A chord of length $6$ subtends an $80^circ$ central angle in a circle. Can we calculate the distance from...
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I know that the following can be answered easily using trigonometric ratios, but is there any way to go about it without relying on trigonometry? (The book from which the problem was taken doesn't cover trig before this point.)
$O$ is the centre of a circle, and $PQ$ and $RS$ are two equal length chords in it. Points $M$ and $N$ are the midpoints of the two chords $PQ$ and $RS$ respectively. The length of $PQ$ is given as $6$cm and the measure of the angle $angle POQ$ (where $O$ is the vertex) is given as $80^circ$ degrees. A diagram of this is in the snapshot I've attached. The question is, how do you calculate the length $|OM|$?
triangle circle
edited Dec 28 '18 at 15:11
Blue
48.5k870154
asked Dec 28 '18 at 13:44
IndulaIndula
506
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show 13 more comments
$begingroup$
I know that the following can be answered easily using trigonometric ratios, but is there any way to go about it without relying on trigonometry? (The book from which the problem was taken doesn't cover trig before this point.)
$O$ is the centre of a circle, and $PQ$ and $RS$ are two equal length chords in it. Points $M$ and $N$ are the midpoints of the two chords $PQ$ and $RS$ respectively. The length of $PQ$ is given as $6$cm and the measure of the angle $angle POQ$ (where $O$ is the vertex) is given as $80^circ$ degrees. A diagram of this is in the snapshot I've attached. The question is, how do you calculate the length $|OM|$?
triangle circle
edited Dec 28 '18 at 15:11
Blue
48.5k870154
asked Dec 28 '18 at 13:44
IndulaIndula
506
$endgroup$
1
$begingroup$
What is the relevance of the RS chord? And what's the problem with trigonometry?
$endgroup$
– rafa11111
Dec 28 '18 at 13:46
1
$begingroup$
Well, the length of $OM$ (in centimeters) is $3cot 40^circ$. Computing this cotangent requires solving a degree 3 equation, because you need to trisect the angle of $120^circ$.
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– egreg
Dec 28 '18 at 13:50
1
$begingroup$
@egreg According to Wolfram, $3 cot 40^circ$ has the minimal polynomial $3x^6 -26x^4 + 33x^2 - 1$.
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– Connor Harris
Dec 28 '18 at 15:34
1
$begingroup$
@ConnorHarris Yes, but this reduces to a degree 3 equation, doesn't it?
$endgroup$
– egreg
Dec 28 '18 at 15:40
1
$begingroup$
And thanks goes to @Blue for that bit of editing which has considerably improved the presentation :)
$endgroup$
– Indula
Dec 30 '18 at 10:25
|
show 13 more comments
$begingroup$
I know that the following can be answered easily using trigonometric ratios, but is there any way to go about it without relying on trigonometry? (The book from which the problem was taken doesn't cover trig before this point.)
$O$ is the centre of a circle, and $PQ$ and $RS$ are two equal length chords in it. Points $M$ and $N$ are the midpoints of the two chords $PQ$ and $RS$ respectively. The length of $PQ$ is given as $6$cm and the measure of the angle $angle POQ$ (where $O$ is the vertex) is given as $80^circ$ degrees. A diagram of this is in the snapshot I've attached. The question is, how do you calculate the length $|OM|$?
triangle circle
edited Dec 28 '18 at 15:11
Blue
48.5k870154
asked Dec 28 '18 at 13:44
IndulaIndula
506
$endgroup$
I know that the following can be answered easily using trigonometric ratios, but is there any way to go about it without relying on trigonometry? (The book from which the problem was taken doesn't cover trig before this point.)
$O$ is the centre of a circle, and $PQ$ and $RS$ are two equal length chords in it. Points $M$ and $N$ are the midpoints of the two chords $PQ$ and $RS$ respectively. The length of $PQ$ is given as $6$cm and the measure of the angle $angle POQ$ (where $O$ is the vertex) is given as $80^circ$ degrees. A diagram of this is in the snapshot I've attached. The question is, how do you calculate the length $|OM|$?
triangle circle
triangle circle
edited Dec 28 '18 at 15:11
Blue
48.5k870154
asked Dec 28 '18 at 13:44
IndulaIndula
506
edited Dec 28 '18 at 15:11
Blue
48.5k870154
asked Dec 28 '18 at 13:44
IndulaIndula
506
edited Dec 28 '18 at 15:11
Blue
48.5k870154
edited Dec 28 '18 at 15:11
Blue
48.5k870154
edited Dec 28 '18 at 15:11
Blue
48.5k870154
48.5k870154
asked Dec 28 '18 at 13:44
IndulaIndula
506
asked Dec 28 '18 at 13:44
IndulaIndula
506
asked Dec 28 '18 at 13:44
IndulaIndula
506
506
1
$begingroup$
What is the relevance of the RS chord? And what's the problem with trigonometry?
$endgroup$
– rafa11111
Dec 28 '18 at 13:46
1
$begingroup$
Well, the length of $OM$ (in centimeters) is $3cot 40^circ$. Computing this cotangent requires solving a degree 3 equation, because you need to trisect the angle of $120^circ$.
$endgroup$
– egreg
Dec 28 '18 at 13:50
1
$begingroup$
@egreg According to Wolfram, $3 cot 40^circ$ has the minimal polynomial $3x^6 -26x^4 + 33x^2 - 1$.
$endgroup$
– Connor Harris
Dec 28 '18 at 15:34
1
$begingroup$
@ConnorHarris Yes, but this reduces to a degree 3 equation, doesn't it?
$endgroup$
– egreg
Dec 28 '18 at 15:40
1
$begingroup$
And thanks goes to @Blue for that bit of editing which has considerably improved the presentation :)
$endgroup$
– Indula
Dec 30 '18 at 10:25
|
show 13 more comments
1
$begingroup$
What is the relevance of the RS chord? And what's the problem with trigonometry?
$endgroup$
– rafa11111
Dec 28 '18 at 13:46
1
$begingroup$
Well, the length of $OM$ (in centimeters) is $3cot 40^circ$. Computing this cotangent requires solving a degree 3 equation, because you need to trisect the angle of $120^circ$.
$endgroup$
– egreg
Dec 28 '18 at 13:50
1
$begingroup$
@egreg According to Wolfram, $3 cot 40^circ$ has the minimal polynomial $3x^6 -26x^4 + 33x^2 - 1$.
$endgroup$
– Connor Harris
Dec 28 '18 at 15:34
1
$begingroup$
@ConnorHarris Yes, but this reduces to a degree 3 equation, doesn't it?
$endgroup$
– egreg
Dec 28 '18 at 15:40
1
$begingroup$
And thanks goes to @Blue for that bit of editing which has considerably improved the presentation :)
$endgroup$
– Indula
Dec 30 '18 at 10:25
1
1
$begingroup$
What is the relevance of the RS chord? And what's the problem with trigonometry?
$endgroup$
– rafa11111
Dec 28 '18 at 13:46
$begingroup$
What is the relevance of the RS chord? And what's the problem with trigonometry?
$endgroup$
– rafa11111
Dec 28 '18 at 13:46
1
1
$begingroup$
Well, the length of $OM$ (in centimeters) is $3cot 40^circ$. Computing this cotangent requires solving a degree 3 equation, because you need to trisect the angle of $120^circ$.
$endgroup$
– egreg
Dec 28 '18 at 13:50
$begingroup$
Well, the length of $OM$ (in centimeters) is $3cot 40^circ$. Computing this cotangent requires solving a degree 3 equation, because you need to trisect the angle of $120^circ$.
$endgroup$
– egreg
Dec 28 '18 at 13:50
1
1
$begingroup$
@egreg According to Wolfram, $3 cot 40^circ$ has the minimal polynomial $3x^6 -26x^4 + 33x^2 - 1$.
$endgroup$
– Connor Harris
Dec 28 '18 at 15:34
$begingroup$
@egreg According to Wolfram, $3 cot 40^circ$ has the minimal polynomial $3x^6 -26x^4 + 33x^2 - 1$.
$endgroup$
– Connor Harris
Dec 28 '18 at 15:34
1
1
$begingroup$
@ConnorHarris Yes, but this reduces to a degree 3 equation, doesn't it?
$endgroup$
– egreg
Dec 28 '18 at 15:40
$begingroup$
@ConnorHarris Yes, but this reduces to a degree 3 equation, doesn't it?
$endgroup$
– egreg
Dec 28 '18 at 15:40
1
1
$begingroup$
And thanks goes to @Blue for that bit of editing which has considerably improved the presentation :)
$endgroup$
– Indula
Dec 30 '18 at 10:25
$begingroup$
And thanks goes to @Blue for that bit of editing which has considerably improved the presentation :)
$endgroup$
– Indula
Dec 30 '18 at 10:25
|
show 13 more comments
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1
$begingroup$
What is the relevance of the RS chord? And what's the problem with trigonometry?
$endgroup$
– rafa11111
Dec 28 '18 at 13:46
1
$begingroup$
Well, the length of $OM$ (in centimeters) is $3cot 40^circ$. Computing this cotangent requires solving a degree 3 equation, because you need to trisect the angle of $120^circ$.
$endgroup$
– egreg
Dec 28 '18 at 13:50
1
$begingroup$
@egreg According to Wolfram, $3 cot 40^circ$ has the minimal polynomial $3x^6 -26x^4 + 33x^2 - 1$.
$endgroup$
– Connor Harris
Dec 28 '18 at 15:34
1
$begingroup$
@ConnorHarris Yes, but this reduces to a degree 3 equation, doesn't it?
$endgroup$
– egreg
Dec 28 '18 at 15:40
1
$begingroup$
And thanks goes to @Blue for that bit of editing which has considerably improved the presentation :)
$endgroup$
– Indula
Dec 30 '18 at 10:25