Mixing
Get an angle from the center of circle, based on other angle in the circle
$begingroup$
I know the radius "r" of a circle. I have a point "P", always in the circle and always "looking" at the center of the circle, with a certain angle, or overture "a". I know the distance between "P" and the center of the circle. I would like to know the angle or overture "b" from the center, so that "b" covers the same arc of the circle as "a".
Here's a schema explaining the problem :
The goal is to retrieve the angle "b" from all the other parameters.
Thanks a lot in advance !
geometry
edited Jan 28 at 21:50
Aretino
25.8k31545
asked Jan 28 at 19:54
XysXys
1031
$endgroup$
add a comment |
$begingroup$
I know the radius "r" of a circle. I have a point "P", always in the circle and always "looking" at the center of the circle, with a certain angle, or overture "a". I know the distance between "P" and the center of the circle. I would like to know the angle or overture "b" from the center, so that "b" covers the same arc of the circle as "a".
Here's a schema explaining the problem :
The goal is to retrieve the angle "b" from all the other parameters.
Thanks a lot in advance !
geometry
edited Jan 28 at 21:50
Aretino
25.8k31545
asked Jan 28 at 19:54
XysXys
1031
$endgroup$
$begingroup$
You can retrieve b from a and p in the particular case where the red line is the angle bissector ; otherwise, it is not possible.
$endgroup$
– Jean Marie
Jan 28 at 20:38
$begingroup$
@JeanMarie I believe that is what the OP meant by "P is always looking at the center of the circle"
$endgroup$
– R. Burton
Jan 28 at 20:42
$begingroup$
Indeed, the red line is the angle bissector. In my words, I would say, the red line cuts the green angle in half :)
$endgroup$
– Xys
Jan 28 at 20:47
$begingroup$
Apply the sine law to the triangle with sides $d$, $r$ and the green one.
$endgroup$
– Aretino
Jan 28 at 21:15
$begingroup$
They call it the aperture.
$endgroup$
– Yves Daoust
Jan 28 at 21:57
add a comment |
$begingroup$
I know the radius "r" of a circle. I have a point "P", always in the circle and always "looking" at the center of the circle, with a certain angle, or overture "a". I know the distance between "P" and the center of the circle. I would like to know the angle or overture "b" from the center, so that "b" covers the same arc of the circle as "a".
Here's a schema explaining the problem :
The goal is to retrieve the angle "b" from all the other parameters.
Thanks a lot in advance !
geometry
edited Jan 28 at 21:50
Aretino
25.8k31545
asked Jan 28 at 19:54
XysXys
1031
$endgroup$
I know the radius "r" of a circle. I have a point "P", always in the circle and always "looking" at the center of the circle, with a certain angle, or overture "a". I know the distance between "P" and the center of the circle. I would like to know the angle or overture "b" from the center, so that "b" covers the same arc of the circle as "a".
Here's a schema explaining the problem :
The goal is to retrieve the angle "b" from all the other parameters.
Thanks a lot in advance !
geometry
geometry
edited Jan 28 at 21:50
Aretino
25.8k31545
asked Jan 28 at 19:54
XysXys
1031
edited Jan 28 at 21:50
Aretino
25.8k31545
asked Jan 28 at 19:54
XysXys
1031
edited Jan 28 at 21:50
Aretino
25.8k31545
edited Jan 28 at 21:50
Aretino
25.8k31545
edited Jan 28 at 21:50
Aretino
25.8k31545
25.8k31545
asked Jan 28 at 19:54
XysXys
1031
asked Jan 28 at 19:54
XysXys
1031
asked Jan 28 at 19:54
XysXys
1031
1031
$begingroup$
You can retrieve b from a and p in the particular case where the red line is the angle bissector ; otherwise, it is not possible.
$endgroup$
– Jean Marie
Jan 28 at 20:38
$begingroup$
@JeanMarie I believe that is what the OP meant by "P is always looking at the center of the circle"
$endgroup$
– R. Burton
Jan 28 at 20:42
$begingroup$
Indeed, the red line is the angle bissector. In my words, I would say, the red line cuts the green angle in half :)
$endgroup$
– Xys
Jan 28 at 20:47
$begingroup$
Apply the sine law to the triangle with sides $d$, $r$ and the green one.
$endgroup$
– Aretino
Jan 28 at 21:15
$begingroup$
They call it the aperture.
$endgroup$
– Yves Daoust
Jan 28 at 21:57
add a comment |
$begingroup$
You can retrieve b from a and p in the particular case where the red line is the angle bissector ; otherwise, it is not possible.
$endgroup$
– Jean Marie
Jan 28 at 20:38
$begingroup$
@JeanMarie I believe that is what the OP meant by "P is always looking at the center of the circle"
$endgroup$
– R. Burton
Jan 28 at 20:42
$begingroup$
Indeed, the red line is the angle bissector. In my words, I would say, the red line cuts the green angle in half :)
$endgroup$
– Xys
Jan 28 at 20:47
$begingroup$
Apply the sine law to the triangle with sides $d$, $r$ and the green one.
$endgroup$
– Aretino
Jan 28 at 21:15
$begingroup$
They call it the aperture.
$endgroup$
– Yves Daoust
Jan 28 at 21:57
$begingroup$
You can retrieve b from a and p in the particular case where the red line is the angle bissector ; otherwise, it is not possible.
$endgroup$
– Jean Marie
Jan 28 at 20:38
$begingroup$
You can retrieve b from a and p in the particular case where the red line is the angle bissector ; otherwise, it is not possible.
$endgroup$
– Jean Marie
Jan 28 at 20:38
$begingroup$
@JeanMarie I believe that is what the OP meant by "P is always looking at the center of the circle"
$endgroup$
– R. Burton
Jan 28 at 20:42
$begingroup$
@JeanMarie I believe that is what the OP meant by "P is always looking at the center of the circle"
$endgroup$
– R. Burton
Jan 28 at 20:42
$begingroup$
Indeed, the red line is the angle bissector. In my words, I would say, the red line cuts the green angle in half :)
$endgroup$
– Xys
Jan 28 at 20:47
$begingroup$
Indeed, the red line is the angle bissector. In my words, I would say, the red line cuts the green angle in half :)
$endgroup$
– Xys
Jan 28 at 20:47
$begingroup$
Apply the sine law to the triangle with sides $d$, $r$ and the green one.
$endgroup$
– Aretino
Jan 28 at 21:15
$begingroup$
Apply the sine law to the triangle with sides $d$, $r$ and the green one.
$endgroup$
– Aretino
Jan 28 at 21:15
$begingroup$
They call it the aperture.
$endgroup$
– Yves Daoust
Jan 28 at 21:57
$begingroup$
They call it the aperture.
$endgroup$
– Yves Daoust
Jan 28 at 21:57
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Let $alpha=a/2$ and $beta=b/2$. Applying the sine law to the triangle with sides $d$, $r$ we get:
$$
{roversinalpha}={doversin(beta-alpha)},
$$
which after expanding $sin(beta-alpha)$ becomes:
$$
sinbeta=tanalphacosbeta+{dover r}tanalpha.
$$
This equation can be solved, for example, plugging it into $sin^2beta+cos^2beta=1$ and solving for $cosbeta$:
$$
cosbeta=cosalphasqrt{1-{d^2over r^2}sin^2alpha}-{dover r}sin^2alpha,
$$
where I discarded the negative solution as $0lebetalepi/2$.
EDIT.
Here's a graph of $b$ vs. $d/r$, comparing (for $a=180°$) the exact solution above (black curve) with the approximate solution $b=(1+d/r)a$ (red curve). The difference is less pronounced for smaller values of $a$.
answered Jan 28 at 21:59
AretinoAretino
25.8k31545
$endgroup$
$begingroup$
Thanks a lot for your answer ! Unfortunately, I'm not sure to understand.. Is there a function that express b directly ? Like b = f(a) ?
$endgroup$
– Xys
Jan 28 at 23:40
$begingroup$
The final formula given by @Aretino can be expressed under the form: $b/2=$acos$left(cosalphasqrt{1-{d^2over r^2}sin^2alpha}-{dover r}sin^2alpharight)$
$endgroup$
– Jean Marie
Jan 29 at 0:01
$begingroup$
@JeanMarie Thanks !
$endgroup$
– Xys
Jan 29 at 9:45
$begingroup$
It seems to me that if d=r, then b=2a. Of course if d=0, then b=a. Then Isn't the solution just : b = (1 + d/r) a ? I can't prove it, but the values seems to confirm my little theorem.
$endgroup$
– Xys
Jan 29 at 11:41
$begingroup$
The solution cannot be written in the simple form you propose: just try both formulas for some values of $d/r$ to be convinced. Of course you could use your formula if an approximate result is enough.
$endgroup$
– Aretino
Jan 29 at 13:26
|
show 2 more comments
$begingroup$
Take a look at the figure below that you will easily recognize :
Let us use 2 properties : a) the sine law in triangle POQ :
$$dfrac{r}{sin(a/2)}=underbrace{dfrac{c}{sin(pi-b/2)}}_{= dfrac{c}{sin(b/2)}}tag{1}$$
b) orthogonal projection on axis $POH$ expressing that $PH=PO+OH$ :
$$c cos(a/2)=d+rcos(b/2)tag{2}$$
It suffices now to extract the unknown $c$ from (2) and to plug it into (1) giving :
$$sin(a/2)(d+r cos(b/2))=r cos(a/2)sin(b/2)tag{3}$$
As you want to express $b$ as a function of $a$, a good option here is to take the classical formulas (https://en.wikipedia.org/wiki/Tangent_half-angle_formula) :
$$cos(b/2)=frac{1-t^2}{1+t^2}, sin(b/2)=frac{2t}{1+t^2}, text{with} t:=tan(b/4)$$
in order to transform (3) into a quadratic in $t$. Solving it will give you two roots $t_1$ and $t_2$, out of which you will extract the solutions, under the constraint that $b/2<pi/2$.
answered Jan 28 at 21:40
Jean MarieJean Marie
31k42255
$endgroup$
$begingroup$
$sin(pi-b/2)=sin(b/2)$.
$endgroup$
– Aretino
Jan 28 at 21:47
$begingroup$
@Aretino Thanks !
$endgroup$
– Jean Marie
Jan 28 at 21:54
$begingroup$
Does the final solution change if $POH$ does not bisect angle $b$?
$endgroup$
– Chase Ryan Taylor
Jan 28 at 22:01
$begingroup$
@Chase Ryan Taylor Yes ; another answer is that we need a supplementary information to be able to conclude.
$endgroup$
– Jean Marie
Jan 28 at 22:40
$begingroup$
Thanks a lot for your answers guys ! As I said to Aretino, unfortunately, I'm not sure to understand.. Is there a function that express b directly ? Like b = f(a) ?
$endgroup$
– Xys
Jan 28 at 23:42
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Let $alpha=a/2$ and $beta=b/2$. Applying the sine law to the triangle with sides $d$, $r$ we get:
$$
{roversinalpha}={doversin(beta-alpha)},
$$
which after expanding $sin(beta-alpha)$ becomes:
$$
sinbeta=tanalphacosbeta+{dover r}tanalpha.
$$
This equation can be solved, for example, plugging it into $sin^2beta+cos^2beta=1$ and solving for $cosbeta$:
$$
cosbeta=cosalphasqrt{1-{d^2over r^2}sin^2alpha}-{dover r}sin^2alpha,
$$
where I discarded the negative solution as $0lebetalepi/2$.
EDIT.
Here's a graph of $b$ vs. $d/r$, comparing (for $a=180°$) the exact solution above (black curve) with the approximate solution $b=(1+d/r)a$ (red curve). The difference is less pronounced for smaller values of $a$.
answered Jan 28 at 21:59
AretinoAretino
25.8k31545
$endgroup$
$begingroup$
Thanks a lot for your answer ! Unfortunately, I'm not sure to understand.. Is there a function that express b directly ? Like b = f(a) ?
$endgroup$
– Xys
Jan 28 at 23:40
$begingroup$
The final formula given by @Aretino can be expressed under the form: $b/2=$acos$left(cosalphasqrt{1-{d^2over r^2}sin^2alpha}-{dover r}sin^2alpharight)$
$endgroup$
– Jean Marie
Jan 29 at 0:01
$begingroup$
@JeanMarie Thanks !
$endgroup$
– Xys
Jan 29 at 9:45
$begingroup$
It seems to me that if d=r, then b=2a. Of course if d=0, then b=a. Then Isn't the solution just : b = (1 + d/r) a ? I can't prove it, but the values seems to confirm my little theorem.
$endgroup$
– Xys
Jan 29 at 11:41
$begingroup$
The solution cannot be written in the simple form you propose: just try both formulas for some values of $d/r$ to be convinced. Of course you could use your formula if an approximate result is enough.
$endgroup$
– Aretino
Jan 29 at 13:26
|
show 2 more comments
$begingroup$
Let $alpha=a/2$ and $beta=b/2$. Applying the sine law to the triangle with sides $d$, $r$ we get:
$$
{roversinalpha}={doversin(beta-alpha)},
$$
which after expanding $sin(beta-alpha)$ becomes:
$$
sinbeta=tanalphacosbeta+{dover r}tanalpha.
$$
This equation can be solved, for example, plugging it into $sin^2beta+cos^2beta=1$ and solving for $cosbeta$:
$$
cosbeta=cosalphasqrt{1-{d^2over r^2}sin^2alpha}-{dover r}sin^2alpha,
$$
where I discarded the negative solution as $0lebetalepi/2$.
EDIT.
Here's a graph of $b$ vs. $d/r$, comparing (for $a=180°$) the exact solution above (black curve) with the approximate solution $b=(1+d/r)a$ (red curve). The difference is less pronounced for smaller values of $a$.
answered Jan 28 at 21:59
AretinoAretino
25.8k31545
$endgroup$
$begingroup$
Thanks a lot for your answer ! Unfortunately, I'm not sure to understand.. Is there a function that express b directly ? Like b = f(a) ?
$endgroup$
– Xys
Jan 28 at 23:40
$begingroup$
The final formula given by @Aretino can be expressed under the form: $b/2=$acos$left(cosalphasqrt{1-{d^2over r^2}sin^2alpha}-{dover r}sin^2alpharight)$
$endgroup$
– Jean Marie
Jan 29 at 0:01
$begingroup$
@JeanMarie Thanks !
$endgroup$
– Xys
Jan 29 at 9:45
$begingroup$
It seems to me that if d=r, then b=2a. Of course if d=0, then b=a. Then Isn't the solution just : b = (1 + d/r) a ? I can't prove it, but the values seems to confirm my little theorem.
$endgroup$
– Xys
Jan 29 at 11:41
$begingroup$
The solution cannot be written in the simple form you propose: just try both formulas for some values of $d/r$ to be convinced. Of course you could use your formula if an approximate result is enough.
$endgroup$
– Aretino
Jan 29 at 13:26
|
show 2 more comments
$begingroup$
Let $alpha=a/2$ and $beta=b/2$. Applying the sine law to the triangle with sides $d$, $r$ we get:
$$
{roversinalpha}={doversin(beta-alpha)},
$$
which after expanding $sin(beta-alpha)$ becomes:
$$
sinbeta=tanalphacosbeta+{dover r}tanalpha.
$$
This equation can be solved, for example, plugging it into $sin^2beta+cos^2beta=1$ and solving for $cosbeta$:
$$
cosbeta=cosalphasqrt{1-{d^2over r^2}sin^2alpha}-{dover r}sin^2alpha,
$$
where I discarded the negative solution as $0lebetalepi/2$.
EDIT.
Here's a graph of $b$ vs. $d/r$, comparing (for $a=180°$) the exact solution above (black curve) with the approximate solution $b=(1+d/r)a$ (red curve). The difference is less pronounced for smaller values of $a$.
answered Jan 28 at 21:59
AretinoAretino
25.8k31545
$endgroup$
Let $alpha=a/2$ and $beta=b/2$. Applying the sine law to the triangle with sides $d$, $r$ we get:
$$
{roversinalpha}={doversin(beta-alpha)},
$$
which after expanding $sin(beta-alpha)$ becomes:
$$
sinbeta=tanalphacosbeta+{dover r}tanalpha.
$$
This equation can be solved, for example, plugging it into $sin^2beta+cos^2beta=1$ and solving for $cosbeta$:
$$
cosbeta=cosalphasqrt{1-{d^2over r^2}sin^2alpha}-{dover r}sin^2alpha,
$$
where I discarded the negative solution as $0lebetalepi/2$.
EDIT.
Here's a graph of $b$ vs. $d/r$, comparing (for $a=180°$) the exact solution above (black curve) with the approximate solution $b=(1+d/r)a$ (red curve). The difference is less pronounced for smaller values of $a$.
answered Jan 28 at 21:59
AretinoAretino
25.8k31545
edited Jan 29 at 14:10
answered Jan 28 at 21:59
AretinoAretino
25.8k31545
answered Jan 28 at 21:59
AretinoAretino
25.8k31545
answered Jan 28 at 21:59
AretinoAretino
25.8k31545
25.8k31545
$begingroup$
Thanks a lot for your answer ! Unfortunately, I'm not sure to understand.. Is there a function that express b directly ? Like b = f(a) ?
$endgroup$
– Xys
Jan 28 at 23:40
$begingroup$
The final formula given by @Aretino can be expressed under the form: $b/2=$acos$left(cosalphasqrt{1-{d^2over r^2}sin^2alpha}-{dover r}sin^2alpharight)$
$endgroup$
– Jean Marie
Jan 29 at 0:01
$begingroup$
@JeanMarie Thanks !
$endgroup$
– Xys
Jan 29 at 9:45
$begingroup$
It seems to me that if d=r, then b=2a. Of course if d=0, then b=a. Then Isn't the solution just : b = (1 + d/r) a ? I can't prove it, but the values seems to confirm my little theorem.
$endgroup$
– Xys
Jan 29 at 11:41
$begingroup$
The solution cannot be written in the simple form you propose: just try both formulas for some values of $d/r$ to be convinced. Of course you could use your formula if an approximate result is enough.
$endgroup$
– Aretino
Jan 29 at 13:26
|
show 2 more comments
$begingroup$
Thanks a lot for your answer ! Unfortunately, I'm not sure to understand.. Is there a function that express b directly ? Like b = f(a) ?
$endgroup$
– Xys
Jan 28 at 23:40
$begingroup$
The final formula given by @Aretino can be expressed under the form: $b/2=$acos$left(cosalphasqrt{1-{d^2over r^2}sin^2alpha}-{dover r}sin^2alpharight)$
$endgroup$
– Jean Marie
Jan 29 at 0:01
$begingroup$
@JeanMarie Thanks !
$endgroup$
– Xys
Jan 29 at 9:45
$begingroup$
It seems to me that if d=r, then b=2a. Of course if d=0, then b=a. Then Isn't the solution just : b = (1 + d/r) a ? I can't prove it, but the values seems to confirm my little theorem.
$endgroup$
– Xys
Jan 29 at 11:41
$begingroup$
The solution cannot be written in the simple form you propose: just try both formulas for some values of $d/r$ to be convinced. Of course you could use your formula if an approximate result is enough.
$endgroup$
– Aretino
Jan 29 at 13:26
$begingroup$
Thanks a lot for your answer ! Unfortunately, I'm not sure to understand.. Is there a function that express b directly ? Like b = f(a) ?
$endgroup$
– Xys
Jan 28 at 23:40
$begingroup$
Thanks a lot for your answer ! Unfortunately, I'm not sure to understand.. Is there a function that express b directly ? Like b = f(a) ?
$endgroup$
– Xys
Jan 28 at 23:40
$begingroup$
The final formula given by @Aretino can be expressed under the form: $b/2=$acos$left(cosalphasqrt{1-{d^2over r^2}sin^2alpha}-{dover r}sin^2alpharight)$
$endgroup$
– Jean Marie
Jan 29 at 0:01
$begingroup$
The final formula given by @Aretino can be expressed under the form: $b/2=$acos$left(cosalphasqrt{1-{d^2over r^2}sin^2alpha}-{dover r}sin^2alpharight)$
$endgroup$
– Jean Marie
Jan 29 at 0:01
$begingroup$
@JeanMarie Thanks !
$endgroup$
– Xys
Jan 29 at 9:45
$begingroup$
@JeanMarie Thanks !
$endgroup$
– Xys
Jan 29 at 9:45
$begingroup$
It seems to me that if d=r, then b=2a. Of course if d=0, then b=a. Then Isn't the solution just : b = (1 + d/r) a ? I can't prove it, but the values seems to confirm my little theorem.
$endgroup$
– Xys
Jan 29 at 11:41
$begingroup$
It seems to me that if d=r, then b=2a. Of course if d=0, then b=a. Then Isn't the solution just : b = (1 + d/r) a ? I can't prove it, but the values seems to confirm my little theorem.
$endgroup$
– Xys
Jan 29 at 11:41
$begingroup$
The solution cannot be written in the simple form you propose: just try both formulas for some values of $d/r$ to be convinced. Of course you could use your formula if an approximate result is enough.
$endgroup$
– Aretino
Jan 29 at 13:26
$begingroup$
The solution cannot be written in the simple form you propose: just try both formulas for some values of $d/r$ to be convinced. Of course you could use your formula if an approximate result is enough.
$endgroup$
– Aretino
Jan 29 at 13:26
|
show 2 more comments
$begingroup$
Take a look at the figure below that you will easily recognize :
Let us use 2 properties : a) the sine law in triangle POQ :
$$dfrac{r}{sin(a/2)}=underbrace{dfrac{c}{sin(pi-b/2)}}_{= dfrac{c}{sin(b/2)}}tag{1}$$
b) orthogonal projection on axis $POH$ expressing that $PH=PO+OH$ :
$$c cos(a/2)=d+rcos(b/2)tag{2}$$
It suffices now to extract the unknown $c$ from (2) and to plug it into (1) giving :
$$sin(a/2)(d+r cos(b/2))=r cos(a/2)sin(b/2)tag{3}$$
As you want to express $b$ as a function of $a$, a good option here is to take the classical formulas (https://en.wikipedia.org/wiki/Tangent_half-angle_formula) :
$$cos(b/2)=frac{1-t^2}{1+t^2}, sin(b/2)=frac{2t}{1+t^2}, text{with} t:=tan(b/4)$$
in order to transform (3) into a quadratic in $t$. Solving it will give you two roots $t_1$ and $t_2$, out of which you will extract the solutions, under the constraint that $b/2<pi/2$.
answered Jan 28 at 21:40
Jean MarieJean Marie
31k42255
$endgroup$
$begingroup$
$sin(pi-b/2)=sin(b/2)$.
$endgroup$
– Aretino
Jan 28 at 21:47
$begingroup$
@Aretino Thanks !
$endgroup$
– Jean Marie
Jan 28 at 21:54
$begingroup$
Does the final solution change if $POH$ does not bisect angle $b$?
$endgroup$
– Chase Ryan Taylor
Jan 28 at 22:01
$begingroup$
@Chase Ryan Taylor Yes ; another answer is that we need a supplementary information to be able to conclude.
$endgroup$
– Jean Marie
Jan 28 at 22:40
$begingroup$
Thanks a lot for your answers guys ! As I said to Aretino, unfortunately, I'm not sure to understand.. Is there a function that express b directly ? Like b = f(a) ?
$endgroup$
– Xys
Jan 28 at 23:42
add a comment |
$begingroup$
Take a look at the figure below that you will easily recognize :
Let us use 2 properties : a) the sine law in triangle POQ :
$$dfrac{r}{sin(a/2)}=underbrace{dfrac{c}{sin(pi-b/2)}}_{= dfrac{c}{sin(b/2)}}tag{1}$$
b) orthogonal projection on axis $POH$ expressing that $PH=PO+OH$ :
$$c cos(a/2)=d+rcos(b/2)tag{2}$$
It suffices now to extract the unknown $c$ from (2) and to plug it into (1) giving :
$$sin(a/2)(d+r cos(b/2))=r cos(a/2)sin(b/2)tag{3}$$
As you want to express $b$ as a function of $a$, a good option here is to take the classical formulas (https://en.wikipedia.org/wiki/Tangent_half-angle_formula) :
$$cos(b/2)=frac{1-t^2}{1+t^2}, sin(b/2)=frac{2t}{1+t^2}, text{with} t:=tan(b/4)$$
in order to transform (3) into a quadratic in $t$. Solving it will give you two roots $t_1$ and $t_2$, out of which you will extract the solutions, under the constraint that $b/2<pi/2$.
answered Jan 28 at 21:40
Jean MarieJean Marie
31k42255
$endgroup$
$begingroup$
$sin(pi-b/2)=sin(b/2)$.
$endgroup$
– Aretino
Jan 28 at 21:47
$begingroup$
@Aretino Thanks !
$endgroup$
– Jean Marie
Jan 28 at 21:54
$begingroup$
Does the final solution change if $POH$ does not bisect angle $b$?
$endgroup$
– Chase Ryan Taylor
Jan 28 at 22:01
$begingroup$
@Chase Ryan Taylor Yes ; another answer is that we need a supplementary information to be able to conclude.
$endgroup$
– Jean Marie
Jan 28 at 22:40
$begingroup$
Thanks a lot for your answers guys ! As I said to Aretino, unfortunately, I'm not sure to understand.. Is there a function that express b directly ? Like b = f(a) ?
$endgroup$
– Xys
Jan 28 at 23:42
add a comment |
$begingroup$
Take a look at the figure below that you will easily recognize :
Let us use 2 properties : a) the sine law in triangle POQ :
$$dfrac{r}{sin(a/2)}=underbrace{dfrac{c}{sin(pi-b/2)}}_{= dfrac{c}{sin(b/2)}}tag{1}$$
b) orthogonal projection on axis $POH$ expressing that $PH=PO+OH$ :
$$c cos(a/2)=d+rcos(b/2)tag{2}$$
It suffices now to extract the unknown $c$ from (2) and to plug it into (1) giving :
$$sin(a/2)(d+r cos(b/2))=r cos(a/2)sin(b/2)tag{3}$$
As you want to express $b$ as a function of $a$, a good option here is to take the classical formulas (https://en.wikipedia.org/wiki/Tangent_half-angle_formula) :
$$cos(b/2)=frac{1-t^2}{1+t^2}, sin(b/2)=frac{2t}{1+t^2}, text{with} t:=tan(b/4)$$
in order to transform (3) into a quadratic in $t$. Solving it will give you two roots $t_1$ and $t_2$, out of which you will extract the solutions, under the constraint that $b/2<pi/2$.
answered Jan 28 at 21:40
Jean MarieJean Marie
31k42255
$endgroup$
Take a look at the figure below that you will easily recognize :
Let us use 2 properties : a) the sine law in triangle POQ :
$$dfrac{r}{sin(a/2)}=underbrace{dfrac{c}{sin(pi-b/2)}}_{= dfrac{c}{sin(b/2)}}tag{1}$$
b) orthogonal projection on axis $POH$ expressing that $PH=PO+OH$ :
$$c cos(a/2)=d+rcos(b/2)tag{2}$$
It suffices now to extract the unknown $c$ from (2) and to plug it into (1) giving :
$$sin(a/2)(d+r cos(b/2))=r cos(a/2)sin(b/2)tag{3}$$
As you want to express $b$ as a function of $a$, a good option here is to take the classical formulas (https://en.wikipedia.org/wiki/Tangent_half-angle_formula) :
$$cos(b/2)=frac{1-t^2}{1+t^2}, sin(b/2)=frac{2t}{1+t^2}, text{with} t:=tan(b/4)$$
in order to transform (3) into a quadratic in $t$. Solving it will give you two roots $t_1$ and $t_2$, out of which you will extract the solutions, under the constraint that $b/2<pi/2$.
answered Jan 28 at 21:40
Jean MarieJean Marie
31k42255
edited Jan 28 at 22:44
answered Jan 28 at 21:40
Jean MarieJean Marie
31k42255
answered Jan 28 at 21:40
Jean MarieJean Marie
31k42255
answered Jan 28 at 21:40
Jean MarieJean Marie
31k42255
31k42255
$begingroup$
$sin(pi-b/2)=sin(b/2)$.
$endgroup$
– Aretino
Jan 28 at 21:47
$begingroup$
@Aretino Thanks !
$endgroup$
– Jean Marie
Jan 28 at 21:54
$begingroup$
Does the final solution change if $POH$ does not bisect angle $b$?
$endgroup$
– Chase Ryan Taylor
Jan 28 at 22:01
$begingroup$
@Chase Ryan Taylor Yes ; another answer is that we need a supplementary information to be able to conclude.
$endgroup$
– Jean Marie
Jan 28 at 22:40
$begingroup$
Thanks a lot for your answers guys ! As I said to Aretino, unfortunately, I'm not sure to understand.. Is there a function that express b directly ? Like b = f(a) ?
$endgroup$
– Xys
Jan 28 at 23:42
add a comment |
$begingroup$
$sin(pi-b/2)=sin(b/2)$.
$endgroup$
– Aretino
Jan 28 at 21:47
$begingroup$
@Aretino Thanks !
$endgroup$
– Jean Marie
Jan 28 at 21:54
$begingroup$
Does the final solution change if $POH$ does not bisect angle $b$?
$endgroup$
– Chase Ryan Taylor
Jan 28 at 22:01
$begingroup$
@Chase Ryan Taylor Yes ; another answer is that we need a supplementary information to be able to conclude.
$endgroup$
– Jean Marie
Jan 28 at 22:40
$begingroup$
Thanks a lot for your answers guys ! As I said to Aretino, unfortunately, I'm not sure to understand.. Is there a function that express b directly ? Like b = f(a) ?
$endgroup$
– Xys
Jan 28 at 23:42
$begingroup$
$sin(pi-b/2)=sin(b/2)$.
$endgroup$
– Aretino
Jan 28 at 21:47
$begingroup$
$sin(pi-b/2)=sin(b/2)$.
$endgroup$
– Aretino
Jan 28 at 21:47
$begingroup$
@Aretino Thanks !
$endgroup$
– Jean Marie
Jan 28 at 21:54
$begingroup$
@Aretino Thanks !
$endgroup$
– Jean Marie
Jan 28 at 21:54
$begingroup$
Does the final solution change if $POH$ does not bisect angle $b$?
$endgroup$
– Chase Ryan Taylor
Jan 28 at 22:01
$begingroup$
Does the final solution change if $POH$ does not bisect angle $b$?
$endgroup$
– Chase Ryan Taylor
Jan 28 at 22:01
$begingroup$
@Chase Ryan Taylor Yes ; another answer is that we need a supplementary information to be able to conclude.
$endgroup$
– Jean Marie
Jan 28 at 22:40
$begingroup$
@Chase Ryan Taylor Yes ; another answer is that we need a supplementary information to be able to conclude.
$endgroup$
– Jean Marie
Jan 28 at 22:40
$begingroup$
Thanks a lot for your answers guys ! As I said to Aretino, unfortunately, I'm not sure to understand.. Is there a function that express b directly ? Like b = f(a) ?
$endgroup$
– Xys
Jan 28 at 23:42
$begingroup$
Thanks a lot for your answers guys ! As I said to Aretino, unfortunately, I'm not sure to understand.. Is there a function that express b directly ? Like b = f(a) ?
$endgroup$
– Xys
Jan 28 at 23:42
add a comment |
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$begingroup$
You can retrieve b from a and p in the particular case where the red line is the angle bissector ; otherwise, it is not possible.
$endgroup$
– Jean Marie
Jan 28 at 20:38
$begingroup$
@JeanMarie I believe that is what the OP meant by "P is always looking at the center of the circle"
$endgroup$
– R. Burton
Jan 28 at 20:42
$begingroup$
Indeed, the red line is the angle bissector. In my words, I would say, the red line cuts the green angle in half :)
$endgroup$
– Xys
Jan 28 at 20:47
$begingroup$
Apply the sine law to the triangle with sides $d$, $r$ and the green one.
$endgroup$
– Aretino
Jan 28 at 21:15
$begingroup$
They call it the aperture.
$endgroup$
– Yves Daoust
Jan 28 at 21:57