Hint:jschwa1 said:I'm trying to find the ratio of the area of this right triangle to the area of the semicircle. The solution is supposed to be .409, but I can't figure out how to get it.
I'm trying to find the ratio of the area of this right triangle to the area of the semicircle.
The solution is supposed to be 0.409.
Denis said:[can] you just let the diameter = 1 and go from there? Boring!
soroban said:\(\displaystyle \sec33^o \:=\:\frac{AD}{a\sec33^o}\)
Yikes! Is this a bombshell?
Subhotosh Khan said:But why a$$-u-me something - that we really don't need to?
mmm4444bot said:I asked the guy at Beth's Cafe for some scratch paper; Subhotosh's hint haunted my hamburger.
I thought that I had an insight, but I resorted to my usual 24-step program of analytical geometry, to solve the exercise, instead.
So, the "hint" that I come up with is that the base of the triangle is: 2r cos(33°)^2, where r is the radius.
Wow. I'd like to see Subhotosh's work. I can't believe that's a hint! :wink:
mmm4444bot said:Subhotosh, I might have misworded my request.
I'm interested in seeing how you arrived at: hypotenuse = D cos(33°).
My hint for the base expression took too much effort, to be called a hint.
(I'm concerned that I'm missing something obvious.)
In the figure, if you create another triangle that is inscribed in the circle by drawing a line segment from the current triangle's top vertex to the point on the right where the diameter intersects the circle, you create another right triangle and can use Subhotosh's hint.
Subhotosh Khan said:jschwa1 explained it above -