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Planar sines are special cases of spherical sines?
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Show that the triangles are congruentProve that: $S_XYZgeq frac14S_ABC$Points $A$, $B$, and $C$ are on the circumference of a circle with radius 2What is the angle $widehatBAC$?Spanish Math OlympiadInterior angle of spherical polygon given the coordinates of vertices in spherical coordinatessolving spherical right triangleA triangle inscribed in a circle of radius $2$ has angles $45^circ$ and $60^circ$. What's its area?How do we evaluate the degree of $x$ using sine law?What is the relationship between the side length and the angle of a spherical lune? Is there a sine theorem?
$begingroup$
In the figure $BC$, $AC$ and $BA$ are all large arcs, the spherical angle $∠ACB$ is right angle, and $Delta ABC$ is spherical right triangle.
$O$ is spherical center.
Let the radius of the sphere $r= 1$.
$∠BEO=π/2, ∠BDO=π/2$,
$Letθ=∠BDE$
$sinθ= BE/ BD$
$∵$
$sin(BC/r)=BE/r= sin(BC)=BE$
$∴$
$BC=arcsin(BE)$
$∵$
$sin(BA/r)=BD/r= sin(BA) =BD$
$∴$
$BA= arcsin(BD)$
Let $ρ=∠BAC$
$sin ρ=BC/BA= arcsin(BE)/arcsin(BD)$
When BE and $BD$ are very small,
$arcsin(BE)=BE, arcsin(BD)=BD$
$∴$
$sinρ=sinθ=BE/ BD$
So planar sines are special cases of spherical sines.
That's my idea. I don't know if it's right.
geometry analytic-geometry
asked Apr 9 at 2:53
E.weiE.wei
134
$endgroup$
|
show 20 more comments
$begingroup$
In the figure $BC$, $AC$ and $BA$ are all large arcs, the spherical angle $∠ACB$ is right angle, and $Delta ABC$ is spherical right triangle.
$O$ is spherical center.
Let the radius of the sphere $r= 1$.
$∠BEO=π/2, ∠BDO=π/2$,
$Letθ=∠BDE$
$sinθ= BE/ BD$
$∵$
$sin(BC/r)=BE/r= sin(BC)=BE$
$∴$
$BC=arcsin(BE)$
$∵$
$sin(BA/r)=BD/r= sin(BA) =BD$
$∴$
$BA= arcsin(BD)$
Let $ρ=∠BAC$
$sin ρ=BC/BA= arcsin(BE)/arcsin(BD)$
When BE and $BD$ are very small,
$arcsin(BE)=BE, arcsin(BD)=BD$
$∴$
$sinρ=sinθ=BE/ BD$
So planar sines are special cases of spherical sines.
That's my idea. I don't know if it's right.
geometry analytic-geometry
asked Apr 9 at 2:53
E.weiE.wei
134
$endgroup$
$begingroup$
@Aretino Where is it? I don't seem to write that.
$endgroup$
– E.wei
Apr 10 at 0:44
$begingroup$
When you introduce $rho := angle ABC$, it's not clear whether you mean the planar angle or the spherical angle; likewise, when you write $sin rho = BC/BA$, it's not clear if you're working with segments or arcs. Either way, this isn't correct. The planar $triangle ABC$ is not a right triangle, so $sinrho$ is not the ratio of sides. The spherical $triangle ABC$ has a right angle, but $sinrho$ is not the ratio of side-arcs; rather, the Spherical Law of Sines says in this case that $sinrho = sinstackrelfrownBC/sinstackrelfrownBA$.
$endgroup$
– Blue
Apr 10 at 2:00
$begingroup$
@Blue As I said at the beginning of the question, $BC$, $AB$, $AC$ are big arcs. $△ABC$ is not a plane triangle, of course. It is a spherical right triangle. By definition, a sine can only be a ratio of side length (because it's a sphere, so it's a ratio of arc length). As shown in the figure, $sinBC/sinB$ is the $BE/BD$ of plane $△BED$. This is different from what I said.
$endgroup$
– E.wei
Apr 10 at 7:14
1
$begingroup$
Sorry, I made a mistake in writing the question. I have revised the question.
$endgroup$
– E.wei
Apr 10 at 23:45
1
$begingroup$
I don't know how you got that idea, E.wei.
$endgroup$
– Gerry Myerson
Apr 12 at 0:06
|
show 20 more comments
$begingroup$
In the figure $BC$, $AC$ and $BA$ are all large arcs, the spherical angle $∠ACB$ is right angle, and $Delta ABC$ is spherical right triangle.
$O$ is spherical center.
Let the radius of the sphere $r= 1$.
$∠BEO=π/2, ∠BDO=π/2$,
$Letθ=∠BDE$
$sinθ= BE/ BD$
$∵$
$sin(BC/r)=BE/r= sin(BC)=BE$
$∴$
$BC=arcsin(BE)$
$∵$
$sin(BA/r)=BD/r= sin(BA) =BD$
$∴$
$BA= arcsin(BD)$
Let $ρ=∠BAC$
$sin ρ=BC/BA= arcsin(BE)/arcsin(BD)$
When BE and $BD$ are very small,
$arcsin(BE)=BE, arcsin(BD)=BD$
$∴$
$sinρ=sinθ=BE/ BD$
So planar sines are special cases of spherical sines.
That's my idea. I don't know if it's right.
geometry analytic-geometry
asked Apr 9 at 2:53
E.weiE.wei
134
$endgroup$
In the figure $BC$, $AC$ and $BA$ are all large arcs, the spherical angle $∠ACB$ is right angle, and $Delta ABC$ is spherical right triangle.
$O$ is spherical center.
Let the radius of the sphere $r= 1$.
$∠BEO=π/2, ∠BDO=π/2$,
$Letθ=∠BDE$
$sinθ= BE/ BD$
$∵$
$sin(BC/r)=BE/r= sin(BC)=BE$
$∴$
$BC=arcsin(BE)$
$∵$
$sin(BA/r)=BD/r= sin(BA) =BD$
$∴$
$BA= arcsin(BD)$
Let $ρ=∠BAC$
$sin ρ=BC/BA= arcsin(BE)/arcsin(BD)$
When BE and $BD$ are very small,
$arcsin(BE)=BE, arcsin(BD)=BD$
$∴$
$sinρ=sinθ=BE/ BD$
So planar sines are special cases of spherical sines.
That's my idea. I don't know if it's right.
geometry analytic-geometry
geometry analytic-geometry
asked Apr 9 at 2:53
E.weiE.wei
134
asked Apr 9 at 2:53
E.weiE.wei
134
edited Apr 10 at 23:42
E.wei
asked Apr 9 at 2:53
E.weiE.wei
134
asked Apr 9 at 2:53
E.weiE.wei
134
asked Apr 9 at 2:53
E.weiE.wei
134
134
$begingroup$
@Aretino Where is it? I don't seem to write that.
$endgroup$
– E.wei
Apr 10 at 0:44
$begingroup$
When you introduce $rho := angle ABC$, it's not clear whether you mean the planar angle or the spherical angle; likewise, when you write $sin rho = BC/BA$, it's not clear if you're working with segments or arcs. Either way, this isn't correct. The planar $triangle ABC$ is not a right triangle, so $sinrho$ is not the ratio of sides. The spherical $triangle ABC$ has a right angle, but $sinrho$ is not the ratio of side-arcs; rather, the Spherical Law of Sines says in this case that $sinrho = sinstackrelfrownBC/sinstackrelfrownBA$.
$endgroup$
– Blue
Apr 10 at 2:00
$begingroup$
@Blue As I said at the beginning of the question, $BC$, $AB$, $AC$ are big arcs. $△ABC$ is not a plane triangle, of course. It is a spherical right triangle. By definition, a sine can only be a ratio of side length (because it's a sphere, so it's a ratio of arc length). As shown in the figure, $sinBC/sinB$ is the $BE/BD$ of plane $△BED$. This is different from what I said.
$endgroup$
– E.wei
Apr 10 at 7:14
1
$begingroup$
Sorry, I made a mistake in writing the question. I have revised the question.
$endgroup$
– E.wei
Apr 10 at 23:45
1
$begingroup$
I don't know how you got that idea, E.wei.
$endgroup$
– Gerry Myerson
Apr 12 at 0:06
|
show 20 more comments
$begingroup$
@Aretino Where is it? I don't seem to write that.
$endgroup$
– E.wei
Apr 10 at 0:44
$begingroup$
When you introduce $rho := angle ABC$, it's not clear whether you mean the planar angle or the spherical angle; likewise, when you write $sin rho = BC/BA$, it's not clear if you're working with segments or arcs. Either way, this isn't correct. The planar $triangle ABC$ is not a right triangle, so $sinrho$ is not the ratio of sides. The spherical $triangle ABC$ has a right angle, but $sinrho$ is not the ratio of side-arcs; rather, the Spherical Law of Sines says in this case that $sinrho = sinstackrelfrownBC/sinstackrelfrownBA$.
$endgroup$
– Blue
Apr 10 at 2:00
$begingroup$
@Blue As I said at the beginning of the question, $BC$, $AB$, $AC$ are big arcs. $△ABC$ is not a plane triangle, of course. It is a spherical right triangle. By definition, a sine can only be a ratio of side length (because it's a sphere, so it's a ratio of arc length). As shown in the figure, $sinBC/sinB$ is the $BE/BD$ of plane $△BED$. This is different from what I said.
$endgroup$
– E.wei
Apr 10 at 7:14
1
$begingroup$
Sorry, I made a mistake in writing the question. I have revised the question.
$endgroup$
– E.wei
Apr 10 at 23:45
1
$begingroup$
I don't know how you got that idea, E.wei.
$endgroup$
– Gerry Myerson
Apr 12 at 0:06
$begingroup$
@Aretino Where is it? I don't seem to write that.
$endgroup$
– E.wei
Apr 10 at 0:44
$begingroup$
@Aretino Where is it? I don't seem to write that.
$endgroup$
– E.wei
Apr 10 at 0:44
$begingroup$
When you introduce $rho := angle ABC$, it's not clear whether you mean the planar angle or the spherical angle; likewise, when you write $sin rho = BC/BA$, it's not clear if you're working with segments or arcs. Either way, this isn't correct. The planar $triangle ABC$ is not a right triangle, so $sinrho$ is not the ratio of sides. The spherical $triangle ABC$ has a right angle, but $sinrho$ is not the ratio of side-arcs; rather, the Spherical Law of Sines says in this case that $sinrho = sinstackrelfrownBC/sinstackrelfrownBA$.
$endgroup$
– Blue
Apr 10 at 2:00
$begingroup$
When you introduce $rho := angle ABC$, it's not clear whether you mean the planar angle or the spherical angle; likewise, when you write $sin rho = BC/BA$, it's not clear if you're working with segments or arcs. Either way, this isn't correct. The planar $triangle ABC$ is not a right triangle, so $sinrho$ is not the ratio of sides. The spherical $triangle ABC$ has a right angle, but $sinrho$ is not the ratio of side-arcs; rather, the Spherical Law of Sines says in this case that $sinrho = sinstackrelfrownBC/sinstackrelfrownBA$.
$endgroup$
– Blue
Apr 10 at 2:00
$begingroup$
@Blue As I said at the beginning of the question, $BC$, $AB$, $AC$ are big arcs. $△ABC$ is not a plane triangle, of course. It is a spherical right triangle. By definition, a sine can only be a ratio of side length (because it's a sphere, so it's a ratio of arc length). As shown in the figure, $sinBC/sinB$ is the $BE/BD$ of plane $△BED$. This is different from what I said.
$endgroup$
– E.wei
Apr 10 at 7:14
$begingroup$
@Blue As I said at the beginning of the question, $BC$, $AB$, $AC$ are big arcs. $△ABC$ is not a plane triangle, of course. It is a spherical right triangle. By definition, a sine can only be a ratio of side length (because it's a sphere, so it's a ratio of arc length). As shown in the figure, $sinBC/sinB$ is the $BE/BD$ of plane $△BED$. This is different from what I said.
$endgroup$
– E.wei
Apr 10 at 7:14
1
1
$begingroup$
Sorry, I made a mistake in writing the question. I have revised the question.
$endgroup$
– E.wei
Apr 10 at 23:45
$begingroup$
Sorry, I made a mistake in writing the question. I have revised the question.
$endgroup$
– E.wei
Apr 10 at 23:45
1
1
$begingroup$
I don't know how you got that idea, E.wei.
$endgroup$
– Gerry Myerson
Apr 12 at 0:06
$begingroup$
I don't know how you got that idea, E.wei.
$endgroup$
– Gerry Myerson
Apr 12 at 0:06
|
show 20 more comments
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$begingroup$
@Aretino Where is it? I don't seem to write that.
$endgroup$
– E.wei
Apr 10 at 0:44
$begingroup$
When you introduce $rho := angle ABC$, it's not clear whether you mean the planar angle or the spherical angle; likewise, when you write $sin rho = BC/BA$, it's not clear if you're working with segments or arcs. Either way, this isn't correct. The planar $triangle ABC$ is not a right triangle, so $sinrho$ is not the ratio of sides. The spherical $triangle ABC$ has a right angle, but $sinrho$ is not the ratio of side-arcs; rather, the Spherical Law of Sines says in this case that $sinrho = sinstackrelfrownBC/sinstackrelfrownBA$.
$endgroup$
– Blue
Apr 10 at 2:00
$begingroup$
@Blue As I said at the beginning of the question, $BC$, $AB$, $AC$ are big arcs. $△ABC$ is not a plane triangle, of course. It is a spherical right triangle. By definition, a sine can only be a ratio of side length (because it's a sphere, so it's a ratio of arc length). As shown in the figure, $sinBC/sinB$ is the $BE/BD$ of plane $△BED$. This is different from what I said.
$endgroup$
– E.wei
Apr 10 at 7:14
1
$begingroup$
Sorry, I made a mistake in writing the question. I have revised the question.
$endgroup$
– E.wei
Apr 10 at 23:45
1
$begingroup$
I don't know how you got that idea, E.wei.
$endgroup$
– Gerry Myerson
Apr 12 at 0:06