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