Learn the Arc Length of a Circle Formula with Plainmath's Expert Help

Find the degree measure of an arc whose measure is 7pi/12 radian

What is the area of a sector with a central angle of$185$and a diameter of $6.4$?

The angles in a linear pair are $\_\_\_\_\_\_\_$

The angle between the minute and hour hands of a clock at 8:30 is
1) ${80}^{\circ <!--\; \circ \; -->}$
2) ${75}^{\circ <!--\; \circ \; -->}$
3) ${60}^{\circ <!--\; \circ \; -->}$
4) ${105}^{\circ <!--\; \circ \; -->}$

What is the equation of a circle with center (2,-1) that passes through the point (3,4)?

The midpoint of a chord of length $2a$ is at a distance d from the midpoint of the minor arc it cuts out from the circle. Show that the diameter of the circle is $\frac{a^2+d^2}{d}$ .
I know I have to find similar triangles, I cannot see them...

I need to find the locus of points (on an Argand diagram) such that:
(i) $\arg <!--\; \; -->(z-<!--\; -\; -->(-<!--\; -\; -->1-<!--\; -\; -->4i))+\arg <!--\; \; -->(z-<!--\; -\; -->(5+8i))=0$
(ii) $\arg <!--\; \; -->(z-<!--\; -\; -->(-<!--\; -\; -->1-<!--\; -\; -->4i))+\arg <!--\; \; -->(z-<!--\; -\; -->(5+8i))=\frac{\mathrm{\pi <!--\; \pi \; -->}}{2}$
I could not see a way to solve these problems other than plotting arbitrary points and trying to observe a general pattern.
I am aware that $\arg <!--\; \; -->(z-<!--\; -\; -->(-<!--\; -\; -->1-<!--\; -\; -->4i))-<!--\; -\; -->\arg <!--\; \; -->(z-<!--\; -\; -->(5+8i))=\frac{\mathrm{\pi <!--\; \pi \; -->}}{2}$ is a semicircle, and for other angles, say $\frac{\mathrm{\pi <!--\; \pi \; -->}}{3}$ or $\frac{\mathrm{\pi <!--\; \pi \; -->}}{4}$, part of the arc of a circle, but this is only because I knew that this equation represents the locus of points that made a certain angle between the two complex numbers. I was unable to find a similar representation, however, for (i) and (ii).
I am also interested in whether problems (i) and (ii) can be generalised to any angle between 0 to $\mathrm{\pi <!--\; \pi \; -->}$. Any help here would be greatly appreciated.

So I have to evaluate
${\int <!--\; \int \; -->}_{-<!--\; -\; -->\mathrm{\infty <!--\; \infty \; -->}}^{\mathrm{\infty <!--\; \infty \; -->}}\frac{e^{ax}}{\cosh <!--\; \; -->x}dx$
I tried takeing the analytic expansion, and integrating over the real axis. I took this as being a half circle from $-<!--\; -\; -->\mathrm{\infty <!--\; \infty \; -->}$ to $\mathrm{\infty <!--\; \infty \; -->}$, minus the arc of the circle. over the arc I proved that the integral is zero, and I have left only with a sum of the residues of the function. With the help of a previous question I calculated the values of the residues, but I could not manage to converge the sum.

The vertex of angle $\mathrm{\angle <!--\; \angle \; -->}BAC$ lies inside of a circle. Prove that the value of angle $\mathrm{\angle <!--\; \angle \; -->}BAC$ is equal to half the sum of angle measures of the arcs of the circle confined inside angle itself and inside the angle symmetric to it through vertex $A$.
So I don't understand how to prove this. I've already drawn a diagram but I can't figure out how to prove this. Please help! Thank you!

Is it true that if a holomorphic function in the unit disk converges uniformly to the 0 function some connected arc of the unit circle, this function is globally null?
If that is true, this would stand for a uniqueness fact and so how to recover some holomorphic function from its boundary value on an arc (if the uniform convergence still holds on the arc)

There's a circle of diameter $d$ that is on a wall, and touches a block. Find the value of $d$.
I'm really unsure how to go about solving this. I wanted to first approach this by using the arc length, but I'm really unsure how to proceed. Anyone have ideas?

Would the sum not change for different quadrilaterals like a triangle with sides that are arcs of great circles?

Two pairs of points are randomly chosen on a circle. Find the probability that the line joining the two points in one pair intersects that in the other pair.
I've been thinking over this problem, assuming one pair and finding that the other pair has to be entirely in one of the two arcs of the circle that the first pair of points divides it into.
But I've not been able to find an explicit answer. If I assume one point to be (a,b), I'm not able to manage the other three points.

The arc and lines form the sides of the shape. The sides touch each other at end point in such a way that each end point can touch only one shape and the shape is closed
eg:- line--arc--line--line--line--arc--arc

Given the coordinates of a single point on a circle and a length of an arc $L$, how do I find the coordinates of another point?

Or, to put in another form: I have the radius $r$, the length of the arc $L$ and $(x_1,y_1)$ the coordinates. I need to express $(x_2,y_2)$ using only $r,L,x_1$, and $y_1$.

I'm at a dead end on this.

Given a circle with radius 1. Take an arbitrary starting point. Then go around the circle an infinite number of times, always drawing a point when you are one unit further. So the next point is always an arc length 1 step further. Easy to prove that all points will be different! But my question is: is each point randomly close approached? I believe this is true, but I can’t prove it!

Let C be an arbitrary arc of the unit circle. Give a geometrical interpretation for$\int <!--\; \int \; -->C\mathrm{▽<!--\; ▽\; -->}f\cdot <!--\; \cdot \; -->dr=0$ where $f(x,y)=x^2+y^2.$

Perhaps a rather elementary question, but I simply couldn't figure out the calculations on this one. Say one takes a circle centeblack at the origin with radius $R$. He or she then proceeds to place $N$ circles with radius $r$ ($R>r$) on the larger circles circumference equidistantly, so every $2\mathrm{\pi <!--\; \pi \; -->}/N$ in the angular sense. What is then the relationship between $R$ and $r$ such that all neighboring circles exactly touch?
I've been trying to write down some equations with arc lengths and such for $N=4$, but I can't seem to get anything sensible out of it.

So I know the length L of the curve $y=\sqrt{R^2-<!--\; -\; -->x^2}$ from $x=0$ to $x=a$ where $|a|<R$ is given by:
$L={\int <!--\; \int \; -->}_0^a\frac{R}{\sqrt{R^2-<!--\; -\; -->x^2}}dx$
Now I must set up the arc length integral and simplify it so that it is in the form listed above.
$L={\int <!--\; \int \; -->}_0^a\sqrt{1+{\left(\frac{dy}{dx}\right)}^2}dx$
and
$\frac{dy}{dx}=-<!--\; -\; -->\frac{x}{\sqrt{R^2-<!--\; -\; -->x^2}}$
${\left(\frac{dy}{dx}\right)}^2=\frac{x^2}{R^2-<!--\; -\; -->x^2}$
so
$L={\int <!--\; \int \; -->}_0^a\sqrt{1+\frac{x^2}{R^2-<!--\; -\; -->x^2}}dx$
I am unsure where to go from here to simplify into the first integral, any help would be greatly appreciated. Thanks

I am trying to compute this integral ${\int <!--\; \int \; -->}_0^{\mathrm{\infty <!--\; \infty \; -->}}\frac{x}{1+x^6}dx$ using the residue theorem. To do so, I am integrating $f(z)=\frac{z}{1+z^6}$ in the frontier of this sector of a circle: $\{z:|z|<R,0<arg(z)<\mathrm{\pi <!--\; \pi \; -->}/3\}$.

I kwo how to deal with the integrals over the horizontal segment of the sector and over the arc of the circle. My problem is the "diagonal segment". When I parametrize it, I do not get something easy to integrate. How could I approach this?

(The path of integration was suggested in my book, so I do not think that is the problem).