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

Freddy Chaney
2022-09-26
Answered

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Matthias Calhoun

Answered 2022-09-27
Author has **11** answers

Yes it would change.

It will be greater than ${360}^{o}$, but it can vary.

The maximum possible is ${1080}^{o}$

It will be greater than ${360}^{o}$, but it can vary.

The maximum possible is ${1080}^{o}$

asked 2022-07-13

I'm not great at mathematics so I'm sure this is trivial to most. I have been searching around however and not been able to find how to figure out the incircle of a circle sector, or, in other words, the point inside a sector that is furthest away from the radii and the arc. The simpler solution the better

asked 2022-09-05

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.

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.

asked 2022-08-14

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:

$L={\int}_{0}^{a}\frac{R}{\sqrt{{R}^{2}-{x}^{2}}}dx$

Now I must set up the arc length integral and simplify it so that it is in the form listed above.

$L={\int}_{0}^{a}\sqrt{1+{\left(\frac{dy}{dx}\right)}^{2}}dx$

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

$L={\int}_{0}^{a}\sqrt{1+\frac{{x}^{2}}{{R}^{2}-{x}^{2}}}dx$

I am unsure where to go from here to simplify into the first integral, any help would be greatly appreciated. Thanks

$L={\int}_{0}^{a}\frac{R}{\sqrt{{R}^{2}-{x}^{2}}}dx$

Now I must set up the arc length integral and simplify it so that it is in the form listed above.

$L={\int}_{0}^{a}\sqrt{1+{\left(\frac{dy}{dx}\right)}^{2}}dx$

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

$L={\int}_{0}^{a}\sqrt{1+\frac{{x}^{2}}{{R}^{2}-{x}^{2}}}dx$

I am unsure where to go from here to simplify into the first integral, any help would be greatly appreciated. Thanks

asked 2022-06-22

$arg(\frac{z-a}{z-b})=c$

My understanding is as follows. The angle c between the lines za and zb is constant. za and zb meet at z and the angle between the lines perpendicular to za and zb is 2c, which is only the case if z lies on a circle passing through z, a and b with centre where the lines perpendicular to za and zb intersect. So the locus is an arc of a circle through a and b except a and b (because $arg(\frac{z-a}{z-b})$ is undefined there). Is this reasoning correct?

Is there another way the statement could this be expressed (e.g. in terms of moduli)? Also, what is an efficient way of finding the centre of the circle?

My understanding is as follows. The angle c between the lines za and zb is constant. za and zb meet at z and the angle between the lines perpendicular to za and zb is 2c, which is only the case if z lies on a circle passing through z, a and b with centre where the lines perpendicular to za and zb intersect. So the locus is an arc of a circle through a and b except a and b (because $arg(\frac{z-a}{z-b})$ is undefined there). Is this reasoning correct?

Is there another way the statement could this be expressed (e.g. in terms of moduli)? Also, what is an efficient way of finding the centre of the circle?

asked 2022-08-17

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 $2\pi /N$ 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.

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.

asked 2022-05-01

How to find the center of a circle with given an arbitrary arc. we only have the arc nothing else. Is there any known equation or way to complete the circle.

asked 2022-05-08

The length of an arc of a circle is 12 cm. The corresponding sector area is 108 cm${}^{2}$. Find the radius of the circle.

I have not attempted this question and do not understand how to solve this.

I have not attempted this question and do not understand how to solve this.