# A convex quadrilateral has an area of 30 and side lengths 5, 6, 9, and 7, in that order. Denote by `theta` the measure of the acute angle formed by the diagonals of the quadrilateral. Then tan `theta`can be written in the form m/n, where m and n are relatively prime positive integers. Find m + n.

Following the outlined steps, m + n = 47.

Denote the vertices as `A , B , C , D ` in the given order and denote the point of diagonals intersection as `O . ` Also, denote the parts of diagonals adjacent to vertices as `a , b, c , d , ` respectively. Suppose the acute angle is the angle `BOC ; ` if not, its cosine will appear negative and we'll correct.

The area can be expressed as

`1 / 2 ( ab + bc + cd + da ) sin theta = 30 .`

The side lengths (their squares) can be expressed using the cosine law:

`a^2 + b^2 + 2ab cos theta = 25 , `
`b^2 + c^2 - 2bc cos theta = 36 ,`
`c^2 + d^2 + 2cd cos theta = 81 ,`
`d^2 + a^2 - 2da cos theta = 49`

(We used the facts that the cosine of a complementing angle is `-cos theta ` while the sine is the same).

Now add the first and the third equations and subtract from them the second and the fourth:

`a^2 + b^2 - b^2 - c^2 + c^2 + d^2 - d^2 - a^2 +`
`+ 2 ( ab + bc + cd + da ) cos theta = 25 - 36 + 81 - 49 ,`

so, ` 2 ( ab + bc + cd + da ) cos theta = 21 .`

Now we can easily find `tan theta :`

`tan theta = ( sin theta ) / ( cos theta ) = ( 30 * 2 ) / ( 21 / 2 ) = 120 / 21 = 40 / 7 .`

Finally, the answer is 40 + 7 = 47.

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