2ab\cos{A}\cos{B} + b^2\cos^2{A} = 0

    (a\cos{B} - b\cos{A})^2 = 0

    a\cos{B} = b\cos{A}

    DB = AD

    what does that say about \Delta ACB ?
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