- #1
Studiot said:What exactly are the two right angle symbols doing at the ends of the two lower radii?
Studiot said:Thank you, I am aware of circle geometry, but it does not answer my question since no tangents appear in the diagram.
zgozvrm said:The right angle indicators are not needed for this problem.
As I already stated, I believe they are there just to reinforce the fact that those are indeed radii of the arc.
Besides, tangents are not necessary; a line (or line segment) is perpendicular to a curve if it is perpendicular to the tangent at the point of intersection.
SammyS said:
I know the original post is 4 weeks old, but here's my two cents worth.
The distance, d, from the point where the vertical line intersects the circle, to the point on the left (or on the right, if you prefer) where the circle intersects the horizontal line is:
[tex]\textstyle d=\sqrt{({{L}\over{2}})^2+h^2}[/tex].
Bisect the angle [tex]\textstyle \beta[/tex].
[tex]
\displaystyle \sin\left({{\beta}\over{2}}\right)={{d/2}\over{R}}={{\sqrt{({{L}\over{2}})^2+h^2}}\over{2R}}={{\sqrt{L^2+4h^2}}\over{4R}}
[/tex]
Using the double angle formula:
[tex]\textstyle \cos\beta=1-2\sin^2\left({{\beta}\over{2}}\right)[/tex]
[tex]
\displaystyle = 1-2\left[\,{{\sqrt{L^2+4h^2}}\over{4R}}\ \right]^2
[/tex]
Substitute [tex] {{L^2+4h^2}\over{8h}}[/tex] for R and simplify. It does work out.
Andrea2 said:the lengt of the cord L is given by 2Rsin(b), so we compute sin(b)=(8Lh)/(L^2+4h^2). So cos(b)=sqrt(1-sin^2(b)), but the result i obtain in this way is that the sqrt of the denominator is correct, but the numerator is different because i obtain sqrt(L^4+4h^4-56L^2h^2) that is different by (L^2-4h^2)^2...i think there is a mistake in the text of the problem, but I'm not sure of my results...
Andrea2 said:But why my in my first post i didn't found the correct answer?
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