How to make correct system of equations to solve for the angles in this triangle?












0












$begingroup$


enter image description here



I'm trying to solve this triangle for $X$. Thereby, I've tried to make correct system of equations. What would be the correct equations?



Here are the equations I can find




  • In $triangle ABC$, recalling that $angle ACD = y$


$$48 + 24 + x + 12 + y = 180$$




  • In $triangle ADC$


$$84+x + y = 180$$



However, I'm getting the same equations.



Regards






triangle







share|cite|improve this question











$endgroup$












  • $begingroup$
    Which equations can you find?
    $endgroup$
    – Arthur
    Dec 1 '18 at 10:34










  • $begingroup$
    @Arthur See the update, please.
    $endgroup$
    – Hamilton
    Dec 1 '18 at 10:57












  • $begingroup$
    Geogebra suggests the $x=80$.
    $endgroup$
    – Michael Hoppe
    Dec 1 '18 at 11:44






  • 1




    $begingroup$
    No, the information is sufficient: just try to construct.
    $endgroup$
    – Michael Hoppe
    Dec 1 '18 at 12:10












  • $begingroup$
    I was wondering can someone find $x$ without using trignometry? And is it better to ask this as a seperate question?
    $endgroup$
    – Fareed AF
    Dec 8 '18 at 14:13
















0












$begingroup$


enter image description here



I'm trying to solve this triangle for $X$. Thereby, I've tried to make correct system of equations. What would be the correct equations?



Here are the equations I can find




  • In $triangle ABC$, recalling that $angle ACD = y$


$$48 + 24 + x + 12 + y = 180$$




  • In $triangle ADC$


$$84+x + y = 180$$



However, I'm getting the same equations.



Regards






triangle







share|cite|improve this question











$endgroup$












  • $begingroup$
    Which equations can you find?
    $endgroup$
    – Arthur
    Dec 1 '18 at 10:34










  • $begingroup$
    @Arthur See the update, please.
    $endgroup$
    – Hamilton
    Dec 1 '18 at 10:57












  • $begingroup$
    Geogebra suggests the $x=80$.
    $endgroup$
    – Michael Hoppe
    Dec 1 '18 at 11:44






  • 1




    $begingroup$
    No, the information is sufficient: just try to construct.
    $endgroup$
    – Michael Hoppe
    Dec 1 '18 at 12:10












  • $begingroup$
    I was wondering can someone find $x$ without using trignometry? And is it better to ask this as a seperate question?
    $endgroup$
    – Fareed AF
    Dec 8 '18 at 14:13














0












0








0


1



$begingroup$


enter image description here



I'm trying to solve this triangle for $X$. Thereby, I've tried to make correct system of equations. What would be the correct equations?



Here are the equations I can find




  • In $triangle ABC$, recalling that $angle ACD = y$


$$48 + 24 + x + 12 + y = 180$$




  • In $triangle ADC$


$$84+x + y = 180$$



However, I'm getting the same equations.



Regards






triangle







share|cite|improve this question











$endgroup$




enter image description here



I'm trying to solve this triangle for $X$. Thereby, I've tried to make correct system of equations. What would be the correct equations?



Here are the equations I can find




  • In $triangle ABC$, recalling that $angle ACD = y$


$$48 + 24 + x + 12 + y = 180$$




  • In $triangle ADC$


$$84+x + y = 180$$



However, I'm getting the same equations.



Regards






triangle




triangle






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 1 '18 at 10:57







Hamilton

















asked Dec 1 '18 at 10:09









HamiltonHamilton

1798




1798












  • $begingroup$
    Which equations can you find?
    $endgroup$
    – Arthur
    Dec 1 '18 at 10:34










  • $begingroup$
    @Arthur See the update, please.
    $endgroup$
    – Hamilton
    Dec 1 '18 at 10:57












  • $begingroup$
    Geogebra suggests the $x=80$.
    $endgroup$
    – Michael Hoppe
    Dec 1 '18 at 11:44






  • 1




    $begingroup$
    No, the information is sufficient: just try to construct.
    $endgroup$
    – Michael Hoppe
    Dec 1 '18 at 12:10












  • $begingroup$
    I was wondering can someone find $x$ without using trignometry? And is it better to ask this as a seperate question?
    $endgroup$
    – Fareed AF
    Dec 8 '18 at 14:13


















  • $begingroup$
    Which equations can you find?
    $endgroup$
    – Arthur
    Dec 1 '18 at 10:34










  • $begingroup$
    @Arthur See the update, please.
    $endgroup$
    – Hamilton
    Dec 1 '18 at 10:57












  • $begingroup$
    Geogebra suggests the $x=80$.
    $endgroup$
    – Michael Hoppe
    Dec 1 '18 at 11:44






  • 1




    $begingroup$
    No, the information is sufficient: just try to construct.
    $endgroup$
    – Michael Hoppe
    Dec 1 '18 at 12:10












  • $begingroup$
    I was wondering can someone find $x$ without using trignometry? And is it better to ask this as a seperate question?
    $endgroup$
    – Fareed AF
    Dec 8 '18 at 14:13
















$begingroup$
Which equations can you find?
$endgroup$
– Arthur
Dec 1 '18 at 10:34




$begingroup$
Which equations can you find?
$endgroup$
– Arthur
Dec 1 '18 at 10:34












$begingroup$
@Arthur See the update, please.
$endgroup$
– Hamilton
Dec 1 '18 at 10:57






$begingroup$
@Arthur See the update, please.
$endgroup$
– Hamilton
Dec 1 '18 at 10:57














$begingroup$
Geogebra suggests the $x=80$.
$endgroup$
– Michael Hoppe
Dec 1 '18 at 11:44




$begingroup$
Geogebra suggests the $x=80$.
$endgroup$
– Michael Hoppe
Dec 1 '18 at 11:44




1




1




$begingroup$
No, the information is sufficient: just try to construct.
$endgroup$
– Michael Hoppe
Dec 1 '18 at 12:10






$begingroup$
No, the information is sufficient: just try to construct.
$endgroup$
– Michael Hoppe
Dec 1 '18 at 12:10














$begingroup$
I was wondering can someone find $x$ without using trignometry? And is it better to ask this as a seperate question?
$endgroup$
– Fareed AF
Dec 8 '18 at 14:13




$begingroup$
I was wondering can someone find $x$ without using trignometry? And is it better to ask this as a seperate question?
$endgroup$
– Fareed AF
Dec 8 '18 at 14:13










2 Answers
2






active

oldest

votes


















0












$begingroup$

The key insight is to observe that $angle(BDA)=angle(BDC)=138$. From here in triangle $BDC$ we have $AB/sin(138)=BD/sin(24)$ and from triangle BCD we get
$BC/sin(138)=BD/sin(12)$. Divingig both equations gives
$$BC=frac{ABsin(24)}{sin(12)}.$$



Now use law of cosine in $ABC$ to get
$$AC^2=AB^2+BC^2-2ABcdot BCcos(48)
=BC^2left(frac{sin^2(12)}{sin^2(24)}+1
-2frac{sin(12)}{sin(24)}{cos(48)}right).$$



Finally use law of sine in triangle $ABC$:
$$frac{sin(x+24)}{BC}=frac{sin(48)}{AC}$$
plug in $AB$ to get
$$sin(x+48)=frac{sin(48)}{sqrt{dfrac{sin^2(12)}{sin^2(24)}+1
-2dfrac{sin(12)}{sin(24)}{cos(48)}}}$$






share|cite|improve this answer











$endgroup$





















    0












    $begingroup$

    I use degrees, but omit the notation.



    With the use of the law of sines we have $$frac{AD}{sin 18}=frac{AB}{sin 138}=frac{BD}{sin 24}.$$



    One lenght can be chosen arbitrarily, because convenient triangles are similar. Set $AD=1.$ We get the lenghts $AB, BD.$



    Similarly we obtain the sides of $triangle BCD.$



    It remains the side $AC,$ common to two triangles. If I have not mistaken, $$AC=frac{CDsin 84}{sin X}=frac{BC sin 48}{sin(24+X)}$$ From this one gets easily $tan X.$






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Why did you use law of sines?
      $endgroup$
      – Hamilton
      Dec 1 '18 at 11:06











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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    0












    $begingroup$

    The key insight is to observe that $angle(BDA)=angle(BDC)=138$. From here in triangle $BDC$ we have $AB/sin(138)=BD/sin(24)$ and from triangle BCD we get
    $BC/sin(138)=BD/sin(12)$. Divingig both equations gives
    $$BC=frac{ABsin(24)}{sin(12)}.$$



    Now use law of cosine in $ABC$ to get
    $$AC^2=AB^2+BC^2-2ABcdot BCcos(48)
    =BC^2left(frac{sin^2(12)}{sin^2(24)}+1
    -2frac{sin(12)}{sin(24)}{cos(48)}right).$$



    Finally use law of sine in triangle $ABC$:
    $$frac{sin(x+24)}{BC}=frac{sin(48)}{AC}$$
    plug in $AB$ to get
    $$sin(x+48)=frac{sin(48)}{sqrt{dfrac{sin^2(12)}{sin^2(24)}+1
    -2dfrac{sin(12)}{sin(24)}{cos(48)}}}$$






    share|cite|improve this answer











    $endgroup$


















      0












      $begingroup$

      The key insight is to observe that $angle(BDA)=angle(BDC)=138$. From here in triangle $BDC$ we have $AB/sin(138)=BD/sin(24)$ and from triangle BCD we get
      $BC/sin(138)=BD/sin(12)$. Divingig both equations gives
      $$BC=frac{ABsin(24)}{sin(12)}.$$



      Now use law of cosine in $ABC$ to get
      $$AC^2=AB^2+BC^2-2ABcdot BCcos(48)
      =BC^2left(frac{sin^2(12)}{sin^2(24)}+1
      -2frac{sin(12)}{sin(24)}{cos(48)}right).$$



      Finally use law of sine in triangle $ABC$:
      $$frac{sin(x+24)}{BC}=frac{sin(48)}{AC}$$
      plug in $AB$ to get
      $$sin(x+48)=frac{sin(48)}{sqrt{dfrac{sin^2(12)}{sin^2(24)}+1
      -2dfrac{sin(12)}{sin(24)}{cos(48)}}}$$






      share|cite|improve this answer











      $endgroup$
















        0












        0








        0





        $begingroup$

        The key insight is to observe that $angle(BDA)=angle(BDC)=138$. From here in triangle $BDC$ we have $AB/sin(138)=BD/sin(24)$ and from triangle BCD we get
        $BC/sin(138)=BD/sin(12)$. Divingig both equations gives
        $$BC=frac{ABsin(24)}{sin(12)}.$$



        Now use law of cosine in $ABC$ to get
        $$AC^2=AB^2+BC^2-2ABcdot BCcos(48)
        =BC^2left(frac{sin^2(12)}{sin^2(24)}+1
        -2frac{sin(12)}{sin(24)}{cos(48)}right).$$



        Finally use law of sine in triangle $ABC$:
        $$frac{sin(x+24)}{BC}=frac{sin(48)}{AC}$$
        plug in $AB$ to get
        $$sin(x+48)=frac{sin(48)}{sqrt{dfrac{sin^2(12)}{sin^2(24)}+1
        -2dfrac{sin(12)}{sin(24)}{cos(48)}}}$$






        share|cite|improve this answer











        $endgroup$



        The key insight is to observe that $angle(BDA)=angle(BDC)=138$. From here in triangle $BDC$ we have $AB/sin(138)=BD/sin(24)$ and from triangle BCD we get
        $BC/sin(138)=BD/sin(12)$. Divingig both equations gives
        $$BC=frac{ABsin(24)}{sin(12)}.$$



        Now use law of cosine in $ABC$ to get
        $$AC^2=AB^2+BC^2-2ABcdot BCcos(48)
        =BC^2left(frac{sin^2(12)}{sin^2(24)}+1
        -2frac{sin(12)}{sin(24)}{cos(48)}right).$$



        Finally use law of sine in triangle $ABC$:
        $$frac{sin(x+24)}{BC}=frac{sin(48)}{AC}$$
        plug in $AB$ to get
        $$sin(x+48)=frac{sin(48)}{sqrt{dfrac{sin^2(12)}{sin^2(24)}+1
        -2dfrac{sin(12)}{sin(24)}{cos(48)}}}$$







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Dec 1 '18 at 16:01

























        answered Dec 1 '18 at 14:59









        Michael HoppeMichael Hoppe

        10.8k31834




        10.8k31834























            0












            $begingroup$

            I use degrees, but omit the notation.



            With the use of the law of sines we have $$frac{AD}{sin 18}=frac{AB}{sin 138}=frac{BD}{sin 24}.$$



            One lenght can be chosen arbitrarily, because convenient triangles are similar. Set $AD=1.$ We get the lenghts $AB, BD.$



            Similarly we obtain the sides of $triangle BCD.$



            It remains the side $AC,$ common to two triangles. If I have not mistaken, $$AC=frac{CDsin 84}{sin X}=frac{BC sin 48}{sin(24+X)}$$ From this one gets easily $tan X.$






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              Why did you use law of sines?
              $endgroup$
              – Hamilton
              Dec 1 '18 at 11:06
















            0












            $begingroup$

            I use degrees, but omit the notation.



            With the use of the law of sines we have $$frac{AD}{sin 18}=frac{AB}{sin 138}=frac{BD}{sin 24}.$$



            One lenght can be chosen arbitrarily, because convenient triangles are similar. Set $AD=1.$ We get the lenghts $AB, BD.$



            Similarly we obtain the sides of $triangle BCD.$



            It remains the side $AC,$ common to two triangles. If I have not mistaken, $$AC=frac{CDsin 84}{sin X}=frac{BC sin 48}{sin(24+X)}$$ From this one gets easily $tan X.$






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              Why did you use law of sines?
              $endgroup$
              – Hamilton
              Dec 1 '18 at 11:06














            0












            0








            0





            $begingroup$

            I use degrees, but omit the notation.



            With the use of the law of sines we have $$frac{AD}{sin 18}=frac{AB}{sin 138}=frac{BD}{sin 24}.$$



            One lenght can be chosen arbitrarily, because convenient triangles are similar. Set $AD=1.$ We get the lenghts $AB, BD.$



            Similarly we obtain the sides of $triangle BCD.$



            It remains the side $AC,$ common to two triangles. If I have not mistaken, $$AC=frac{CDsin 84}{sin X}=frac{BC sin 48}{sin(24+X)}$$ From this one gets easily $tan X.$






            share|cite|improve this answer









            $endgroup$



            I use degrees, but omit the notation.



            With the use of the law of sines we have $$frac{AD}{sin 18}=frac{AB}{sin 138}=frac{BD}{sin 24}.$$



            One lenght can be chosen arbitrarily, because convenient triangles are similar. Set $AD=1.$ We get the lenghts $AB, BD.$



            Similarly we obtain the sides of $triangle BCD.$



            It remains the side $AC,$ common to two triangles. If I have not mistaken, $$AC=frac{CDsin 84}{sin X}=frac{BC sin 48}{sin(24+X)}$$ From this one gets easily $tan X.$







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Dec 1 '18 at 11:02









            user376343user376343

            3,3833826




            3,3833826












            • $begingroup$
              Why did you use law of sines?
              $endgroup$
              – Hamilton
              Dec 1 '18 at 11:06


















            • $begingroup$
              Why did you use law of sines?
              $endgroup$
              – Hamilton
              Dec 1 '18 at 11:06
















            $begingroup$
            Why did you use law of sines?
            $endgroup$
            – Hamilton
            Dec 1 '18 at 11:06




            $begingroup$
            Why did you use law of sines?
            $endgroup$
            – Hamilton
            Dec 1 '18 at 11:06


















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