Is it possible to find a closed form for $x$?
$begingroup$
To solve the problem, I followed the following steps:
Is it possible to find a closed form for $x$?
$$frac{sin(x)}{sin(beta-x)}=frac{sin(alpha),sin(theta-gamma)}{sin(gamma),sin(theta-alpha)}$$
where, $$x:= angle OBC,beta:=angle ABC ,alpha:=angle OAC, gamma:=angle OCA ,theta:=angle BAC=angle ACB$$
Mathematica says that,
$xapprox 0.033921 approx 1.94353^circ,$
Is there another method to find $x$?
trigonometry contest-math euclidean-geometry triangle closed-form
asked Dec 3 '18 at 16:09
BeginnerBeginner
33810
$endgroup$
add a comment |
$begingroup$
To solve the problem, I followed the following steps:
Is it possible to find a closed form for $x$?
$$frac{sin(x)}{sin(beta-x)}=frac{sin(alpha),sin(theta-gamma)}{sin(gamma),sin(theta-alpha)}$$
where, $$x:= angle OBC,beta:=angle ABC ,alpha:=angle OAC, gamma:=angle OCA ,theta:=angle BAC=angle ACB$$
Mathematica says that,
$xapprox 0.033921 approx 1.94353^circ,$
Is there another method to find $x$?
trigonometry contest-math euclidean-geometry triangle closed-form
asked Dec 3 '18 at 16:09
BeginnerBeginner
33810
$endgroup$
$begingroup$
I would like to thank @Batominovski for helping me take the steps above.
$endgroup$
– Beginner
Dec 3 '18 at 16:11
$begingroup$
You can use use here compound angle formula and eleminate as far as possible the variables.
$endgroup$
– priyanka kumari
Dec 3 '18 at 16:14
add a comment |
$begingroup$
To solve the problem, I followed the following steps:
Is it possible to find a closed form for $x$?
$$frac{sin(x)}{sin(beta-x)}=frac{sin(alpha),sin(theta-gamma)}{sin(gamma),sin(theta-alpha)}$$
where, $$x:= angle OBC,beta:=angle ABC ,alpha:=angle OAC, gamma:=angle OCA ,theta:=angle BAC=angle ACB$$
Mathematica says that,
$xapprox 0.033921 approx 1.94353^circ,$
Is there another method to find $x$?
trigonometry contest-math euclidean-geometry triangle closed-form
asked Dec 3 '18 at 16:09
BeginnerBeginner
33810
$endgroup$
To solve the problem, I followed the following steps:
Is it possible to find a closed form for $x$?
$$frac{sin(x)}{sin(beta-x)}=frac{sin(alpha),sin(theta-gamma)}{sin(gamma),sin(theta-alpha)}$$
where, $$x:= angle OBC,beta:=angle ABC ,alpha:=angle OAC, gamma:=angle OCA ,theta:=angle BAC=angle ACB$$
Mathematica says that,
$xapprox 0.033921 approx 1.94353^circ,$
Is there another method to find $x$?
trigonometry contest-math euclidean-geometry triangle closed-form
trigonometry contest-math euclidean-geometry triangle closed-form
asked Dec 3 '18 at 16:09
BeginnerBeginner
33810
asked Dec 3 '18 at 16:09
BeginnerBeginner
33810
edited Dec 3 '18 at 21:12
Beginner
asked Dec 3 '18 at 16:09
BeginnerBeginner
33810
asked Dec 3 '18 at 16:09
BeginnerBeginner
33810
asked Dec 3 '18 at 16:09
BeginnerBeginner
33810
33810
$begingroup$
I would like to thank @Batominovski for helping me take the steps above.
$endgroup$
– Beginner
Dec 3 '18 at 16:11
$begingroup$
You can use use here compound angle formula and eleminate as far as possible the variables.
$endgroup$
– priyanka kumari
Dec 3 '18 at 16:14
add a comment |
$begingroup$
I would like to thank @Batominovski for helping me take the steps above.
$endgroup$
– Beginner
Dec 3 '18 at 16:11
$begingroup$
You can use use here compound angle formula and eleminate as far as possible the variables.
$endgroup$
– priyanka kumari
Dec 3 '18 at 16:14
$begingroup$
I would like to thank @Batominovski for helping me take the steps above.
$endgroup$
– Beginner
Dec 3 '18 at 16:11
$begingroup$
I would like to thank @Batominovski for helping me take the steps above.
$endgroup$
– Beginner
Dec 3 '18 at 16:11
$begingroup$
You can use use here compound angle formula and eleminate as far as possible the variables.
$endgroup$
– priyanka kumari
Dec 3 '18 at 16:14
$begingroup$
You can use use here compound angle formula and eleminate as far as possible the variables.
$endgroup$
– priyanka kumari
Dec 3 '18 at 16:14
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
If everything except $x$ is known,
$$ frac{sin(x)}{sin(beta-x)}=c$$
reduces to
$$ tan(x) = frac{c sin(beta)}{1+ c cos(beta)}$$
answered Dec 3 '18 at 16:17
Robert IsraelRobert Israel
320k23209459
$endgroup$
1
$begingroup$
Teacher, Are the steps correct I have taken to solve the problem?
$endgroup$
– Beginner
Dec 3 '18 at 16:20
1
$begingroup$
What steps have you taken?
$endgroup$
– Robert Israel
Dec 3 '18 at 16:59
$begingroup$
>$$frac{sin(x)}{sin(beta-x)}=frac{sin(alpha),sin(theta-gamma)}{sin(gamma),sin(theta-alpha)}$$
$endgroup$
– Beginner
Dec 3 '18 at 18:20
add a comment |
$begingroup$
Given that you haven't shown your steps, we can't know whether you've used a logical method but I think you've found the angle incorrectly. Nevertheless, a way you could find it could be:
- Find. $angle BAC$, $angle BCA$ using the fact that the triangle is isosceles
- Find $angle BAO$ and $angle BCO$
- Find $angle AOB$ and $angle COB$
- Relate $angle BAO$, $angle AOB$, $angle BCO$ and $angle AOC$ with each other using the sine rule.
You can get a solution of the form $frac{sin(x+c_1)}{sin(x+c_2)}=frac{sin(c_3)}{sin(c_4)}$, where $c_{1-4}$ are constants. However this equation doesn't seem to easily manipulate into a simple solution for $x$. I inputted the solution into WolframAlpha and it finds a closed form, albeit a nasty expression with $9$ terms.
answered Dec 3 '18 at 19:19
JamJam
4,95021431
$endgroup$
$begingroup$
Can you give me a link?
$endgroup$
– Beginner
Dec 3 '18 at 19:24
$begingroup$
I'm afraid not. But if you write out your attempt at the proof, we can show you where you've gone wrong :)
$endgroup$
– Jam
Dec 3 '18 at 19:31
$begingroup$
what is $c_1,...c_4$
$endgroup$
– Beginner
Dec 3 '18 at 19:37
add a comment |
Your Answer
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
If everything except $x$ is known,
$$ frac{sin(x)}{sin(beta-x)}=c$$
reduces to
$$ tan(x) = frac{c sin(beta)}{1+ c cos(beta)}$$
answered Dec 3 '18 at 16:17
Robert IsraelRobert Israel
320k23209459
$endgroup$
1
$begingroup$
Teacher, Are the steps correct I have taken to solve the problem?
$endgroup$
– Beginner
Dec 3 '18 at 16:20
1
$begingroup$
What steps have you taken?
$endgroup$
– Robert Israel
Dec 3 '18 at 16:59
$begingroup$
>$$frac{sin(x)}{sin(beta-x)}=frac{sin(alpha),sin(theta-gamma)}{sin(gamma),sin(theta-alpha)}$$
$endgroup$
– Beginner
Dec 3 '18 at 18:20
add a comment |
$begingroup$
If everything except $x$ is known,
$$ frac{sin(x)}{sin(beta-x)}=c$$
reduces to
$$ tan(x) = frac{c sin(beta)}{1+ c cos(beta)}$$
answered Dec 3 '18 at 16:17
Robert IsraelRobert Israel
320k23209459
$endgroup$
1
$begingroup$
Teacher, Are the steps correct I have taken to solve the problem?
$endgroup$
– Beginner
Dec 3 '18 at 16:20
1
$begingroup$
What steps have you taken?
$endgroup$
– Robert Israel
Dec 3 '18 at 16:59
$begingroup$
>$$frac{sin(x)}{sin(beta-x)}=frac{sin(alpha),sin(theta-gamma)}{sin(gamma),sin(theta-alpha)}$$
$endgroup$
– Beginner
Dec 3 '18 at 18:20
add a comment |
$begingroup$
If everything except $x$ is known,
$$ frac{sin(x)}{sin(beta-x)}=c$$
reduces to
$$ tan(x) = frac{c sin(beta)}{1+ c cos(beta)}$$
answered Dec 3 '18 at 16:17
Robert IsraelRobert Israel
320k23209459
$endgroup$
If everything except $x$ is known,
$$ frac{sin(x)}{sin(beta-x)}=c$$
reduces to
$$ tan(x) = frac{c sin(beta)}{1+ c cos(beta)}$$
answered Dec 3 '18 at 16:17
Robert IsraelRobert Israel
320k23209459
answered Dec 3 '18 at 16:17
Robert IsraelRobert Israel
320k23209459
answered Dec 3 '18 at 16:17
Robert IsraelRobert Israel
320k23209459
answered Dec 3 '18 at 16:17
Robert IsraelRobert Israel
320k23209459
320k23209459
1
$begingroup$
Teacher, Are the steps correct I have taken to solve the problem?
$endgroup$
– Beginner
Dec 3 '18 at 16:20
1
$begingroup$
What steps have you taken?
$endgroup$
– Robert Israel
Dec 3 '18 at 16:59
$begingroup$
>$$frac{sin(x)}{sin(beta-x)}=frac{sin(alpha),sin(theta-gamma)}{sin(gamma),sin(theta-alpha)}$$
$endgroup$
– Beginner
Dec 3 '18 at 18:20
add a comment |
1
$begingroup$
Teacher, Are the steps correct I have taken to solve the problem?
$endgroup$
– Beginner
Dec 3 '18 at 16:20
1
$begingroup$
What steps have you taken?
$endgroup$
– Robert Israel
Dec 3 '18 at 16:59
$begingroup$
>$$frac{sin(x)}{sin(beta-x)}=frac{sin(alpha),sin(theta-gamma)}{sin(gamma),sin(theta-alpha)}$$
$endgroup$
– Beginner
Dec 3 '18 at 18:20
1
1
$begingroup$
Teacher, Are the steps correct I have taken to solve the problem?
$endgroup$
– Beginner
Dec 3 '18 at 16:20
$begingroup$
Teacher, Are the steps correct I have taken to solve the problem?
$endgroup$
– Beginner
Dec 3 '18 at 16:20
1
1
$begingroup$
What steps have you taken?
$endgroup$
– Robert Israel
Dec 3 '18 at 16:59
$begingroup$
What steps have you taken?
$endgroup$
– Robert Israel
Dec 3 '18 at 16:59
$begingroup$
>$$frac{sin(x)}{sin(beta-x)}=frac{sin(alpha),sin(theta-gamma)}{sin(gamma),sin(theta-alpha)}$$
$endgroup$
– Beginner
Dec 3 '18 at 18:20
$begingroup$
>$$frac{sin(x)}{sin(beta-x)}=frac{sin(alpha),sin(theta-gamma)}{sin(gamma),sin(theta-alpha)}$$
$endgroup$
– Beginner
Dec 3 '18 at 18:20
add a comment |
$begingroup$
Given that you haven't shown your steps, we can't know whether you've used a logical method but I think you've found the angle incorrectly. Nevertheless, a way you could find it could be:
- Find. $angle BAC$, $angle BCA$ using the fact that the triangle is isosceles
- Find $angle BAO$ and $angle BCO$
- Find $angle AOB$ and $angle COB$
- Relate $angle BAO$, $angle AOB$, $angle BCO$ and $angle AOC$ with each other using the sine rule.
You can get a solution of the form $frac{sin(x+c_1)}{sin(x+c_2)}=frac{sin(c_3)}{sin(c_4)}$, where $c_{1-4}$ are constants. However this equation doesn't seem to easily manipulate into a simple solution for $x$. I inputted the solution into WolframAlpha and it finds a closed form, albeit a nasty expression with $9$ terms.
answered Dec 3 '18 at 19:19
JamJam
4,95021431
$endgroup$
$begingroup$
Can you give me a link?
$endgroup$
– Beginner
Dec 3 '18 at 19:24
$begingroup$
I'm afraid not. But if you write out your attempt at the proof, we can show you where you've gone wrong :)
$endgroup$
– Jam
Dec 3 '18 at 19:31
$begingroup$
what is $c_1,...c_4$
$endgroup$
– Beginner
Dec 3 '18 at 19:37
add a comment |
$begingroup$
Given that you haven't shown your steps, we can't know whether you've used a logical method but I think you've found the angle incorrectly. Nevertheless, a way you could find it could be:
- Find. $angle BAC$, $angle BCA$ using the fact that the triangle is isosceles
- Find $angle BAO$ and $angle BCO$
- Find $angle AOB$ and $angle COB$
- Relate $angle BAO$, $angle AOB$, $angle BCO$ and $angle AOC$ with each other using the sine rule.
You can get a solution of the form $frac{sin(x+c_1)}{sin(x+c_2)}=frac{sin(c_3)}{sin(c_4)}$, where $c_{1-4}$ are constants. However this equation doesn't seem to easily manipulate into a simple solution for $x$. I inputted the solution into WolframAlpha and it finds a closed form, albeit a nasty expression with $9$ terms.
answered Dec 3 '18 at 19:19
JamJam
4,95021431
$endgroup$
$begingroup$
Can you give me a link?
$endgroup$
– Beginner
Dec 3 '18 at 19:24
$begingroup$
I'm afraid not. But if you write out your attempt at the proof, we can show you where you've gone wrong :)
$endgroup$
– Jam
Dec 3 '18 at 19:31
$begingroup$
what is $c_1,...c_4$
$endgroup$
– Beginner
Dec 3 '18 at 19:37
add a comment |
$begingroup$
Given that you haven't shown your steps, we can't know whether you've used a logical method but I think you've found the angle incorrectly. Nevertheless, a way you could find it could be:
- Find. $angle BAC$, $angle BCA$ using the fact that the triangle is isosceles
- Find $angle BAO$ and $angle BCO$
- Find $angle AOB$ and $angle COB$
- Relate $angle BAO$, $angle AOB$, $angle BCO$ and $angle AOC$ with each other using the sine rule.
You can get a solution of the form $frac{sin(x+c_1)}{sin(x+c_2)}=frac{sin(c_3)}{sin(c_4)}$, where $c_{1-4}$ are constants. However this equation doesn't seem to easily manipulate into a simple solution for $x$. I inputted the solution into WolframAlpha and it finds a closed form, albeit a nasty expression with $9$ terms.
answered Dec 3 '18 at 19:19
JamJam
4,95021431
$endgroup$
Given that you haven't shown your steps, we can't know whether you've used a logical method but I think you've found the angle incorrectly. Nevertheless, a way you could find it could be:
- Find. $angle BAC$, $angle BCA$ using the fact that the triangle is isosceles
- Find $angle BAO$ and $angle BCO$
- Find $angle AOB$ and $angle COB$
- Relate $angle BAO$, $angle AOB$, $angle BCO$ and $angle AOC$ with each other using the sine rule.
You can get a solution of the form $frac{sin(x+c_1)}{sin(x+c_2)}=frac{sin(c_3)}{sin(c_4)}$, where $c_{1-4}$ are constants. However this equation doesn't seem to easily manipulate into a simple solution for $x$. I inputted the solution into WolframAlpha and it finds a closed form, albeit a nasty expression with $9$ terms.
answered Dec 3 '18 at 19:19
JamJam
4,95021431
answered Dec 3 '18 at 19:19
JamJam
4,95021431
answered Dec 3 '18 at 19:19
JamJam
4,95021431
answered Dec 3 '18 at 19:19
JamJam
4,95021431
4,95021431
$begingroup$
Can you give me a link?
$endgroup$
– Beginner
Dec 3 '18 at 19:24
$begingroup$
I'm afraid not. But if you write out your attempt at the proof, we can show you where you've gone wrong :)
$endgroup$
– Jam
Dec 3 '18 at 19:31
$begingroup$
what is $c_1,...c_4$
$endgroup$
– Beginner
Dec 3 '18 at 19:37
add a comment |
$begingroup$
Can you give me a link?
$endgroup$
– Beginner
Dec 3 '18 at 19:24
$begingroup$
I'm afraid not. But if you write out your attempt at the proof, we can show you where you've gone wrong :)
$endgroup$
– Jam
Dec 3 '18 at 19:31
$begingroup$
what is $c_1,...c_4$
$endgroup$
– Beginner
Dec 3 '18 at 19:37
$begingroup$
Can you give me a link?
$endgroup$
– Beginner
Dec 3 '18 at 19:24
$begingroup$
Can you give me a link?
$endgroup$
– Beginner
Dec 3 '18 at 19:24
$begingroup$
I'm afraid not. But if you write out your attempt at the proof, we can show you where you've gone wrong :)
$endgroup$
– Jam
Dec 3 '18 at 19:31
$begingroup$
I'm afraid not. But if you write out your attempt at the proof, we can show you where you've gone wrong :)
$endgroup$
– Jam
Dec 3 '18 at 19:31
$begingroup$
what is $c_1,...c_4$
$endgroup$
– Beginner
Dec 3 '18 at 19:37
$begingroup$
what is $c_1,...c_4$
$endgroup$
– Beginner
Dec 3 '18 at 19:37
add a comment |
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$begingroup$
I would like to thank @Batominovski for helping me take the steps above.
$endgroup$
– Beginner
Dec 3 '18 at 16:11
$begingroup$
You can use use here compound angle formula and eleminate as far as possible the variables.
$endgroup$
– priyanka kumari
Dec 3 '18 at 16:14