Is it mathematically right to convert triangle it to such a circle like that?











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enter image description here



Given that



$$hat{ABC} = 26^circ, hat{ACB} = 52^circ, |AC| = 11, [AB] perp [AD] $$



Evaluate $|BD| = x $.




I want to change my point of view against these type of questions. What if we consider that this is a circle? Below I drew a diagram



enter image description here



Is it mathematically right to convert triangle it to such a circle like that? If yes, how can we take it from there?



Regards










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  • 1




    You can draw a circle around triangle any time, but it does not necesery lead you to the solution.
    – greedoid
    Nov 27 at 20:49






  • 2




    I would use the word "inscribe" and not "convert" (i.e. inscribe the triangle in the circle). While there isn't anything wrong with it, does it really help you get to a solution? My guess would be no, but always open to new ideas.
    – Aditya Dua
    Nov 27 at 20:52










  • Hints: What are the internal angles of a triangle? What's the sum of angles in a straight line?
    – Jam
    Nov 27 at 20:56










  • Oh so, instead of directly solving the question, I need your dear assistance to change my perspective because I'm really having issues whenever I attempt to solve a question in triangles. For instance, doing what would make the question clear? I also found all angles.
    – Enzo
    Nov 27 at 20:59








  • 2




    @Jam Sure thing! I know Sine's theorem which states that $dfrac{a}{sin A} = dfrac{b}{sin A} = dfrac{c}{sin C}$. However, I'm also looking for strategy/method that makes it more clear than it is in this situtation.
    – Enzo
    Nov 27 at 21:04

















up vote
1
down vote

favorite













enter image description here



Given that



$$hat{ABC} = 26^circ, hat{ACB} = 52^circ, |AC| = 11, [AB] perp [AD] $$



Evaluate $|BD| = x $.




I want to change my point of view against these type of questions. What if we consider that this is a circle? Below I drew a diagram



enter image description here



Is it mathematically right to convert triangle it to such a circle like that? If yes, how can we take it from there?



Regards










share|cite|improve this question


















  • 1




    You can draw a circle around triangle any time, but it does not necesery lead you to the solution.
    – greedoid
    Nov 27 at 20:49






  • 2




    I would use the word "inscribe" and not "convert" (i.e. inscribe the triangle in the circle). While there isn't anything wrong with it, does it really help you get to a solution? My guess would be no, but always open to new ideas.
    – Aditya Dua
    Nov 27 at 20:52










  • Hints: What are the internal angles of a triangle? What's the sum of angles in a straight line?
    – Jam
    Nov 27 at 20:56










  • Oh so, instead of directly solving the question, I need your dear assistance to change my perspective because I'm really having issues whenever I attempt to solve a question in triangles. For instance, doing what would make the question clear? I also found all angles.
    – Enzo
    Nov 27 at 20:59








  • 2




    @Jam Sure thing! I know Sine's theorem which states that $dfrac{a}{sin A} = dfrac{b}{sin A} = dfrac{c}{sin C}$. However, I'm also looking for strategy/method that makes it more clear than it is in this situtation.
    – Enzo
    Nov 27 at 21:04















up vote
1
down vote

favorite









up vote
1
down vote

favorite












enter image description here



Given that



$$hat{ABC} = 26^circ, hat{ACB} = 52^circ, |AC| = 11, [AB] perp [AD] $$



Evaluate $|BD| = x $.




I want to change my point of view against these type of questions. What if we consider that this is a circle? Below I drew a diagram



enter image description here



Is it mathematically right to convert triangle it to such a circle like that? If yes, how can we take it from there?



Regards










share|cite|improve this question














enter image description here



Given that



$$hat{ABC} = 26^circ, hat{ACB} = 52^circ, |AC| = 11, [AB] perp [AD] $$



Evaluate $|BD| = x $.




I want to change my point of view against these type of questions. What if we consider that this is a circle? Below I drew a diagram



enter image description here



Is it mathematically right to convert triangle it to such a circle like that? If yes, how can we take it from there?



Regards







geometry triangle circle






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share|cite|improve this question











share|cite|improve this question




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asked Nov 27 at 20:47









Enzo

996




996








  • 1




    You can draw a circle around triangle any time, but it does not necesery lead you to the solution.
    – greedoid
    Nov 27 at 20:49






  • 2




    I would use the word "inscribe" and not "convert" (i.e. inscribe the triangle in the circle). While there isn't anything wrong with it, does it really help you get to a solution? My guess would be no, but always open to new ideas.
    – Aditya Dua
    Nov 27 at 20:52










  • Hints: What are the internal angles of a triangle? What's the sum of angles in a straight line?
    – Jam
    Nov 27 at 20:56










  • Oh so, instead of directly solving the question, I need your dear assistance to change my perspective because I'm really having issues whenever I attempt to solve a question in triangles. For instance, doing what would make the question clear? I also found all angles.
    – Enzo
    Nov 27 at 20:59








  • 2




    @Jam Sure thing! I know Sine's theorem which states that $dfrac{a}{sin A} = dfrac{b}{sin A} = dfrac{c}{sin C}$. However, I'm also looking for strategy/method that makes it more clear than it is in this situtation.
    – Enzo
    Nov 27 at 21:04
















  • 1




    You can draw a circle around triangle any time, but it does not necesery lead you to the solution.
    – greedoid
    Nov 27 at 20:49






  • 2




    I would use the word "inscribe" and not "convert" (i.e. inscribe the triangle in the circle). While there isn't anything wrong with it, does it really help you get to a solution? My guess would be no, but always open to new ideas.
    – Aditya Dua
    Nov 27 at 20:52










  • Hints: What are the internal angles of a triangle? What's the sum of angles in a straight line?
    – Jam
    Nov 27 at 20:56










  • Oh so, instead of directly solving the question, I need your dear assistance to change my perspective because I'm really having issues whenever I attempt to solve a question in triangles. For instance, doing what would make the question clear? I also found all angles.
    – Enzo
    Nov 27 at 20:59








  • 2




    @Jam Sure thing! I know Sine's theorem which states that $dfrac{a}{sin A} = dfrac{b}{sin A} = dfrac{c}{sin C}$. However, I'm also looking for strategy/method that makes it more clear than it is in this situtation.
    – Enzo
    Nov 27 at 21:04










1




1




You can draw a circle around triangle any time, but it does not necesery lead you to the solution.
– greedoid
Nov 27 at 20:49




You can draw a circle around triangle any time, but it does not necesery lead you to the solution.
– greedoid
Nov 27 at 20:49




2




2




I would use the word "inscribe" and not "convert" (i.e. inscribe the triangle in the circle). While there isn't anything wrong with it, does it really help you get to a solution? My guess would be no, but always open to new ideas.
– Aditya Dua
Nov 27 at 20:52




I would use the word "inscribe" and not "convert" (i.e. inscribe the triangle in the circle). While there isn't anything wrong with it, does it really help you get to a solution? My guess would be no, but always open to new ideas.
– Aditya Dua
Nov 27 at 20:52












Hints: What are the internal angles of a triangle? What's the sum of angles in a straight line?
– Jam
Nov 27 at 20:56




Hints: What are the internal angles of a triangle? What's the sum of angles in a straight line?
– Jam
Nov 27 at 20:56












Oh so, instead of directly solving the question, I need your dear assistance to change my perspective because I'm really having issues whenever I attempt to solve a question in triangles. For instance, doing what would make the question clear? I also found all angles.
– Enzo
Nov 27 at 20:59






Oh so, instead of directly solving the question, I need your dear assistance to change my perspective because I'm really having issues whenever I attempt to solve a question in triangles. For instance, doing what would make the question clear? I also found all angles.
– Enzo
Nov 27 at 20:59






2




2




@Jam Sure thing! I know Sine's theorem which states that $dfrac{a}{sin A} = dfrac{b}{sin A} = dfrac{c}{sin C}$. However, I'm also looking for strategy/method that makes it more clear than it is in this situtation.
– Enzo
Nov 27 at 21:04






@Jam Sure thing! I know Sine's theorem which states that $dfrac{a}{sin A} = dfrac{b}{sin A} = dfrac{c}{sin C}$. However, I'm also looking for strategy/method that makes it more clear than it is in this situtation.
– Enzo
Nov 27 at 21:04












2 Answers
2






active

oldest

votes

















up vote
1
down vote













If we want to use the sine law, to get $x$ we need to look at triangle $BAD$. $$frac x{sin 90^circ}=frac{AD}{sin 26^circ}$$
We don't know $AD$, but we can once again apply the sine law in the $DAC$ triangle. $$frac{AD}{sin 52^circ}=frac{AC}{sin angle{ADC}}$$
You know $AC=11$ and you can get the angle $angle ADC$ since it's a supplement of $angle ADB$, and you know all angles in that triangle.






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  • From $frac{AD}{sin 52^circ}=dfrac{AC}{sin angle{ADC}}$, do we get that $dfrac{AD}{sin 52^circ}=dfrac{11}{sin 116^circ}$?
    – Enzo
    Nov 27 at 21:51












  • No. What's the $ADC$ angle? If you know it, you can get $AD$.
    – Andrei
    Nov 27 at 21:54










  • Isn't it $hat{ADC} = 116$?
    – Enzo
    Nov 27 at 21:58












  • Correct. Now find $AD$ from my second equation. Then use it in the first equation, to find $x$.
    – Andrei
    Nov 27 at 21:59


















up vote
0
down vote













Knowing $angle ABD$ and $angle BAD$ you can solve for $angle ADB$ $$angle ADB=180°-26°-90°=64°$$ The angles $angle ADB$ and $angle ADC$ are supplementary, so $angle ADC=116°$. Knowing angles $angle ADC, angle ACD$, and the length of $AC$, the law of sines can be used to solve for the length of $AD$.
$$frac{11}{sin(116)}=frac{AD}{sin(52)}Rightarrow AD=frac{11sin(52)}{sin(116)}approx9,64$$ The law of sines can be used again to find $x$. $$frac{9.64}{sin(26)}=frac{x}{sin(90)}rightarrow x=frac{9.64sin(90)}{sin(26)}$$ $x$ is 22.



That is how I would do it. Sorry about the format, I am not familiar with the equation makers.






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    2 Answers
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    2 Answers
    2






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    active

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    up vote
    1
    down vote













    If we want to use the sine law, to get $x$ we need to look at triangle $BAD$. $$frac x{sin 90^circ}=frac{AD}{sin 26^circ}$$
    We don't know $AD$, but we can once again apply the sine law in the $DAC$ triangle. $$frac{AD}{sin 52^circ}=frac{AC}{sin angle{ADC}}$$
    You know $AC=11$ and you can get the angle $angle ADC$ since it's a supplement of $angle ADB$, and you know all angles in that triangle.






    share|cite|improve this answer























    • From $frac{AD}{sin 52^circ}=dfrac{AC}{sin angle{ADC}}$, do we get that $dfrac{AD}{sin 52^circ}=dfrac{11}{sin 116^circ}$?
      – Enzo
      Nov 27 at 21:51












    • No. What's the $ADC$ angle? If you know it, you can get $AD$.
      – Andrei
      Nov 27 at 21:54










    • Isn't it $hat{ADC} = 116$?
      – Enzo
      Nov 27 at 21:58












    • Correct. Now find $AD$ from my second equation. Then use it in the first equation, to find $x$.
      – Andrei
      Nov 27 at 21:59















    up vote
    1
    down vote













    If we want to use the sine law, to get $x$ we need to look at triangle $BAD$. $$frac x{sin 90^circ}=frac{AD}{sin 26^circ}$$
    We don't know $AD$, but we can once again apply the sine law in the $DAC$ triangle. $$frac{AD}{sin 52^circ}=frac{AC}{sin angle{ADC}}$$
    You know $AC=11$ and you can get the angle $angle ADC$ since it's a supplement of $angle ADB$, and you know all angles in that triangle.






    share|cite|improve this answer























    • From $frac{AD}{sin 52^circ}=dfrac{AC}{sin angle{ADC}}$, do we get that $dfrac{AD}{sin 52^circ}=dfrac{11}{sin 116^circ}$?
      – Enzo
      Nov 27 at 21:51












    • No. What's the $ADC$ angle? If you know it, you can get $AD$.
      – Andrei
      Nov 27 at 21:54










    • Isn't it $hat{ADC} = 116$?
      – Enzo
      Nov 27 at 21:58












    • Correct. Now find $AD$ from my second equation. Then use it in the first equation, to find $x$.
      – Andrei
      Nov 27 at 21:59













    up vote
    1
    down vote










    up vote
    1
    down vote









    If we want to use the sine law, to get $x$ we need to look at triangle $BAD$. $$frac x{sin 90^circ}=frac{AD}{sin 26^circ}$$
    We don't know $AD$, but we can once again apply the sine law in the $DAC$ triangle. $$frac{AD}{sin 52^circ}=frac{AC}{sin angle{ADC}}$$
    You know $AC=11$ and you can get the angle $angle ADC$ since it's a supplement of $angle ADB$, and you know all angles in that triangle.






    share|cite|improve this answer














    If we want to use the sine law, to get $x$ we need to look at triangle $BAD$. $$frac x{sin 90^circ}=frac{AD}{sin 26^circ}$$
    We don't know $AD$, but we can once again apply the sine law in the $DAC$ triangle. $$frac{AD}{sin 52^circ}=frac{AC}{sin angle{ADC}}$$
    You know $AC=11$ and you can get the angle $angle ADC$ since it's a supplement of $angle ADB$, and you know all angles in that triangle.







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited Nov 27 at 22:39









    math818

    33




    33










    answered Nov 27 at 21:48









    Andrei

    10.8k21025




    10.8k21025












    • From $frac{AD}{sin 52^circ}=dfrac{AC}{sin angle{ADC}}$, do we get that $dfrac{AD}{sin 52^circ}=dfrac{11}{sin 116^circ}$?
      – Enzo
      Nov 27 at 21:51












    • No. What's the $ADC$ angle? If you know it, you can get $AD$.
      – Andrei
      Nov 27 at 21:54










    • Isn't it $hat{ADC} = 116$?
      – Enzo
      Nov 27 at 21:58












    • Correct. Now find $AD$ from my second equation. Then use it in the first equation, to find $x$.
      – Andrei
      Nov 27 at 21:59


















    • From $frac{AD}{sin 52^circ}=dfrac{AC}{sin angle{ADC}}$, do we get that $dfrac{AD}{sin 52^circ}=dfrac{11}{sin 116^circ}$?
      – Enzo
      Nov 27 at 21:51












    • No. What's the $ADC$ angle? If you know it, you can get $AD$.
      – Andrei
      Nov 27 at 21:54










    • Isn't it $hat{ADC} = 116$?
      – Enzo
      Nov 27 at 21:58












    • Correct. Now find $AD$ from my second equation. Then use it in the first equation, to find $x$.
      – Andrei
      Nov 27 at 21:59
















    From $frac{AD}{sin 52^circ}=dfrac{AC}{sin angle{ADC}}$, do we get that $dfrac{AD}{sin 52^circ}=dfrac{11}{sin 116^circ}$?
    – Enzo
    Nov 27 at 21:51






    From $frac{AD}{sin 52^circ}=dfrac{AC}{sin angle{ADC}}$, do we get that $dfrac{AD}{sin 52^circ}=dfrac{11}{sin 116^circ}$?
    – Enzo
    Nov 27 at 21:51














    No. What's the $ADC$ angle? If you know it, you can get $AD$.
    – Andrei
    Nov 27 at 21:54




    No. What's the $ADC$ angle? If you know it, you can get $AD$.
    – Andrei
    Nov 27 at 21:54












    Isn't it $hat{ADC} = 116$?
    – Enzo
    Nov 27 at 21:58






    Isn't it $hat{ADC} = 116$?
    – Enzo
    Nov 27 at 21:58














    Correct. Now find $AD$ from my second equation. Then use it in the first equation, to find $x$.
    – Andrei
    Nov 27 at 21:59




    Correct. Now find $AD$ from my second equation. Then use it in the first equation, to find $x$.
    – Andrei
    Nov 27 at 21:59










    up vote
    0
    down vote













    Knowing $angle ABD$ and $angle BAD$ you can solve for $angle ADB$ $$angle ADB=180°-26°-90°=64°$$ The angles $angle ADB$ and $angle ADC$ are supplementary, so $angle ADC=116°$. Knowing angles $angle ADC, angle ACD$, and the length of $AC$, the law of sines can be used to solve for the length of $AD$.
    $$frac{11}{sin(116)}=frac{AD}{sin(52)}Rightarrow AD=frac{11sin(52)}{sin(116)}approx9,64$$ The law of sines can be used again to find $x$. $$frac{9.64}{sin(26)}=frac{x}{sin(90)}rightarrow x=frac{9.64sin(90)}{sin(26)}$$ $x$ is 22.



    That is how I would do it. Sorry about the format, I am not familiar with the equation makers.






    share|cite|improve this answer



























      up vote
      0
      down vote













      Knowing $angle ABD$ and $angle BAD$ you can solve for $angle ADB$ $$angle ADB=180°-26°-90°=64°$$ The angles $angle ADB$ and $angle ADC$ are supplementary, so $angle ADC=116°$. Knowing angles $angle ADC, angle ACD$, and the length of $AC$, the law of sines can be used to solve for the length of $AD$.
      $$frac{11}{sin(116)}=frac{AD}{sin(52)}Rightarrow AD=frac{11sin(52)}{sin(116)}approx9,64$$ The law of sines can be used again to find $x$. $$frac{9.64}{sin(26)}=frac{x}{sin(90)}rightarrow x=frac{9.64sin(90)}{sin(26)}$$ $x$ is 22.



      That is how I would do it. Sorry about the format, I am not familiar with the equation makers.






      share|cite|improve this answer

























        up vote
        0
        down vote










        up vote
        0
        down vote









        Knowing $angle ABD$ and $angle BAD$ you can solve for $angle ADB$ $$angle ADB=180°-26°-90°=64°$$ The angles $angle ADB$ and $angle ADC$ are supplementary, so $angle ADC=116°$. Knowing angles $angle ADC, angle ACD$, and the length of $AC$, the law of sines can be used to solve for the length of $AD$.
        $$frac{11}{sin(116)}=frac{AD}{sin(52)}Rightarrow AD=frac{11sin(52)}{sin(116)}approx9,64$$ The law of sines can be used again to find $x$. $$frac{9.64}{sin(26)}=frac{x}{sin(90)}rightarrow x=frac{9.64sin(90)}{sin(26)}$$ $x$ is 22.



        That is how I would do it. Sorry about the format, I am not familiar with the equation makers.






        share|cite|improve this answer














        Knowing $angle ABD$ and $angle BAD$ you can solve for $angle ADB$ $$angle ADB=180°-26°-90°=64°$$ The angles $angle ADB$ and $angle ADC$ are supplementary, so $angle ADC=116°$. Knowing angles $angle ADC, angle ACD$, and the length of $AC$, the law of sines can be used to solve for the length of $AD$.
        $$frac{11}{sin(116)}=frac{AD}{sin(52)}Rightarrow AD=frac{11sin(52)}{sin(116)}approx9,64$$ The law of sines can be used again to find $x$. $$frac{9.64}{sin(26)}=frac{x}{sin(90)}rightarrow x=frac{9.64sin(90)}{sin(26)}$$ $x$ is 22.



        That is how I would do it. Sorry about the format, I am not familiar with the equation makers.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Nov 27 at 23:35









        Dr. Mathva

        926315




        926315










        answered Nov 27 at 22:34









        math818

        33




        33






























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