Can We find the area of a Quadrilateral whose four angles and two opposite sides are given ?

Well, for squares and rhombi ,yes.

For parallelograms, since we can find the area of triangle when three angles and a side is given, yes,

But for an irregular quadrilateral , I know the Bretschneider formula exists but I don't know if any formula for my query exists.

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    $\begingroup$ Not for a rectangle. You're given the four 90° angles and two opposite sides which are the width of the rectangle. Now how long is the rectangle? Uh-oh. $\endgroup$ – Oscar Lanzi Sep 20 at 11:52
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    $\begingroup$ Four angles and two opposite sites is generically sufficient for a ruler and compass construction so the quadrilateral is well defined. A rectangle is an example of a non generic situation where the area is not well defined. I'm not sure whether one can get a nice closed formula for the general case. $\endgroup$ – quarague Sep 20 at 11:53
  • $\begingroup$ @OscarLanzi but we can find area of triangle when three angles and a side is given.Does diagonal of rectangle does not bisect the angles ? $\endgroup$ – George carlin Sep 20 at 12:01
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    $\begingroup$ @Georgecarlin obviously not: take a rectangle of side lengths 3 and 4; the hypotenuse is then length 5 and so the interior angles of the triangle cannot be 45, 45 and 90 degrees. The interior non-right angles will sum to 90 degrees but they don't have to be equal $\endgroup$ – postmortes Sep 20 at 12:06
  • $\begingroup$ The area will be unique (though there may be a second spurious solution) unless the pairs of angles on the known sides are supplementary in the sense that the resulting shape would be a trapezium/trapezoid with the unknown sides parallel (including a parallelogram or rectangle). So if not then a formula might possible, though I suspect it is likely to be messy $\endgroup$ – Henry Sep 20 at 12:50

I will try to explain this without pictures, so lets see how this goes.

Let $\alpha, \beta, \gamma, \delta$ denote the $4$ angles at the corners $A,B,C,D$. We also know $\overline{AB}$ and $\overline{CD}$. If $\alpha+\beta=\pi$ the quadrilateral is not uniquely defined, so what follows assumes that the angles at the endpoints of a known side are not supplementary.

As the sum of the $4$ angles is $2\pi$ assume without loss of generality that $\alpha+\beta > \pi$. Draw the quadrilateral $ABCD$ and then extend the two side $BC$ and $AD$ until they meet in the point $F$. Then the area of $ABCD$ is the difference of the area of the triangle $FCD$ minus the area of the triangle $FBA$. For both triangles we know one side and the two adjacent angles which is enough to uniquely define them and compute their area.

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    $\begingroup$ $\alpha+\beta>\pi$ is not true of rectangles or more generally parallelogram. Even trapezoids can make trouble depending on which sides and angles are given. You need to identify these exceptions for which the problem has not a unique solution. See the comments to the question. $\endgroup$ – Oscar Lanzi Sep 20 at 12:34
  • $\begingroup$ @OscarLanzi All the exception that you describe are covered in the case $\alpha+\beta=\pi$ which I address by saying the quadrilateral is not uniquely defined in this case. $\endgroup$ – quarague Sep 20 at 12:41
  • $\begingroup$ Modified the paragraph structure to clarify this point. The method itself is the way I would solve it, if the assumption of nonsupplimentary angles holds. $\endgroup$ – Oscar Lanzi Sep 20 at 12:51

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