circle geometry

[vytautasr]

Could someone please help me with this??

[Andrea2]

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...

[Studiot]

I think this diagram is missing something.

What exactly are the two right angle symbols doing at the ends of the two lower radii?

[zgozvrm]

Simple:

Angle [itex]\beta[/tex] is part of a right triangle.
The adjacent side is equal to R-h
The hypotenuse is equal to R

Therefore, [itex]\cos \theta = \frac{R-h}{R}[/tex]

Substitute the value for R that you were given...

[zgozvrm]

What exactly are the two right angle symbols doing at the ends of the two lower radii?

It looks a bit unorthodox, but every radius of a circle is perpendicular to the circle at the point where it intersects. More clearly, it is perpendicular to the tangent of the circle at that point.

As far as I can tell, they are there simply to reinforce the fact that the two lower lines are, in fact, radii of the circle (arc)

[Studiot]

Thank you, I am aware of circle geometry, but it does not answer my question since no tangents appear in the diagram.

[zgozvrm]

Thank you, I am aware of circle geometry, but it does not answer my question since no tangents appear in the diagram.

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]

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.

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:

\textstyle d=\sqrt{({{L}\over{2}})^2+h^2}.

Bisect the angle \textstyle \beta.


\displaystyle \sin\left({{\beta}\over{2}}\right)={{d/2}\over{R}}={{\sqrt{({{L}\over{2}})^2+h^2}}\over{2 R}}={{\sqrt{L^2+4h^2}}\over{4R}}


Using the double angle formula:
\textstyle \cos\beta=1-2\sin^2\left({{\beta}\over{2}}\right)


\displaystyle = 1-2\left[\,{{\sqrt{L^2+4h^2}}\over{4R}}\ \right]^2


Substitute {{L^2+4h^2}\over{8h}} for R and simplify. It does work out.

[zgozvrm]

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:

\textstyle d=\sqrt{({{L}\over{2}})^2+h^2}.

Bisect the angle \textstyle \beta.


\displaystyle \sin\left({{\beta}\over{2}}\right)={{d/2}\over{R}}={{\sqrt{({{L}\over{2}})^2+h^2}}\over{2 R}}={{\sqrt{L^2+4h^2}}\over{4R}}


Using the double angle formula:
\textstyle \cos\beta=1-2\sin^2\left({{\beta}\over{2}}\right)


\displaystyle = 1-2\left[\,{{\sqrt{L^2+4h^2}}\over{4R}}\ \right]^2


Substitute {{L^2+4h^2}\over{8h}} for R and simplify. It does work out.




Correct. But the way I read it, you are given the fact that

R = \frac{L^2 +4h^2}{8h}

so, we know that the vertical line from the bottom of the diagram (the center of the circle) to the horizontal line segment is equal to (R - h) and, since the horizontal line segment is of length L (between the points where it intersects the circle). That gives us:

\cos(\beta) = \frac{R - h}{R}

Substitute for R and simplify:

\frac{R - h}{R} = \frac{\frac{L^2 + 4h^2}{8h} - h}{\frac{L^2 + 4h^2}{8h}} = \frac{\frac{L^2 + 4h^2}{8h} - \frac{8h^2}{8h}}{\frac{L^2 + 4h^2}{8h}} = \frac{\frac{L^2 + 4h^2 - 8h^2}{8h}}{\frac{L^2 + 4h^2}{8h}} = \frac{L^2 + 4h^2 - 8h^2}{L^2 + 4h^2} = \frac{L^2 - 4h^2}{L^2 + 4h^2}


Note that we are both assuming that [itex]\angle \beta[/tex], between the vertical line and the radius on the left, is the same as the angle between the vertical line and the radius on the right (or that the vertical line is perpendicular to the horizontal line). Neither of these are shown on the diagram.

[Andrea2]

But why my in my first post i didn't found the correct answer?

[zgozvrm]

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...

But why my in my first post i didn't found the correct answer?

You started out with L = 2R \sin(\beta)[/tex] and were given

R = \frac{L^2 + 4h^2}{8h}

but, when you simplified your equation, you forgot the "2"; you [i]should have come up with

\sin(\beta) = \frac{L}{2R} = \frac{4Lh}{L^2 + 4h^2}