A chord of length $6$ subtends an $80^circ$ central angle in a circle. Can we calculate the distance from...












2












$begingroup$


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 $|OM|$?



Diagram of the circle










share|cite|improve this question











$endgroup$








  • 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 $120^circ$.
    $endgroup$
    – 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


















2












$begingroup$


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 $|OM|$?



Diagram of the circle










share|cite|improve this question











$endgroup$








  • 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 $120^circ$.
    $endgroup$
    – 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
















2












2








2


1



$begingroup$


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 $|OM|$?



Diagram of the circle










share|cite|improve this question











$endgroup$




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 $|OM|$?



Diagram of the circle







triangles circles






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 28 '18 at 15:11









Blue

49.7k870158




49.7k870158










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
















  • 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












0






active

oldest

votes












Your Answer








StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);

StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});

function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});


}
});














draft saved

draft discarded


















StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3054894%2fa-chord-of-length-6-subtends-an-80-circ-central-angle-in-a-circle-can-we-c%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown

























0






active

oldest

votes








0






active

oldest

votes









active

oldest

votes






active

oldest

votes
















draft saved

draft discarded




















































Thanks for contributing an answer to Mathematics Stack Exchange!


  • Please be sure to answer the question. Provide details and share your research!

But avoid



  • Asking for help, clarification, or responding to other answers.

  • Making statements based on opinion; back them up with references or personal experience.


Use MathJax to format equations. MathJax reference.


To learn more, see our tips on writing great answers.




draft saved


draft discarded














StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3054894%2fa-chord-of-length-6-subtends-an-80-circ-central-angle-in-a-circle-can-we-c%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown





















































Required, but never shown














Required, but never shown












Required, but never shown







Required, but never shown

































Required, but never shown














Required, but never shown












Required, but never shown







Required, but never shown







Popular posts from this blog

Bundesstraße 106

Verónica Boquete

Ida-Boy-Ed-Garten