Finding angle of sector which forms a cone
$begingroup$
To find the angle where it is in rad,
am I right to say that $10*(angle in rad)=2pi*(radius of cone)$
calculus
$endgroup$
add a comment |
$begingroup$
To find the angle where it is in rad,
am I right to say that $10*(angle in rad)=2pi*(radius of cone)$
calculus
$endgroup$
add a comment |
$begingroup$
To find the angle where it is in rad,
am I right to say that $10*(angle in rad)=2pi*(radius of cone)$
calculus
$endgroup$
To find the angle where it is in rad,
am I right to say that $10*(angle in rad)=2pi*(radius of cone)$
calculus
calculus
edited Jul 27 '16 at 17:40
Widawensen
4,51321446
4,51321446
asked Mar 27 '14 at 11:31
Jake MitchJake Mitch
847
847
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
The radius of the sector is the lateral height $l$ of the cone, and the length of the arc is the circumference of the base, that is, $2pi r$. So the angle you ask for is
$$theta=frac{2pi r}{2pi l}=frac rl$$
measured in radians.
$endgroup$
$begingroup$
Why do you divide by $2pi l$ instead of just $l$?
$endgroup$
– Christoph
Mar 27 '14 at 11:56
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
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
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f728900%2ffinding-angle-of-sector-which-forms-a-cone%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The radius of the sector is the lateral height $l$ of the cone, and the length of the arc is the circumference of the base, that is, $2pi r$. So the angle you ask for is
$$theta=frac{2pi r}{2pi l}=frac rl$$
measured in radians.
$endgroup$
$begingroup$
Why do you divide by $2pi l$ instead of just $l$?
$endgroup$
– Christoph
Mar 27 '14 at 11:56
add a comment |
$begingroup$
The radius of the sector is the lateral height $l$ of the cone, and the length of the arc is the circumference of the base, that is, $2pi r$. So the angle you ask for is
$$theta=frac{2pi r}{2pi l}=frac rl$$
measured in radians.
$endgroup$
$begingroup$
Why do you divide by $2pi l$ instead of just $l$?
$endgroup$
– Christoph
Mar 27 '14 at 11:56
add a comment |
$begingroup$
The radius of the sector is the lateral height $l$ of the cone, and the length of the arc is the circumference of the base, that is, $2pi r$. So the angle you ask for is
$$theta=frac{2pi r}{2pi l}=frac rl$$
measured in radians.
$endgroup$
The radius of the sector is the lateral height $l$ of the cone, and the length of the arc is the circumference of the base, that is, $2pi r$. So the angle you ask for is
$$theta=frac{2pi r}{2pi l}=frac rl$$
measured in radians.
answered Mar 27 '14 at 11:40
ajotatxeajotatxe
53.8k23990
53.8k23990
$begingroup$
Why do you divide by $2pi l$ instead of just $l$?
$endgroup$
– Christoph
Mar 27 '14 at 11:56
add a comment |
$begingroup$
Why do you divide by $2pi l$ instead of just $l$?
$endgroup$
– Christoph
Mar 27 '14 at 11:56
$begingroup$
Why do you divide by $2pi l$ instead of just $l$?
$endgroup$
– Christoph
Mar 27 '14 at 11:56
$begingroup$
Why do you divide by $2pi l$ instead of just $l$?
$endgroup$
– Christoph
Mar 27 '14 at 11:56
add a comment |
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.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f728900%2ffinding-angle-of-sector-which-forms-a-cone%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
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