How do we evaluate the degree of $x$ using sine law?












2












$begingroup$



enter image description here



Given that



ABC is a triangle, $|EC| = |BC| = |BD| $, $angle CBA= 80^circ,angle ACB= 60^circ, angle EDA= x^circ $



Evaluate $x$




I want to solve this for $x$ using law of sines if possible.



My attempt:



From the property of triangle, the sum of the angles will be equal to $180$.



$$angle BAC = 180 - 80 - 60 = 40^circ $$



In $triangle ABC$,



$$frac{sin 40}{|BC|} = frac{sin 80}{|AC|} implies frac{|AC|}{|BC|} = frac{sin 80}{sin40}$$



Could you help me take it from there?



Regards






geometry trigonometry







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$endgroup$

















    2












    $begingroup$



    enter image description here



    Given that



    ABC is a triangle, $|EC| = |BC| = |BD| $, $angle CBA= 80^circ,angle ACB= 60^circ, angle EDA= x^circ $



    Evaluate $x$




    I want to solve this for $x$ using law of sines if possible.



    My attempt:



    From the property of triangle, the sum of the angles will be equal to $180$.



    $$angle BAC = 180 - 80 - 60 = 40^circ $$



    In $triangle ABC$,



    $$frac{sin 40}{|BC|} = frac{sin 80}{|AC|} implies frac{|AC|}{|BC|} = frac{sin 80}{sin40}$$



    Could you help me take it from there?



    Regards






    geometry trigonometry







    share|cite|improve this question









    $endgroup$















      2












      2








      2


      1



      $begingroup$



      enter image description here



      Given that



      ABC is a triangle, $|EC| = |BC| = |BD| $, $angle CBA= 80^circ,angle ACB= 60^circ, angle EDA= x^circ $



      Evaluate $x$




      I want to solve this for $x$ using law of sines if possible.



      My attempt:



      From the property of triangle, the sum of the angles will be equal to $180$.



      $$angle BAC = 180 - 80 - 60 = 40^circ $$



      In $triangle ABC$,



      $$frac{sin 40}{|BC|} = frac{sin 80}{|AC|} implies frac{|AC|}{|BC|} = frac{sin 80}{sin40}$$



      Could you help me take it from there?



      Regards






      geometry trigonometry







      share|cite|improve this question









      $endgroup$





      enter image description here



      Given that



      ABC is a triangle, $|EC| = |BC| = |BD| $, $angle CBA= 80^circ,angle ACB= 60^circ, angle EDA= x^circ $



      Evaluate $x$




      I want to solve this for $x$ using law of sines if possible.



      My attempt:



      From the property of triangle, the sum of the angles will be equal to $180$.



      $$angle BAC = 180 - 80 - 60 = 40^circ $$



      In $triangle ABC$,



      $$frac{sin 40}{|BC|} = frac{sin 80}{|AC|} implies frac{|AC|}{|BC|} = frac{sin 80}{sin40}$$



      Could you help me take it from there?



      Regards






      geometry trigonometry




      geometry trigonometry






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Nov 30 '18 at 12:11









      HamiltonHamilton

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      1798






















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          $begingroup$

          You don't need trigonometry to solve for $x$. Observe that $BCE$ is an equilateral triangle, so $BE=BC$, so $BDE$ is an isosceles triangle. From this, it's easy to see that $x=100°$.






          share|cite|improve this answer











          $endgroup$





















            2












            $begingroup$

            Since $|BC|=|EC|$ and $angle BCE=60$ then $triangle BCE$ is equilateral.



            Hence $|BE|=|BC|=|BD|$, so that $triangle BDE$ is isosceles with $angle DBE=80°-60°=20°$.



            Hence, $angle BDE=(180°-20°)/2=80°$ so that $angle ADE=180°-80°=100°$.






            share|cite|improve this answer









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              $begingroup$

              You don't need trigonometry to solve for $x$. Observe that $BCE$ is an equilateral triangle, so $BE=BC$, so $BDE$ is an isosceles triangle. From this, it's easy to see that $x=100°$.






              share|cite|improve this answer











              $endgroup$


















                3












                $begingroup$

                You don't need trigonometry to solve for $x$. Observe that $BCE$ is an equilateral triangle, so $BE=BC$, so $BDE$ is an isosceles triangle. From this, it's easy to see that $x=100°$.






                share|cite|improve this answer











                $endgroup$
















                  3












                  3








                  3





                  $begingroup$

                  You don't need trigonometry to solve for $x$. Observe that $BCE$ is an equilateral triangle, so $BE=BC$, so $BDE$ is an isosceles triangle. From this, it's easy to see that $x=100°$.






                  share|cite|improve this answer











                  $endgroup$



                  You don't need trigonometry to solve for $x$. Observe that $BCE$ is an equilateral triangle, so $BE=BC$, so $BDE$ is an isosceles triangle. From this, it's easy to see that $x=100°$.







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited Nov 30 '18 at 13:46

























                  answered Nov 30 '18 at 12:20









                  GNUSupporter 8964民主女神 地下教會GNUSupporter 8964民主女神 地下教會

                  12.7k72445




                  12.7k72445























                      2












                      $begingroup$

                      Since $|BC|=|EC|$ and $angle BCE=60$ then $triangle BCE$ is equilateral.



                      Hence $|BE|=|BC|=|BD|$, so that $triangle BDE$ is isosceles with $angle DBE=80°-60°=20°$.



                      Hence, $angle BDE=(180°-20°)/2=80°$ so that $angle ADE=180°-80°=100°$.






                      share|cite|improve this answer









                      $endgroup$


















                        2












                        $begingroup$

                        Since $|BC|=|EC|$ and $angle BCE=60$ then $triangle BCE$ is equilateral.



                        Hence $|BE|=|BC|=|BD|$, so that $triangle BDE$ is isosceles with $angle DBE=80°-60°=20°$.



                        Hence, $angle BDE=(180°-20°)/2=80°$ so that $angle ADE=180°-80°=100°$.






                        share|cite|improve this answer









                        $endgroup$
















                          2












                          2








                          2





                          $begingroup$

                          Since $|BC|=|EC|$ and $angle BCE=60$ then $triangle BCE$ is equilateral.



                          Hence $|BE|=|BC|=|BD|$, so that $triangle BDE$ is isosceles with $angle DBE=80°-60°=20°$.



                          Hence, $angle BDE=(180°-20°)/2=80°$ so that $angle ADE=180°-80°=100°$.






                          share|cite|improve this answer









                          $endgroup$



                          Since $|BC|=|EC|$ and $angle BCE=60$ then $triangle BCE$ is equilateral.



                          Hence $|BE|=|BC|=|BD|$, so that $triangle BDE$ is isosceles with $angle DBE=80°-60°=20°$.



                          Hence, $angle BDE=(180°-20°)/2=80°$ so that $angle ADE=180°-80°=100°$.







                          share|cite|improve this answer












                          share|cite|improve this answer



                          share|cite|improve this answer










                          answered Nov 30 '18 at 12:33









                          AntinousAntinous

                          5,71042051




                          5,71042051






























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