Can we find angles of the new triangle?












2












$begingroup$


In an equilateral triangle $Delta ABC$ with side $a$, an interior point $D$ is chosen and joined with vertices to form line segments $AD$,$BD$ and $CD$.



Also we know that $angle ADC=x$, $angle BDC =y$ and $angle ADB=z$



Now if a new triangle with line segments $AD$, $BD$ and $CD$ are formed, Is it possible to find the angles of this new triangle?



enter image description here



My try:



Let $AD=p$, $BD=q$ and $CD=r$



By cosine rule we have:



$$cos z=frac{p^2+q^2-a^2}{2pq}$$



$$cos x=frac{p^2+r^2-a^2}{2pr}$$



$$cos y=frac{r^2+q^2-a^2}{2rq}$$



which are three equations in three unknowns $p,q,r$.



Can we solve these?






geometry trigonometry triangle







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    Your title talks about finding angles while the body seems to assume the angles are known and you want to find $p,q,r$. You can go either direction.
    $endgroup$
    – Ross Millikan
    Jan 25 at 15:57










  • $begingroup$
    But $x+y+z=360$ right from the diagram, how can $x,y,z$ be angles of new triangle?
    $endgroup$
    – Umesh shankar
    Jan 25 at 16:02
















2












$begingroup$


In an equilateral triangle $Delta ABC$ with side $a$, an interior point $D$ is chosen and joined with vertices to form line segments $AD$,$BD$ and $CD$.



Also we know that $angle ADC=x$, $angle BDC =y$ and $angle ADB=z$



Now if a new triangle with line segments $AD$, $BD$ and $CD$ are formed, Is it possible to find the angles of this new triangle?



enter image description here



My try:



Let $AD=p$, $BD=q$ and $CD=r$



By cosine rule we have:



$$cos z=frac{p^2+q^2-a^2}{2pq}$$



$$cos x=frac{p^2+r^2-a^2}{2pr}$$



$$cos y=frac{r^2+q^2-a^2}{2rq}$$



which are three equations in three unknowns $p,q,r$.



Can we solve these?






geometry trigonometry triangle







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    Your title talks about finding angles while the body seems to assume the angles are known and you want to find $p,q,r$. You can go either direction.
    $endgroup$
    – Ross Millikan
    Jan 25 at 15:57










  • $begingroup$
    But $x+y+z=360$ right from the diagram, how can $x,y,z$ be angles of new triangle?
    $endgroup$
    – Umesh shankar
    Jan 25 at 16:02














2












2








2


1



$begingroup$


In an equilateral triangle $Delta ABC$ with side $a$, an interior point $D$ is chosen and joined with vertices to form line segments $AD$,$BD$ and $CD$.



Also we know that $angle ADC=x$, $angle BDC =y$ and $angle ADB=z$



Now if a new triangle with line segments $AD$, $BD$ and $CD$ are formed, Is it possible to find the angles of this new triangle?



enter image description here



My try:



Let $AD=p$, $BD=q$ and $CD=r$



By cosine rule we have:



$$cos z=frac{p^2+q^2-a^2}{2pq}$$



$$cos x=frac{p^2+r^2-a^2}{2pr}$$



$$cos y=frac{r^2+q^2-a^2}{2rq}$$



which are three equations in three unknowns $p,q,r$.



Can we solve these?






geometry trigonometry triangle







share|cite|improve this question









$endgroup$




In an equilateral triangle $Delta ABC$ with side $a$, an interior point $D$ is chosen and joined with vertices to form line segments $AD$,$BD$ and $CD$.



Also we know that $angle ADC=x$, $angle BDC =y$ and $angle ADB=z$



Now if a new triangle with line segments $AD$, $BD$ and $CD$ are formed, Is it possible to find the angles of this new triangle?



enter image description here



My try:



Let $AD=p$, $BD=q$ and $CD=r$



By cosine rule we have:



$$cos z=frac{p^2+q^2-a^2}{2pq}$$



$$cos x=frac{p^2+r^2-a^2}{2pr}$$



$$cos y=frac{r^2+q^2-a^2}{2rq}$$



which are three equations in three unknowns $p,q,r$.



Can we solve these?






geometry trigonometry triangle




geometry trigonometry triangle






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Jan 25 at 15:52









Umesh shankarUmesh shankar

2,99331220




2,99331220








  • 1




    $begingroup$
    Your title talks about finding angles while the body seems to assume the angles are known and you want to find $p,q,r$. You can go either direction.
    $endgroup$
    – Ross Millikan
    Jan 25 at 15:57










  • $begingroup$
    But $x+y+z=360$ right from the diagram, how can $x,y,z$ be angles of new triangle?
    $endgroup$
    – Umesh shankar
    Jan 25 at 16:02














  • 1




    $begingroup$
    Your title talks about finding angles while the body seems to assume the angles are known and you want to find $p,q,r$. You can go either direction.
    $endgroup$
    – Ross Millikan
    Jan 25 at 15:57










  • $begingroup$
    But $x+y+z=360$ right from the diagram, how can $x,y,z$ be angles of new triangle?
    $endgroup$
    – Umesh shankar
    Jan 25 at 16:02








1




1




$begingroup$
Your title talks about finding angles while the body seems to assume the angles are known and you want to find $p,q,r$. You can go either direction.
$endgroup$
– Ross Millikan
Jan 25 at 15:57




$begingroup$
Your title talks about finding angles while the body seems to assume the angles are known and you want to find $p,q,r$. You can go either direction.
$endgroup$
– Ross Millikan
Jan 25 at 15:57












$begingroup$
But $x+y+z=360$ right from the diagram, how can $x,y,z$ be angles of new triangle?
$endgroup$
– Umesh shankar
Jan 25 at 16:02




$begingroup$
But $x+y+z=360$ right from the diagram, how can $x,y,z$ be angles of new triangle?
$endgroup$
– Umesh shankar
Jan 25 at 16:02










1 Answer
1






active

oldest

votes


















5












$begingroup$

Diagram



Let $T$ be the triangle with sides $AD$, $BD$ and $CD$.



Let $R$ denote a rotation by $frac{pi}{3}$ about $B$. Let $D'=R(D)$. Therefore, $triangle BDD'$ is equilateral. Also, $C=R(A)$. Therefore, $CD'=R(A)R(D)=AD$, as rotation preserves length. Therefore, $triangle CDD'cong T$. As rotation preserves angles too, $angle BCD'=angle BR(A)R(D)=angle BAD$. Therefore, $angle DCD'=angle DCB+angle BAD= 2pi-theta_A-theta_C-frac{pi}{3}=theta_B-frac{pi}{3}$. Therefore, angle opposite to $BD$ in $T = theta_B-frac{pi}{3}$.



Therefore, by symmetry, the angles of $T$ are $theta_A-frac{pi}{3}, theta_B-frac{pi}{3}$ and $theta_C-frac{pi}{3}$.



$blacksquare$






share|cite|improve this answer









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    5












    $begingroup$

    Diagram



    Let $T$ be the triangle with sides $AD$, $BD$ and $CD$.



    Let $R$ denote a rotation by $frac{pi}{3}$ about $B$. Let $D'=R(D)$. Therefore, $triangle BDD'$ is equilateral. Also, $C=R(A)$. Therefore, $CD'=R(A)R(D)=AD$, as rotation preserves length. Therefore, $triangle CDD'cong T$. As rotation preserves angles too, $angle BCD'=angle BR(A)R(D)=angle BAD$. Therefore, $angle DCD'=angle DCB+angle BAD= 2pi-theta_A-theta_C-frac{pi}{3}=theta_B-frac{pi}{3}$. Therefore, angle opposite to $BD$ in $T = theta_B-frac{pi}{3}$.



    Therefore, by symmetry, the angles of $T$ are $theta_A-frac{pi}{3}, theta_B-frac{pi}{3}$ and $theta_C-frac{pi}{3}$.



    $blacksquare$






    share|cite|improve this answer









    $endgroup$


















      5












      $begingroup$

      Diagram



      Let $T$ be the triangle with sides $AD$, $BD$ and $CD$.



      Let $R$ denote a rotation by $frac{pi}{3}$ about $B$. Let $D'=R(D)$. Therefore, $triangle BDD'$ is equilateral. Also, $C=R(A)$. Therefore, $CD'=R(A)R(D)=AD$, as rotation preserves length. Therefore, $triangle CDD'cong T$. As rotation preserves angles too, $angle BCD'=angle BR(A)R(D)=angle BAD$. Therefore, $angle DCD'=angle DCB+angle BAD= 2pi-theta_A-theta_C-frac{pi}{3}=theta_B-frac{pi}{3}$. Therefore, angle opposite to $BD$ in $T = theta_B-frac{pi}{3}$.



      Therefore, by symmetry, the angles of $T$ are $theta_A-frac{pi}{3}, theta_B-frac{pi}{3}$ and $theta_C-frac{pi}{3}$.



      $blacksquare$






      share|cite|improve this answer









      $endgroup$
















        5












        5








        5





        $begingroup$

        Diagram



        Let $T$ be the triangle with sides $AD$, $BD$ and $CD$.



        Let $R$ denote a rotation by $frac{pi}{3}$ about $B$. Let $D'=R(D)$. Therefore, $triangle BDD'$ is equilateral. Also, $C=R(A)$. Therefore, $CD'=R(A)R(D)=AD$, as rotation preserves length. Therefore, $triangle CDD'cong T$. As rotation preserves angles too, $angle BCD'=angle BR(A)R(D)=angle BAD$. Therefore, $angle DCD'=angle DCB+angle BAD= 2pi-theta_A-theta_C-frac{pi}{3}=theta_B-frac{pi}{3}$. Therefore, angle opposite to $BD$ in $T = theta_B-frac{pi}{3}$.



        Therefore, by symmetry, the angles of $T$ are $theta_A-frac{pi}{3}, theta_B-frac{pi}{3}$ and $theta_C-frac{pi}{3}$.



        $blacksquare$






        share|cite|improve this answer









        $endgroup$



        Diagram



        Let $T$ be the triangle with sides $AD$, $BD$ and $CD$.



        Let $R$ denote a rotation by $frac{pi}{3}$ about $B$. Let $D'=R(D)$. Therefore, $triangle BDD'$ is equilateral. Also, $C=R(A)$. Therefore, $CD'=R(A)R(D)=AD$, as rotation preserves length. Therefore, $triangle CDD'cong T$. As rotation preserves angles too, $angle BCD'=angle BR(A)R(D)=angle BAD$. Therefore, $angle DCD'=angle DCB+angle BAD= 2pi-theta_A-theta_C-frac{pi}{3}=theta_B-frac{pi}{3}$. Therefore, angle opposite to $BD$ in $T = theta_B-frac{pi}{3}$.



        Therefore, by symmetry, the angles of $T$ are $theta_A-frac{pi}{3}, theta_B-frac{pi}{3}$ and $theta_C-frac{pi}{3}$.



        $blacksquare$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 25 at 16:49









        Anubhab GhosalAnubhab Ghosal

        1,22319




        1,22319






























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