Question 1042781
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Since the actual size of the rectangle is arbitrary, we can use the fact that the tangent of x is 3/4 to assign the value 4 to the measure of the side of the rectangle that is adjacent to angle x, and the value 3 to the measure of the side opposite angle x.  The other long side of the rectangle must also be 4, and since 4 is 2/3 of 6, assign 2/3 (2/3 times 1) to the segment that is opposite angle y.


Then, relatively speaking, the area of the entire rectangle is 12 square units, the area of the triangle containing angle x is 6 square units, and the area of the triangle containing angle y is 4/3 square units.  The sum of the areas of the two triangles is then 22/3 and 22/3 divided by 12 is 11/18.


Then the shaded area is 1 minus 11/18.  7/18


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

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