Trigonometry - Geometry - side lengths

In convex quadrilateral ABCD, angle A is congruent to angle C, AB=CD=180, and AD is not equal to BC. The perimeter of ABCD is 640. Find cos A.

Hmm...I drew the diagram, and labeled some things, I drew an altitude from D to AB and did some other things, but no success was obtained. Any help? Thanks

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  1. joint BD, and use the cosine law twice, once for each triangle
    we know AB = DC = 180
    let angle A be Ø
    let AD = a and BC = b

    BD^2 = a^2 + 180^2 - 2(180)a cosØ
    BD^2 = b^2 + 180^2 - 2(180)bcosØ

    a^2 + 180^2 - 360a cosØ = b^2 + 180^2 - 360b cosØ
    a^2 - b^2 = 360a cosØ - 360b cosØ
    (a+b)(a-b) = 360cosØ (a-b)
    a+b = 360cosØ
    cosØ = (a+b)/360

    but perimeter = 640
    a+b+180+180=640
    a+b = 280

    then cosØ = 280/360= 7/9

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  2. ohhhhh that makes sense i should have thought of that thanks!

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