Hard trigonometric problem (picture included) [closed]












2












$begingroup$


triangles



I’ve been trying to solve this problem and find the indicated angle but unfortunately I couldn’t find the answer. Could anyone one please help me with the answer and how I can find the angle?
Thanks






geometry







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$endgroup$



closed as off-topic by José Carlos Santos, Claude Leibovici, mrtaurho, Adrian Keister, Abcd Jan 29 at 16:13


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – José Carlos Santos, Claude Leibovici, mrtaurho, Adrian Keister, Abcd

If this question can be reworded to fit the rules in the help center, please edit the question.












  • 1




    $begingroup$
    youtube.com/watch?v=5vhklRWogzo
    $endgroup$
    – Matti P.
    Jan 28 at 10:21










  • $begingroup$
    Use the sine rule in the two triangles
    $endgroup$
    – Shubham Johri
    Jan 28 at 10:33
















2












$begingroup$


triangles



I’ve been trying to solve this problem and find the indicated angle but unfortunately I couldn’t find the answer. Could anyone one please help me with the answer and how I can find the angle?
Thanks






geometry







share|cite|improve this question











$endgroup$



closed as off-topic by José Carlos Santos, Claude Leibovici, mrtaurho, Adrian Keister, Abcd Jan 29 at 16:13


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – José Carlos Santos, Claude Leibovici, mrtaurho, Adrian Keister, Abcd

If this question can be reworded to fit the rules in the help center, please edit the question.












  • 1




    $begingroup$
    youtube.com/watch?v=5vhklRWogzo
    $endgroup$
    – Matti P.
    Jan 28 at 10:21










  • $begingroup$
    Use the sine rule in the two triangles
    $endgroup$
    – Shubham Johri
    Jan 28 at 10:33














2












2








2





$begingroup$


triangles



I’ve been trying to solve this problem and find the indicated angle but unfortunately I couldn’t find the answer. Could anyone one please help me with the answer and how I can find the angle?
Thanks






geometry







share|cite|improve this question











$endgroup$




triangles



I’ve been trying to solve this problem and find the indicated angle but unfortunately I couldn’t find the answer. Could anyone one please help me with the answer and how I can find the angle?
Thanks






geometry




geometry






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 28 at 10:31









N. F. Taussig

44.8k103358




44.8k103358










asked Jan 28 at 10:20









Aliah56Aliah56

111




111




closed as off-topic by José Carlos Santos, Claude Leibovici, mrtaurho, Adrian Keister, Abcd Jan 29 at 16:13


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – José Carlos Santos, Claude Leibovici, mrtaurho, Adrian Keister, Abcd

If this question can be reworded to fit the rules in the help center, please edit the question.







closed as off-topic by José Carlos Santos, Claude Leibovici, mrtaurho, Adrian Keister, Abcd Jan 29 at 16:13


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – José Carlos Santos, Claude Leibovici, mrtaurho, Adrian Keister, Abcd

If this question can be reworded to fit the rules in the help center, please edit the question.








  • 1




    $begingroup$
    youtube.com/watch?v=5vhklRWogzo
    $endgroup$
    – Matti P.
    Jan 28 at 10:21










  • $begingroup$
    Use the sine rule in the two triangles
    $endgroup$
    – Shubham Johri
    Jan 28 at 10:33














  • 1




    $begingroup$
    youtube.com/watch?v=5vhklRWogzo
    $endgroup$
    – Matti P.
    Jan 28 at 10:21










  • $begingroup$
    Use the sine rule in the two triangles
    $endgroup$
    – Shubham Johri
    Jan 28 at 10:33








1




1




$begingroup$
youtube.com/watch?v=5vhklRWogzo
$endgroup$
– Matti P.
Jan 28 at 10:21




$begingroup$
youtube.com/watch?v=5vhklRWogzo
$endgroup$
– Matti P.
Jan 28 at 10:21












$begingroup$
Use the sine rule in the two triangles
$endgroup$
– Shubham Johri
Jan 28 at 10:33




$begingroup$
Use the sine rule in the two triangles
$endgroup$
– Shubham Johri
Jan 28 at 10:33










1 Answer
1






active

oldest

votes


















2












$begingroup$

The youtube video referenced by Matti P. shows the elementary method.



Alternative method with the use of trigonometry. Label the equal sides by $x$, the inside line by $y$ and the searched angle by $alpha$. Using the Sine rule:
$$frac{x}{sin alpha}=frac{y}{sin 80}\
frac{x}{sin (alpha -20)}=frac{y}{sin 20}.$$
Divide the first by the second:
$$frac{sin (alpha-20)}{sin alpha}=frac{sin 20}{sin 80} =frac{2sin 10cos 10}{cos 10}=frac{sin 10}{frac12} Rightarrowalpha =30.$$






share|cite|improve this answer









$endgroup$




















    1 Answer
    1






    active

    oldest

    votes








    1 Answer
    1






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    2












    $begingroup$

    The youtube video referenced by Matti P. shows the elementary method.



    Alternative method with the use of trigonometry. Label the equal sides by $x$, the inside line by $y$ and the searched angle by $alpha$. Using the Sine rule:
    $$frac{x}{sin alpha}=frac{y}{sin 80}\
    frac{x}{sin (alpha -20)}=frac{y}{sin 20}.$$
    Divide the first by the second:
    $$frac{sin (alpha-20)}{sin alpha}=frac{sin 20}{sin 80} =frac{2sin 10cos 10}{cos 10}=frac{sin 10}{frac12} Rightarrowalpha =30.$$






    share|cite|improve this answer









    $endgroup$


















      2












      $begingroup$

      The youtube video referenced by Matti P. shows the elementary method.



      Alternative method with the use of trigonometry. Label the equal sides by $x$, the inside line by $y$ and the searched angle by $alpha$. Using the Sine rule:
      $$frac{x}{sin alpha}=frac{y}{sin 80}\
      frac{x}{sin (alpha -20)}=frac{y}{sin 20}.$$
      Divide the first by the second:
      $$frac{sin (alpha-20)}{sin alpha}=frac{sin 20}{sin 80} =frac{2sin 10cos 10}{cos 10}=frac{sin 10}{frac12} Rightarrowalpha =30.$$






      share|cite|improve this answer









      $endgroup$
















        2












        2








        2





        $begingroup$

        The youtube video referenced by Matti P. shows the elementary method.



        Alternative method with the use of trigonometry. Label the equal sides by $x$, the inside line by $y$ and the searched angle by $alpha$. Using the Sine rule:
        $$frac{x}{sin alpha}=frac{y}{sin 80}\
        frac{x}{sin (alpha -20)}=frac{y}{sin 20}.$$
        Divide the first by the second:
        $$frac{sin (alpha-20)}{sin alpha}=frac{sin 20}{sin 80} =frac{2sin 10cos 10}{cos 10}=frac{sin 10}{frac12} Rightarrowalpha =30.$$






        share|cite|improve this answer









        $endgroup$



        The youtube video referenced by Matti P. shows the elementary method.



        Alternative method with the use of trigonometry. Label the equal sides by $x$, the inside line by $y$ and the searched angle by $alpha$. Using the Sine rule:
        $$frac{x}{sin alpha}=frac{y}{sin 80}\
        frac{x}{sin (alpha -20)}=frac{y}{sin 20}.$$
        Divide the first by the second:
        $$frac{sin (alpha-20)}{sin alpha}=frac{sin 20}{sin 80} =frac{2sin 10cos 10}{cos 10}=frac{sin 10}{frac12} Rightarrowalpha =30.$$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 28 at 12:00









        farruhotafarruhota

        21.7k2842




        21.7k2842















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