Question 969331
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The diagonals of a rhombus are always perpendicular bisectors of each other and always bisect the vertex angles of the rhombus.  Since adjacent angles of any parallelogram are supplementary, and we are given one of the angles as 50 degrees, the other angle has to be 180 minus 50 or 130 degrees.


Since the diagonals are perpendicular, by constructing them you have created four right triangles.  In this case, each of your right triangles has a hypotenuse of 32 and acute angles of 25 degrees and 65 degrees.  The measure of the hypotenuse multiplied times the sine of 25 degrees which is the angle of the triangle opposite the short leg, gives you the measure of the short leg.


Since the diagonals are bisectors, the short leg of one of the right triangles is exactly half of the measure of the short diagonal.


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 2(32\,\sin\,25^\circ)\ =\ 64\,\sin\,25^\circ]


John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
My calculator said it, I believe it, that settles it

*[tex \Large \ \
*[tex \LARGE \ \ \ \ \ \ \ \ \ \