Slant Angle of a Cone in Comparison with the Degree of a Circle Sector












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My friend and I were building paper cones made from circles in which we cut out sectors of the circle and joined the two sides together. We were wondering about the relationship between the angle of the sector of the circle we cut out and the inclined angle of the cone the circle eventually made. While doing the math, when cutting the circle in half, a 180 degree sector, we eventually figured out the slant angle was somewhere around 56.944 degrees using cosine and the radius and slant height of the cone. I happened to notice that number was also very close to 180/pi, or 57.296 degrees. Is there any relationship between these two? Or is it coincidence that they're close in number? Could anyone explain to me what the relationship between these two degrees are?






geometry angle







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    My friend and I were building paper cones made from circles in which we cut out sectors of the circle and joined the two sides together. We were wondering about the relationship between the angle of the sector of the circle we cut out and the inclined angle of the cone the circle eventually made. While doing the math, when cutting the circle in half, a 180 degree sector, we eventually figured out the slant angle was somewhere around 56.944 degrees using cosine and the radius and slant height of the cone. I happened to notice that number was also very close to 180/pi, or 57.296 degrees. Is there any relationship between these two? Or is it coincidence that they're close in number? Could anyone explain to me what the relationship between these two degrees are?






    geometry angle







    share|cite|improve this question









    $endgroup$















      1












      1








      1





      $begingroup$


      My friend and I were building paper cones made from circles in which we cut out sectors of the circle and joined the two sides together. We were wondering about the relationship between the angle of the sector of the circle we cut out and the inclined angle of the cone the circle eventually made. While doing the math, when cutting the circle in half, a 180 degree sector, we eventually figured out the slant angle was somewhere around 56.944 degrees using cosine and the radius and slant height of the cone. I happened to notice that number was also very close to 180/pi, or 57.296 degrees. Is there any relationship between these two? Or is it coincidence that they're close in number? Could anyone explain to me what the relationship between these two degrees are?






      geometry angle







      share|cite|improve this question









      $endgroup$




      My friend and I were building paper cones made from circles in which we cut out sectors of the circle and joined the two sides together. We were wondering about the relationship between the angle of the sector of the circle we cut out and the inclined angle of the cone the circle eventually made. While doing the math, when cutting the circle in half, a 180 degree sector, we eventually figured out the slant angle was somewhere around 56.944 degrees using cosine and the radius and slant height of the cone. I happened to notice that number was also very close to 180/pi, or 57.296 degrees. Is there any relationship between these two? Or is it coincidence that they're close in number? Could anyone explain to me what the relationship between these two degrees are?






      geometry angle




      geometry angle






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      asked Jan 29 '16 at 0:39









      Ben LinkBen Link

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          $begingroup$

          The projection of slant height $L$ is the cone radius $ r$, so semi-vertical angle of cone is:



          $$ sin alpha = frac{r}{L} = dfrac{gamma}{2 pi}$$



          $$ dfrac{gamma}{sin alpha} = 2 pi $$



          is the relation you are asking for. Accordingly the fully developed angle $gamma $ at center of sector is



          $$ 2 pi sin alpha $$



          In your case semi-vertex angle is $ sin^{-1}dfrac12 = sin^{-1} dfrac{pi}{2 pi} = 30^{circ}$ from straight angle $pi$ to full circle $2 pi $ ratio $=dfrac12$.



          If you repeat the experiment with little more care on a larger/stiffer sheet the double angle would be exactly $60^{circ}$, not $180/pi $ degrees or a radian.






          share|cite|improve this answer











          $endgroup$





















            0












            $begingroup$

            The length of a circular arc is directly proportional to its central angle, so when you cut out a sector of angle $theta$ and glue the cut edges together, the circumference of the base of the cone is going to be smaller than that of the original circle by this proportional amount: $c=C-theta R$. Since the circumference of a circle is $2pi$ times its radius, we then have for the radius of the cone’s base $$r = left(1-fractheta{2pi}right)R.$$



            The slant height of the cone is the original circle’s radius $R$, so $$sin alpha=frac r R=1-fractheta{2pi},$$ where $alpha$ is the half-angle at the cone’s vertex.



            If you cut out half of the circle, you’ll have $sinalpha=frac12$, or $alpha=fracpi6$, which is $30$ degrees.






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              $r,theta$ are not (directly) proportional to each other in the given first relation. Given linear relation of negative slope is not correct. Answer happened to be right as $ 180$ degrees stands midway between $0,360.$.
              $endgroup$
              – Narasimham
              Jan 12 '18 at 16:52












            • $begingroup$
              @Narasimham Aside from a quibble about my wording, where’s the error? This derivation is essentially the same as yours, the only difference being that I used the angle $theta$ of the deleted sector, whereas you used the angle of the remaining sector $gamma = 2pi-theta$.
              $endgroup$
              – amd
              Jan 12 '18 at 19:04










            • $begingroup$
              Oh,ok ..then if it is wording and English usage only, ..because what I understood is the sector part of paper that remains after the cutting.
              $endgroup$
              – Narasimham
              Jan 12 '18 at 21:45










            • $begingroup$
              @Narasimham I’ll make some edits to make this clearer. Thanks!
              $endgroup$
              – amd
              Jan 12 '18 at 22:11











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            2 Answers
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            2 Answers
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            0












            $begingroup$

            The projection of slant height $L$ is the cone radius $ r$, so semi-vertical angle of cone is:



            $$ sin alpha = frac{r}{L} = dfrac{gamma}{2 pi}$$



            $$ dfrac{gamma}{sin alpha} = 2 pi $$



            is the relation you are asking for. Accordingly the fully developed angle $gamma $ at center of sector is



            $$ 2 pi sin alpha $$



            In your case semi-vertex angle is $ sin^{-1}dfrac12 = sin^{-1} dfrac{pi}{2 pi} = 30^{circ}$ from straight angle $pi$ to full circle $2 pi $ ratio $=dfrac12$.



            If you repeat the experiment with little more care on a larger/stiffer sheet the double angle would be exactly $60^{circ}$, not $180/pi $ degrees or a radian.






            share|cite|improve this answer











            $endgroup$


















              0












              $begingroup$

              The projection of slant height $L$ is the cone radius $ r$, so semi-vertical angle of cone is:



              $$ sin alpha = frac{r}{L} = dfrac{gamma}{2 pi}$$



              $$ dfrac{gamma}{sin alpha} = 2 pi $$



              is the relation you are asking for. Accordingly the fully developed angle $gamma $ at center of sector is



              $$ 2 pi sin alpha $$



              In your case semi-vertex angle is $ sin^{-1}dfrac12 = sin^{-1} dfrac{pi}{2 pi} = 30^{circ}$ from straight angle $pi$ to full circle $2 pi $ ratio $=dfrac12$.



              If you repeat the experiment with little more care on a larger/stiffer sheet the double angle would be exactly $60^{circ}$, not $180/pi $ degrees or a radian.






              share|cite|improve this answer











              $endgroup$
















                0












                0








                0





                $begingroup$

                The projection of slant height $L$ is the cone radius $ r$, so semi-vertical angle of cone is:



                $$ sin alpha = frac{r}{L} = dfrac{gamma}{2 pi}$$



                $$ dfrac{gamma}{sin alpha} = 2 pi $$



                is the relation you are asking for. Accordingly the fully developed angle $gamma $ at center of sector is



                $$ 2 pi sin alpha $$



                In your case semi-vertex angle is $ sin^{-1}dfrac12 = sin^{-1} dfrac{pi}{2 pi} = 30^{circ}$ from straight angle $pi$ to full circle $2 pi $ ratio $=dfrac12$.



                If you repeat the experiment with little more care on a larger/stiffer sheet the double angle would be exactly $60^{circ}$, not $180/pi $ degrees or a radian.






                share|cite|improve this answer











                $endgroup$



                The projection of slant height $L$ is the cone radius $ r$, so semi-vertical angle of cone is:



                $$ sin alpha = frac{r}{L} = dfrac{gamma}{2 pi}$$



                $$ dfrac{gamma}{sin alpha} = 2 pi $$



                is the relation you are asking for. Accordingly the fully developed angle $gamma $ at center of sector is



                $$ 2 pi sin alpha $$



                In your case semi-vertex angle is $ sin^{-1}dfrac12 = sin^{-1} dfrac{pi}{2 pi} = 30^{circ}$ from straight angle $pi$ to full circle $2 pi $ ratio $=dfrac12$.



                If you repeat the experiment with little more care on a larger/stiffer sheet the double angle would be exactly $60^{circ}$, not $180/pi $ degrees or a radian.







                share|cite|improve this answer














                share|cite|improve this answer



                share|cite|improve this answer








                edited Jan 12 '18 at 21:40

























                answered Jan 12 '18 at 15:49









                NarasimhamNarasimham

                20.9k62158




                20.9k62158























                    0












                    $begingroup$

                    The length of a circular arc is directly proportional to its central angle, so when you cut out a sector of angle $theta$ and glue the cut edges together, the circumference of the base of the cone is going to be smaller than that of the original circle by this proportional amount: $c=C-theta R$. Since the circumference of a circle is $2pi$ times its radius, we then have for the radius of the cone’s base $$r = left(1-fractheta{2pi}right)R.$$



                    The slant height of the cone is the original circle’s radius $R$, so $$sin alpha=frac r R=1-fractheta{2pi},$$ where $alpha$ is the half-angle at the cone’s vertex.



                    If you cut out half of the circle, you’ll have $sinalpha=frac12$, or $alpha=fracpi6$, which is $30$ degrees.






                    share|cite|improve this answer











                    $endgroup$













                    • $begingroup$
                      $r,theta$ are not (directly) proportional to each other in the given first relation. Given linear relation of negative slope is not correct. Answer happened to be right as $ 180$ degrees stands midway between $0,360.$.
                      $endgroup$
                      – Narasimham
                      Jan 12 '18 at 16:52












                    • $begingroup$
                      @Narasimham Aside from a quibble about my wording, where’s the error? This derivation is essentially the same as yours, the only difference being that I used the angle $theta$ of the deleted sector, whereas you used the angle of the remaining sector $gamma = 2pi-theta$.
                      $endgroup$
                      – amd
                      Jan 12 '18 at 19:04










                    • $begingroup$
                      Oh,ok ..then if it is wording and English usage only, ..because what I understood is the sector part of paper that remains after the cutting.
                      $endgroup$
                      – Narasimham
                      Jan 12 '18 at 21:45










                    • $begingroup$
                      @Narasimham I’ll make some edits to make this clearer. Thanks!
                      $endgroup$
                      – amd
                      Jan 12 '18 at 22:11
















                    0












                    $begingroup$

                    The length of a circular arc is directly proportional to its central angle, so when you cut out a sector of angle $theta$ and glue the cut edges together, the circumference of the base of the cone is going to be smaller than that of the original circle by this proportional amount: $c=C-theta R$. Since the circumference of a circle is $2pi$ times its radius, we then have for the radius of the cone’s base $$r = left(1-fractheta{2pi}right)R.$$



                    The slant height of the cone is the original circle’s radius $R$, so $$sin alpha=frac r R=1-fractheta{2pi},$$ where $alpha$ is the half-angle at the cone’s vertex.



                    If you cut out half of the circle, you’ll have $sinalpha=frac12$, or $alpha=fracpi6$, which is $30$ degrees.






                    share|cite|improve this answer











                    $endgroup$













                    • $begingroup$
                      $r,theta$ are not (directly) proportional to each other in the given first relation. Given linear relation of negative slope is not correct. Answer happened to be right as $ 180$ degrees stands midway between $0,360.$.
                      $endgroup$
                      – Narasimham
                      Jan 12 '18 at 16:52












                    • $begingroup$
                      @Narasimham Aside from a quibble about my wording, where’s the error? This derivation is essentially the same as yours, the only difference being that I used the angle $theta$ of the deleted sector, whereas you used the angle of the remaining sector $gamma = 2pi-theta$.
                      $endgroup$
                      – amd
                      Jan 12 '18 at 19:04










                    • $begingroup$
                      Oh,ok ..then if it is wording and English usage only, ..because what I understood is the sector part of paper that remains after the cutting.
                      $endgroup$
                      – Narasimham
                      Jan 12 '18 at 21:45










                    • $begingroup$
                      @Narasimham I’ll make some edits to make this clearer. Thanks!
                      $endgroup$
                      – amd
                      Jan 12 '18 at 22:11














                    0












                    0








                    0





                    $begingroup$

                    The length of a circular arc is directly proportional to its central angle, so when you cut out a sector of angle $theta$ and glue the cut edges together, the circumference of the base of the cone is going to be smaller than that of the original circle by this proportional amount: $c=C-theta R$. Since the circumference of a circle is $2pi$ times its radius, we then have for the radius of the cone’s base $$r = left(1-fractheta{2pi}right)R.$$



                    The slant height of the cone is the original circle’s radius $R$, so $$sin alpha=frac r R=1-fractheta{2pi},$$ where $alpha$ is the half-angle at the cone’s vertex.



                    If you cut out half of the circle, you’ll have $sinalpha=frac12$, or $alpha=fracpi6$, which is $30$ degrees.






                    share|cite|improve this answer











                    $endgroup$



                    The length of a circular arc is directly proportional to its central angle, so when you cut out a sector of angle $theta$ and glue the cut edges together, the circumference of the base of the cone is going to be smaller than that of the original circle by this proportional amount: $c=C-theta R$. Since the circumference of a circle is $2pi$ times its radius, we then have for the radius of the cone’s base $$r = left(1-fractheta{2pi}right)R.$$



                    The slant height of the cone is the original circle’s radius $R$, so $$sin alpha=frac r R=1-fractheta{2pi},$$ where $alpha$ is the half-angle at the cone’s vertex.



                    If you cut out half of the circle, you’ll have $sinalpha=frac12$, or $alpha=fracpi6$, which is $30$ degrees.







                    share|cite|improve this answer














                    share|cite|improve this answer



                    share|cite|improve this answer








                    edited Jan 13 '18 at 21:35

























                    answered Jan 29 '16 at 1:54









                    amdamd

                    30.9k21051




                    30.9k21051












                    • $begingroup$
                      $r,theta$ are not (directly) proportional to each other in the given first relation. Given linear relation of negative slope is not correct. Answer happened to be right as $ 180$ degrees stands midway between $0,360.$.
                      $endgroup$
                      – Narasimham
                      Jan 12 '18 at 16:52












                    • $begingroup$
                      @Narasimham Aside from a quibble about my wording, where’s the error? This derivation is essentially the same as yours, the only difference being that I used the angle $theta$ of the deleted sector, whereas you used the angle of the remaining sector $gamma = 2pi-theta$.
                      $endgroup$
                      – amd
                      Jan 12 '18 at 19:04










                    • $begingroup$
                      Oh,ok ..then if it is wording and English usage only, ..because what I understood is the sector part of paper that remains after the cutting.
                      $endgroup$
                      – Narasimham
                      Jan 12 '18 at 21:45










                    • $begingroup$
                      @Narasimham I’ll make some edits to make this clearer. Thanks!
                      $endgroup$
                      – amd
                      Jan 12 '18 at 22:11


















                    • $begingroup$
                      $r,theta$ are not (directly) proportional to each other in the given first relation. Given linear relation of negative slope is not correct. Answer happened to be right as $ 180$ degrees stands midway between $0,360.$.
                      $endgroup$
                      – Narasimham
                      Jan 12 '18 at 16:52












                    • $begingroup$
                      @Narasimham Aside from a quibble about my wording, where’s the error? This derivation is essentially the same as yours, the only difference being that I used the angle $theta$ of the deleted sector, whereas you used the angle of the remaining sector $gamma = 2pi-theta$.
                      $endgroup$
                      – amd
                      Jan 12 '18 at 19:04










                    • $begingroup$
                      Oh,ok ..then if it is wording and English usage only, ..because what I understood is the sector part of paper that remains after the cutting.
                      $endgroup$
                      – Narasimham
                      Jan 12 '18 at 21:45










                    • $begingroup$
                      @Narasimham I’ll make some edits to make this clearer. Thanks!
                      $endgroup$
                      – amd
                      Jan 12 '18 at 22:11
















                    $begingroup$
                    $r,theta$ are not (directly) proportional to each other in the given first relation. Given linear relation of negative slope is not correct. Answer happened to be right as $ 180$ degrees stands midway between $0,360.$.
                    $endgroup$
                    – Narasimham
                    Jan 12 '18 at 16:52






                    $begingroup$
                    $r,theta$ are not (directly) proportional to each other in the given first relation. Given linear relation of negative slope is not correct. Answer happened to be right as $ 180$ degrees stands midway between $0,360.$.
                    $endgroup$
                    – Narasimham
                    Jan 12 '18 at 16:52














                    $begingroup$
                    @Narasimham Aside from a quibble about my wording, where’s the error? This derivation is essentially the same as yours, the only difference being that I used the angle $theta$ of the deleted sector, whereas you used the angle of the remaining sector $gamma = 2pi-theta$.
                    $endgroup$
                    – amd
                    Jan 12 '18 at 19:04




                    $begingroup$
                    @Narasimham Aside from a quibble about my wording, where’s the error? This derivation is essentially the same as yours, the only difference being that I used the angle $theta$ of the deleted sector, whereas you used the angle of the remaining sector $gamma = 2pi-theta$.
                    $endgroup$
                    – amd
                    Jan 12 '18 at 19:04












                    $begingroup$
                    Oh,ok ..then if it is wording and English usage only, ..because what I understood is the sector part of paper that remains after the cutting.
                    $endgroup$
                    – Narasimham
                    Jan 12 '18 at 21:45




                    $begingroup$
                    Oh,ok ..then if it is wording and English usage only, ..because what I understood is the sector part of paper that remains after the cutting.
                    $endgroup$
                    – Narasimham
                    Jan 12 '18 at 21:45












                    $begingroup$
                    @Narasimham I’ll make some edits to make this clearer. Thanks!
                    $endgroup$
                    – amd
                    Jan 12 '18 at 22:11




                    $begingroup$
                    @Narasimham I’ll make some edits to make this clearer. Thanks!
                    $endgroup$
                    – amd
                    Jan 12 '18 at 22:11


















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