Trigonometric solution for finding the lengths of an obtuse triangle given the base, angle, & area












0












$begingroup$



Given a triangle ABC where $measuredangle BAC = 120^circ$, $BC = sqrt{37}$, and area $3sqrt{3}$, find the lengths AB and AC.




Here is my attempt:
enter image description here



At first I thought of doing it by getting BA by $BA = BP - AP$ then use the two lengths and angle formula for triangle area having $sin alpha$ as my variable, but the problem is that I cannot define BP in terms of $sinalpha$, and get an equation with $cos alpha$ instead where they have different factors that prevent me from applying any trigonometric identity.

I've also tried doing $3sqrt{3} = frac{PC cdot BA}{2}$, but I run into the same problem.



Any other solution I've tried (like the Pythagorean theorem) leaves me with a trigonometric identity equation where I cannot extract anything useful.






geometry trigonometry







share|cite|improve this question









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    0












    $begingroup$



    Given a triangle ABC where $measuredangle BAC = 120^circ$, $BC = sqrt{37}$, and area $3sqrt{3}$, find the lengths AB and AC.




    Here is my attempt:
    enter image description here



    At first I thought of doing it by getting BA by $BA = BP - AP$ then use the two lengths and angle formula for triangle area having $sin alpha$ as my variable, but the problem is that I cannot define BP in terms of $sinalpha$, and get an equation with $cos alpha$ instead where they have different factors that prevent me from applying any trigonometric identity.

    I've also tried doing $3sqrt{3} = frac{PC cdot BA}{2}$, but I run into the same problem.



    Any other solution I've tried (like the Pythagorean theorem) leaves me with a trigonometric identity equation where I cannot extract anything useful.






    geometry trigonometry







    share|cite|improve this question









    $endgroup$















      0












      0








      0





      $begingroup$



      Given a triangle ABC where $measuredangle BAC = 120^circ$, $BC = sqrt{37}$, and area $3sqrt{3}$, find the lengths AB and AC.




      Here is my attempt:
      enter image description here



      At first I thought of doing it by getting BA by $BA = BP - AP$ then use the two lengths and angle formula for triangle area having $sin alpha$ as my variable, but the problem is that I cannot define BP in terms of $sinalpha$, and get an equation with $cos alpha$ instead where they have different factors that prevent me from applying any trigonometric identity.

      I've also tried doing $3sqrt{3} = frac{PC cdot BA}{2}$, but I run into the same problem.



      Any other solution I've tried (like the Pythagorean theorem) leaves me with a trigonometric identity equation where I cannot extract anything useful.






      geometry trigonometry







      share|cite|improve this question









      $endgroup$





      Given a triangle ABC where $measuredangle BAC = 120^circ$, $BC = sqrt{37}$, and area $3sqrt{3}$, find the lengths AB and AC.




      Here is my attempt:
      enter image description here



      At first I thought of doing it by getting BA by $BA = BP - AP$ then use the two lengths and angle formula for triangle area having $sin alpha$ as my variable, but the problem is that I cannot define BP in terms of $sinalpha$, and get an equation with $cos alpha$ instead where they have different factors that prevent me from applying any trigonometric identity.

      I've also tried doing $3sqrt{3} = frac{PC cdot BA}{2}$, but I run into the same problem.



      Any other solution I've tried (like the Pythagorean theorem) leaves me with a trigonometric identity equation where I cannot extract anything useful.






      geometry trigonometry




      geometry trigonometry






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




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      asked Jan 10 at 12:36









      daedsidogdaedsidog

      29017




      29017






















          1 Answer
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          $begingroup$

          Guide:



          Let $AB=y$,



          By cosine rule $$x^2+y^2-2xy cos 120^circ = 37$$



          From the area, we know that



          $$frac12 xy sin 120^circ = 3sqrt3$$



          $$xy=12tag{1}$$



          and $$x^2+y^2+12=37$$



          $$x^2+y^2=25=5^2tag{2}$$



          Can you solve for $x$ and $y$ from $(1)$ and $(2)$? (Further hint: I purpose write $25=5^2$, if you are familiar with certain identities, the answer could be very obvious).






          share|cite|improve this answer









          $endgroup$













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            1












            $begingroup$

            Guide:



            Let $AB=y$,



            By cosine rule $$x^2+y^2-2xy cos 120^circ = 37$$



            From the area, we know that



            $$frac12 xy sin 120^circ = 3sqrt3$$



            $$xy=12tag{1}$$



            and $$x^2+y^2+12=37$$



            $$x^2+y^2=25=5^2tag{2}$$



            Can you solve for $x$ and $y$ from $(1)$ and $(2)$? (Further hint: I purpose write $25=5^2$, if you are familiar with certain identities, the answer could be very obvious).






            share|cite|improve this answer









            $endgroup$


















              1












              $begingroup$

              Guide:



              Let $AB=y$,



              By cosine rule $$x^2+y^2-2xy cos 120^circ = 37$$



              From the area, we know that



              $$frac12 xy sin 120^circ = 3sqrt3$$



              $$xy=12tag{1}$$



              and $$x^2+y^2+12=37$$



              $$x^2+y^2=25=5^2tag{2}$$



              Can you solve for $x$ and $y$ from $(1)$ and $(2)$? (Further hint: I purpose write $25=5^2$, if you are familiar with certain identities, the answer could be very obvious).






              share|cite|improve this answer









              $endgroup$
















                1












                1








                1





                $begingroup$

                Guide:



                Let $AB=y$,



                By cosine rule $$x^2+y^2-2xy cos 120^circ = 37$$



                From the area, we know that



                $$frac12 xy sin 120^circ = 3sqrt3$$



                $$xy=12tag{1}$$



                and $$x^2+y^2+12=37$$



                $$x^2+y^2=25=5^2tag{2}$$



                Can you solve for $x$ and $y$ from $(1)$ and $(2)$? (Further hint: I purpose write $25=5^2$, if you are familiar with certain identities, the answer could be very obvious).






                share|cite|improve this answer









                $endgroup$



                Guide:



                Let $AB=y$,



                By cosine rule $$x^2+y^2-2xy cos 120^circ = 37$$



                From the area, we know that



                $$frac12 xy sin 120^circ = 3sqrt3$$



                $$xy=12tag{1}$$



                and $$x^2+y^2+12=37$$



                $$x^2+y^2=25=5^2tag{2}$$



                Can you solve for $x$ and $y$ from $(1)$ and $(2)$? (Further hint: I purpose write $25=5^2$, if you are familiar with certain identities, the answer could be very obvious).







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Jan 10 at 12:48









                Siong Thye GohSiong Thye Goh

                101k1466118




                101k1466118






























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