$cos(a)=4/5$, $sin(b)=12/13$, what is $sin(c)$?

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$cos(a)=4/5$, $sin(b)=12/13$, what is $sin(c)$?
with $a,b,c < fracpi2$.



Attempt :



Since $sin^2(a) = 1 - cos^2(a)$, I get $sin(a) = frac35$.
But how to get $sin(c)$?



I have tried using $$ fracAsin(a) = fracBsin(b) = fracCsin(c)$$
and get
$$ fracAB = frac1320= fracsin(a)sin(b) $$ so $A=13, B = 20$
. How to find $sin(c)$? Thanks.




trigonometry triangle angle




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  • 4




    Assuming that's a triangle, the angles sum up to $,pi,$, so $;sin(c) = sin(a+b),$.
    – dxiv
    Jul 17 at 1:44











  • @dxiv Yes of course. Thanks.
    – Arief
    Jul 17 at 1:50














up vote
0
down vote

favorite












$cos(a)=4/5$, $sin(b)=12/13$, what is $sin(c)$?
with $a,b,c < fracpi2$.



Attempt :



Since $sin^2(a) = 1 - cos^2(a)$, I get $sin(a) = frac35$.
But how to get $sin(c)$?



I have tried using $$ fracAsin(a) = fracBsin(b) = fracCsin(c)$$
and get
$$ fracAB = frac1320= fracsin(a)sin(b) $$ so $A=13, B = 20$
. How to find $sin(c)$? Thanks.




trigonometry triangle angle




share|cite|improve this question















  • 4




    Assuming that's a triangle, the angles sum up to $,pi,$, so $;sin(c) = sin(a+b),$.
    – dxiv
    Jul 17 at 1:44











  • @dxiv Yes of course. Thanks.
    – Arief
    Jul 17 at 1:50












up vote
0
down vote

favorite









up vote
0
down vote

favorite











$cos(a)=4/5$, $sin(b)=12/13$, what is $sin(c)$?
with $a,b,c < fracpi2$.



Attempt :



Since $sin^2(a) = 1 - cos^2(a)$, I get $sin(a) = frac35$.
But how to get $sin(c)$?



I have tried using $$ fracAsin(a) = fracBsin(b) = fracCsin(c)$$
and get
$$ fracAB = frac1320= fracsin(a)sin(b) $$ so $A=13, B = 20$
. How to find $sin(c)$? Thanks.




trigonometry triangle angle




share|cite|improve this question











$cos(a)=4/5$, $sin(b)=12/13$, what is $sin(c)$?
with $a,b,c < fracpi2$.



Attempt :



Since $sin^2(a) = 1 - cos^2(a)$, I get $sin(a) = frac35$.
But how to get $sin(c)$?



I have tried using $$ fracAsin(a) = fracBsin(b) = fracCsin(c)$$
and get
$$ fracAB = frac1320= fracsin(a)sin(b) $$ so $A=13, B = 20$
. How to find $sin(c)$? Thanks.





trigonometry triangle angle





share|cite|improve this question










share|cite|improve this question




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asked Jul 17 at 1:42









Arief

1,3311522




1,3311522







  • 4




    Assuming that's a triangle, the angles sum up to $,pi,$, so $;sin(c) = sin(a+b),$.
    – dxiv
    Jul 17 at 1:44











  • @dxiv Yes of course. Thanks.
    – Arief
    Jul 17 at 1:50












  • 4




    Assuming that's a triangle, the angles sum up to $,pi,$, so $;sin(c) = sin(a+b),$.
    – dxiv
    Jul 17 at 1:44











  • @dxiv Yes of course. Thanks.
    – Arief
    Jul 17 at 1:50







4




4




Assuming that's a triangle, the angles sum up to $,pi,$, so $;sin(c) = sin(a+b),$.
– dxiv
Jul 17 at 1:44





Assuming that's a triangle, the angles sum up to $,pi,$, so $;sin(c) = sin(a+b),$.
– dxiv
Jul 17 at 1:44













@dxiv Yes of course. Thanks.
– Arief
Jul 17 at 1:50




@dxiv Yes of course. Thanks.
– Arief
Jul 17 at 1:50










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Note that $$c=pi -(a+b)$$ Therefore $$sin c = sin (a+b)=sin a cos b +cos a sin b=(3/5)(5/13)+(4/5)(12/13)=63/65 $$






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    up vote
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    Note that $$c=pi -(a+b)$$ Therefore $$sin c = sin (a+b)=sin a cos b +cos a sin b=(3/5)(5/13)+(4/5)(12/13)=63/65 $$






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      up vote
      0
      down vote













      Note that $$c=pi -(a+b)$$ Therefore $$sin c = sin (a+b)=sin a cos b +cos a sin b=(3/5)(5/13)+(4/5)(12/13)=63/65 $$






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        up vote
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        Note that $$c=pi -(a+b)$$ Therefore $$sin c = sin (a+b)=sin a cos b +cos a sin b=(3/5)(5/13)+(4/5)(12/13)=63/65 $$






        share|cite|improve this answer













        Note that $$c=pi -(a+b)$$ Therefore $$sin c = sin (a+b)=sin a cos b +cos a sin b=(3/5)(5/13)+(4/5)(12/13)=63/65 $$







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        answered Jul 17 at 2:20









        Mohammad Riazi-Kermani

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