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Finding an angle in a triangle?


Finding value of an angle in a triangle.Length of angle bisector using Law of SinesFind the angles of triangleFind the value of an angleIn triangle $ABC$ find angle $angle BAC$ given that…Find possible integer values for $x$ using angle-edge relationship in triangle.Find angle $alpha$ in this triangle question.Find an angle of a triangle on a larger triangle which cut through its midpointFind the third angle of the triangle(South Africa 2014) Finding angles in an obtuse triangle













2












$begingroup$


Given the following picture, how do I find $angle CDA$?



enter image description here



Attempt:



I have found the following angles, which I think may be useful:
$$
angle BAC = arctan(3),quad
angle ABC = arctan(2),quad
angle BCA = arctan(1).
$$


But I don't know how to go on.










share|cite|improve this question











$endgroup$











  • $begingroup$
    Hey and Welcome to MSE! Could you please tell us where are you stuck? What have you tried so far?
    $endgroup$
    – Vinyl_cape_jawa
    Mar 4 at 12:08










  • $begingroup$
    So this is not the original problem? In this case you shouid post the whole question. Maybe there are some other information there we could use.
    $endgroup$
    – Vinyl_cape_jawa
    Mar 4 at 12:14










  • $begingroup$
    @Vinyl_coat_jawa I understand what you're saying, but there is no additional information that I have not already provided.
    $endgroup$
    – Ruby Pa
    Mar 4 at 12:15










  • $begingroup$
    But the how do you know these angles?
    $endgroup$
    – Vinyl_cape_jawa
    Mar 4 at 12:26






  • 1




    $begingroup$
    Where exactly is point $D$ ?
    $endgroup$
    – Antinous
    Mar 4 at 13:51















2












$begingroup$


Given the following picture, how do I find $angle CDA$?



enter image description here



Attempt:



I have found the following angles, which I think may be useful:
$$
angle BAC = arctan(3),quad
angle ABC = arctan(2),quad
angle BCA = arctan(1).
$$


But I don't know how to go on.










share|cite|improve this question











$endgroup$











  • $begingroup$
    Hey and Welcome to MSE! Could you please tell us where are you stuck? What have you tried so far?
    $endgroup$
    – Vinyl_cape_jawa
    Mar 4 at 12:08










  • $begingroup$
    So this is not the original problem? In this case you shouid post the whole question. Maybe there are some other information there we could use.
    $endgroup$
    – Vinyl_cape_jawa
    Mar 4 at 12:14










  • $begingroup$
    @Vinyl_coat_jawa I understand what you're saying, but there is no additional information that I have not already provided.
    $endgroup$
    – Ruby Pa
    Mar 4 at 12:15










  • $begingroup$
    But the how do you know these angles?
    $endgroup$
    – Vinyl_cape_jawa
    Mar 4 at 12:26






  • 1




    $begingroup$
    Where exactly is point $D$ ?
    $endgroup$
    – Antinous
    Mar 4 at 13:51













2












2








2


1



$begingroup$


Given the following picture, how do I find $angle CDA$?



enter image description here



Attempt:



I have found the following angles, which I think may be useful:
$$
angle BAC = arctan(3),quad
angle ABC = arctan(2),quad
angle BCA = arctan(1).
$$


But I don't know how to go on.










share|cite|improve this question











$endgroup$




Given the following picture, how do I find $angle CDA$?



enter image description here



Attempt:



I have found the following angles, which I think may be useful:
$$
angle BAC = arctan(3),quad
angle ABC = arctan(2),quad
angle BCA = arctan(1).
$$


But I don't know how to go on.







trigonometry triangle






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Mar 17 at 23:31









Jack

27.7k1782204




27.7k1782204










asked Mar 4 at 12:00









Ruby PaRuby Pa

376




376











  • $begingroup$
    Hey and Welcome to MSE! Could you please tell us where are you stuck? What have you tried so far?
    $endgroup$
    – Vinyl_cape_jawa
    Mar 4 at 12:08










  • $begingroup$
    So this is not the original problem? In this case you shouid post the whole question. Maybe there are some other information there we could use.
    $endgroup$
    – Vinyl_cape_jawa
    Mar 4 at 12:14










  • $begingroup$
    @Vinyl_coat_jawa I understand what you're saying, but there is no additional information that I have not already provided.
    $endgroup$
    – Ruby Pa
    Mar 4 at 12:15










  • $begingroup$
    But the how do you know these angles?
    $endgroup$
    – Vinyl_cape_jawa
    Mar 4 at 12:26






  • 1




    $begingroup$
    Where exactly is point $D$ ?
    $endgroup$
    – Antinous
    Mar 4 at 13:51
















  • $begingroup$
    Hey and Welcome to MSE! Could you please tell us where are you stuck? What have you tried so far?
    $endgroup$
    – Vinyl_cape_jawa
    Mar 4 at 12:08










  • $begingroup$
    So this is not the original problem? In this case you shouid post the whole question. Maybe there are some other information there we could use.
    $endgroup$
    – Vinyl_cape_jawa
    Mar 4 at 12:14










  • $begingroup$
    @Vinyl_coat_jawa I understand what you're saying, but there is no additional information that I have not already provided.
    $endgroup$
    – Ruby Pa
    Mar 4 at 12:15










  • $begingroup$
    But the how do you know these angles?
    $endgroup$
    – Vinyl_cape_jawa
    Mar 4 at 12:26






  • 1




    $begingroup$
    Where exactly is point $D$ ?
    $endgroup$
    – Antinous
    Mar 4 at 13:51















$begingroup$
Hey and Welcome to MSE! Could you please tell us where are you stuck? What have you tried so far?
$endgroup$
– Vinyl_cape_jawa
Mar 4 at 12:08




$begingroup$
Hey and Welcome to MSE! Could you please tell us where are you stuck? What have you tried so far?
$endgroup$
– Vinyl_cape_jawa
Mar 4 at 12:08












$begingroup$
So this is not the original problem? In this case you shouid post the whole question. Maybe there are some other information there we could use.
$endgroup$
– Vinyl_cape_jawa
Mar 4 at 12:14




$begingroup$
So this is not the original problem? In this case you shouid post the whole question. Maybe there are some other information there we could use.
$endgroup$
– Vinyl_cape_jawa
Mar 4 at 12:14












$begingroup$
@Vinyl_coat_jawa I understand what you're saying, but there is no additional information that I have not already provided.
$endgroup$
– Ruby Pa
Mar 4 at 12:15




$begingroup$
@Vinyl_coat_jawa I understand what you're saying, but there is no additional information that I have not already provided.
$endgroup$
– Ruby Pa
Mar 4 at 12:15












$begingroup$
But the how do you know these angles?
$endgroup$
– Vinyl_cape_jawa
Mar 4 at 12:26




$begingroup$
But the how do you know these angles?
$endgroup$
– Vinyl_cape_jawa
Mar 4 at 12:26




1




1




$begingroup$
Where exactly is point $D$ ?
$endgroup$
– Antinous
Mar 4 at 13:51




$begingroup$
Where exactly is point $D$ ?
$endgroup$
– Antinous
Mar 4 at 13:51










2 Answers
2






active

oldest

votes


















4












$begingroup$

Assuming the points are at the exact grid points where it looks like they are, the line $AD$ is orthogonal to $BC$. The line $BC$ goes $2$ units down for each unit we go to the right, and if we turn that $90^circ$, we get a line which goes $1$ unit up for every $2$ units to the right. That's what $AD$ does.






share|cite|improve this answer









$endgroup$












  • $begingroup$
    @RubyPa I'm not assuming that in any way. I'm showing it. There may be details I've glossed over, but this approach is definitely viable.
    $endgroup$
    – Arthur
    Mar 4 at 12:33











  • $begingroup$
    Ahhh yes I see what you mean. Thanks for your help! :)
    $endgroup$
    – Ruby Pa
    Mar 4 at 12:36


















1












$begingroup$

Another approach:



Note that $angle DAB = arctan(dfrac12)\ $ (two units over four ones). Thus,



$$angle ADC = 180°- (angle DAB+ angle CBA) = 180° - arctan(dfrac12)-arctan(2) = 90°.$$






share|cite|improve this answer











$endgroup$













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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    4












    $begingroup$

    Assuming the points are at the exact grid points where it looks like they are, the line $AD$ is orthogonal to $BC$. The line $BC$ goes $2$ units down for each unit we go to the right, and if we turn that $90^circ$, we get a line which goes $1$ unit up for every $2$ units to the right. That's what $AD$ does.






    share|cite|improve this answer









    $endgroup$












    • $begingroup$
      @RubyPa I'm not assuming that in any way. I'm showing it. There may be details I've glossed over, but this approach is definitely viable.
      $endgroup$
      – Arthur
      Mar 4 at 12:33











    • $begingroup$
      Ahhh yes I see what you mean. Thanks for your help! :)
      $endgroup$
      – Ruby Pa
      Mar 4 at 12:36















    4












    $begingroup$

    Assuming the points are at the exact grid points where it looks like they are, the line $AD$ is orthogonal to $BC$. The line $BC$ goes $2$ units down for each unit we go to the right, and if we turn that $90^circ$, we get a line which goes $1$ unit up for every $2$ units to the right. That's what $AD$ does.






    share|cite|improve this answer









    $endgroup$












    • $begingroup$
      @RubyPa I'm not assuming that in any way. I'm showing it. There may be details I've glossed over, but this approach is definitely viable.
      $endgroup$
      – Arthur
      Mar 4 at 12:33











    • $begingroup$
      Ahhh yes I see what you mean. Thanks for your help! :)
      $endgroup$
      – Ruby Pa
      Mar 4 at 12:36













    4












    4








    4





    $begingroup$

    Assuming the points are at the exact grid points where it looks like they are, the line $AD$ is orthogonal to $BC$. The line $BC$ goes $2$ units down for each unit we go to the right, and if we turn that $90^circ$, we get a line which goes $1$ unit up for every $2$ units to the right. That's what $AD$ does.






    share|cite|improve this answer









    $endgroup$



    Assuming the points are at the exact grid points where it looks like they are, the line $AD$ is orthogonal to $BC$. The line $BC$ goes $2$ units down for each unit we go to the right, and if we turn that $90^circ$, we get a line which goes $1$ unit up for every $2$ units to the right. That's what $AD$ does.







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Mar 4 at 12:31









    ArthurArthur

    122k7122211




    122k7122211











    • $begingroup$
      @RubyPa I'm not assuming that in any way. I'm showing it. There may be details I've glossed over, but this approach is definitely viable.
      $endgroup$
      – Arthur
      Mar 4 at 12:33











    • $begingroup$
      Ahhh yes I see what you mean. Thanks for your help! :)
      $endgroup$
      – Ruby Pa
      Mar 4 at 12:36
















    • $begingroup$
      @RubyPa I'm not assuming that in any way. I'm showing it. There may be details I've glossed over, but this approach is definitely viable.
      $endgroup$
      – Arthur
      Mar 4 at 12:33











    • $begingroup$
      Ahhh yes I see what you mean. Thanks for your help! :)
      $endgroup$
      – Ruby Pa
      Mar 4 at 12:36















    $begingroup$
    @RubyPa I'm not assuming that in any way. I'm showing it. There may be details I've glossed over, but this approach is definitely viable.
    $endgroup$
    – Arthur
    Mar 4 at 12:33





    $begingroup$
    @RubyPa I'm not assuming that in any way. I'm showing it. There may be details I've glossed over, but this approach is definitely viable.
    $endgroup$
    – Arthur
    Mar 4 at 12:33













    $begingroup$
    Ahhh yes I see what you mean. Thanks for your help! :)
    $endgroup$
    – Ruby Pa
    Mar 4 at 12:36




    $begingroup$
    Ahhh yes I see what you mean. Thanks for your help! :)
    $endgroup$
    – Ruby Pa
    Mar 4 at 12:36











    1












    $begingroup$

    Another approach:



    Note that $angle DAB = arctan(dfrac12)\ $ (two units over four ones). Thus,



    $$angle ADC = 180°- (angle DAB+ angle CBA) = 180° - arctan(dfrac12)-arctan(2) = 90°.$$






    share|cite|improve this answer











    $endgroup$

















      1












      $begingroup$

      Another approach:



      Note that $angle DAB = arctan(dfrac12)\ $ (two units over four ones). Thus,



      $$angle ADC = 180°- (angle DAB+ angle CBA) = 180° - arctan(dfrac12)-arctan(2) = 90°.$$






      share|cite|improve this answer











      $endgroup$















        1












        1








        1





        $begingroup$

        Another approach:



        Note that $angle DAB = arctan(dfrac12)\ $ (two units over four ones). Thus,



        $$angle ADC = 180°- (angle DAB+ angle CBA) = 180° - arctan(dfrac12)-arctan(2) = 90°.$$






        share|cite|improve this answer











        $endgroup$



        Another approach:



        Note that $angle DAB = arctan(dfrac12)\ $ (two units over four ones). Thus,



        $$angle ADC = 180°- (angle DAB+ angle CBA) = 180° - arctan(dfrac12)-arctan(2) = 90°.$$







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Mar 29 at 10:56

























        answered Mar 4 at 12:48









        Alex SilvaAlex Silva

        2,78731532




        2,78731532



























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