What are the angles of the right-angled triangle, whose area is 1/(2*pi) of said triangle's circumscribing circle's area?

As the ratio is between two areas, scale doesn't matter so let's make the radius of the circle equal 1.

The circle's area is then pi, and we want the triangle to have area 1/2. It's hypotenuse is the diameter of the circle, which is 2.

Call the legs of the triangle x and y.

Then:

xy=1 so that the area is 1/2

x^2 + y^2 = 2^2 as it's a right triangle

then:

x^2 + 1/x^2 = 4

x^2 - 4 + 1/x^2 = 0

x^4 - 4x^2 + 1 = 0

x^2 = (4 +/- sqrt(16-4)) / 2

    = 2 +/- sqrt(3)
    
x = sqrt(2 +/- sqrt(3))

One of the evaluations is 1.93185165257814 and the other is .517638090205039. As suspected, one can serve as x and the other as y, as it doesn't matter if the legs are interchanged.

The arctan of the larger divided by the smaller is 75°, and of course of the smaller divided by the larger is 15°. Of course, to be complete, the other angle is 90°.

Posted by Charlie on 2012-07-09 12:17:43