How to use the Law of Sines to Find an Angle

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up vote
2
down vote

favorite












I am trying to figure out how to find an angle with the law of sines.



I have a triangle where:



A = $120^circ$



B = unmarked



C = $theta$



a = 45



b = unmarked



c = 36



How can I find the angle for C?



I have tried:



$$fracsin120^circ45 = fracsinBb =fracsintheta36 $$



$$36(fracsintheta36) = 36(fracsin120^circ45)$$
$$sintheta = frac36sin120^circ45$$
$$ theta = frac36(120^circ)45 = 95^circ$$



But, the answer is supposed to be $44^circ$.






algebra-precalculus triangle angle







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  • A sin is missed in your last line!
    – Dr. Sonnhard Graubner
    1 hour ago










  • I just dropped the sin altogether. If that's not what I do, then I don't know how to solve it.
    – LuminousNutria
    1 hour ago






  • 1




    $$sin(theta)=frac3645sin(120^circ)approx 43.853..$$
    – Dr. Sonnhard Graubner
    1 hour ago










  • Hint: $$sin(120^circ)=fracsqrt32$$
    – Dr. Sonnhard Graubner
    1 hour ago







  • 1




    You must press the inv sin bottom on your calculator.
    – Dr. Sonnhard Graubner
    1 hour ago














up vote
2
down vote

favorite












I am trying to figure out how to find an angle with the law of sines.



I have a triangle where:



A = $120^circ$



B = unmarked



C = $theta$



a = 45



b = unmarked



c = 36



How can I find the angle for C?



I have tried:



$$fracsin120^circ45 = fracsinBb =fracsintheta36 $$



$$36(fracsintheta36) = 36(fracsin120^circ45)$$
$$sintheta = frac36sin120^circ45$$
$$ theta = frac36(120^circ)45 = 95^circ$$



But, the answer is supposed to be $44^circ$.






algebra-precalculus triangle angle







share|cite|improve this question























  • A sin is missed in your last line!
    – Dr. Sonnhard Graubner
    1 hour ago










  • I just dropped the sin altogether. If that's not what I do, then I don't know how to solve it.
    – LuminousNutria
    1 hour ago






  • 1




    $$sin(theta)=frac3645sin(120^circ)approx 43.853..$$
    – Dr. Sonnhard Graubner
    1 hour ago










  • Hint: $$sin(120^circ)=fracsqrt32$$
    – Dr. Sonnhard Graubner
    1 hour ago







  • 1




    You must press the inv sin bottom on your calculator.
    – Dr. Sonnhard Graubner
    1 hour ago












up vote
2
down vote

favorite









up vote
2
down vote

favorite











I am trying to figure out how to find an angle with the law of sines.



I have a triangle where:



A = $120^circ$



B = unmarked



C = $theta$



a = 45



b = unmarked



c = 36



How can I find the angle for C?



I have tried:



$$fracsin120^circ45 = fracsinBb =fracsintheta36 $$



$$36(fracsintheta36) = 36(fracsin120^circ45)$$
$$sintheta = frac36sin120^circ45$$
$$ theta = frac36(120^circ)45 = 95^circ$$



But, the answer is supposed to be $44^circ$.






algebra-precalculus triangle angle







share|cite|improve this question















I am trying to figure out how to find an angle with the law of sines.



I have a triangle where:



A = $120^circ$



B = unmarked



C = $theta$



a = 45



b = unmarked



c = 36



How can I find the angle for C?



I have tried:



$$fracsin120^circ45 = fracsinBb =fracsintheta36 $$



$$36(fracsintheta36) = 36(fracsin120^circ45)$$
$$sintheta = frac36sin120^circ45$$
$$ theta = frac36(120^circ)45 = 95^circ$$



But, the answer is supposed to be $44^circ$.






algebra-precalculus triangle angle




algebra-precalculus triangle angle






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited 2 hours ago

























asked 2 hours ago









LuminousNutria

235




235











  • A sin is missed in your last line!
    – Dr. Sonnhard Graubner
    1 hour ago










  • I just dropped the sin altogether. If that's not what I do, then I don't know how to solve it.
    – LuminousNutria
    1 hour ago






  • 1




    $$sin(theta)=frac3645sin(120^circ)approx 43.853..$$
    – Dr. Sonnhard Graubner
    1 hour ago










  • Hint: $$sin(120^circ)=fracsqrt32$$
    – Dr. Sonnhard Graubner
    1 hour ago







  • 1




    You must press the inv sin bottom on your calculator.
    – Dr. Sonnhard Graubner
    1 hour ago
















  • A sin is missed in your last line!
    – Dr. Sonnhard Graubner
    1 hour ago










  • I just dropped the sin altogether. If that's not what I do, then I don't know how to solve it.
    – LuminousNutria
    1 hour ago






  • 1




    $$sin(theta)=frac3645sin(120^circ)approx 43.853..$$
    – Dr. Sonnhard Graubner
    1 hour ago










  • Hint: $$sin(120^circ)=fracsqrt32$$
    – Dr. Sonnhard Graubner
    1 hour ago







  • 1




    You must press the inv sin bottom on your calculator.
    – Dr. Sonnhard Graubner
    1 hour ago















A sin is missed in your last line!
– Dr. Sonnhard Graubner
1 hour ago




A sin is missed in your last line!
– Dr. Sonnhard Graubner
1 hour ago












I just dropped the sin altogether. If that's not what I do, then I don't know how to solve it.
– LuminousNutria
1 hour ago




I just dropped the sin altogether. If that's not what I do, then I don't know how to solve it.
– LuminousNutria
1 hour ago




1




1




$$sin(theta)=frac3645sin(120^circ)approx 43.853..$$
– Dr. Sonnhard Graubner
1 hour ago




$$sin(theta)=frac3645sin(120^circ)approx 43.853..$$
– Dr. Sonnhard Graubner
1 hour ago












Hint: $$sin(120^circ)=fracsqrt32$$
– Dr. Sonnhard Graubner
1 hour ago





Hint: $$sin(120^circ)=fracsqrt32$$
– Dr. Sonnhard Graubner
1 hour ago





1




1




You must press the inv sin bottom on your calculator.
– Dr. Sonnhard Graubner
1 hour ago




You must press the inv sin bottom on your calculator.
– Dr. Sonnhard Graubner
1 hour ago










2 Answers
2






active

oldest

votes

















up vote
3
down vote



accepted










The following step is completely wrong



$$sintheta = frac36 sin120^circ45 iff colorredtheta = frac36(120^circ)45 = 95^circ$$



we have that $sin120^circ=sqrt 3/2$ and then



$$sintheta = frac36sin120^circ45 iff sin theta = frac36sqrt 32cdot 45 iff theta = arcsin left(frac2sqrt 39right)$$






share|cite|improve this answer






















  • I'm sorry, my class hasn't covered arcsin yet. Could you please show how to get the answer in degrees?
    – LuminousNutria
    1 hour ago










  • Note that $arcsin x$ is the inverse function of $sin x$, we need that to find the value for $theta$ here. You should have itnon your calculator and you should also be able to obtain the result in degree directly or convrting by radians.
    – gimusi
    1 hour ago










  • Note also that in that case we can use $arcsin x$ to obtain the result because we are sure that the angle is acute otherwise, when the angle is obtuse, we need to take $pi - arcsin x$.
    – gimusi
    1 hour ago

















up vote
1
down vote













Most of your steps are correct, but it is from the second last step you made a fundamental mistake.



Using arcsine ($sine^-1$ on calculator):



$$sintheta = frac36sin120^circ45 $$
$$theta = arcsine(frac36sin120^circ45) $$
$$theta = 43.8537^circ... $$
$$theta approx 44^circ $$



Just as @Dr. Sonnhard Graubner pointed out. The mistake you made was that you interpreted arsine and sin$theta$ canceling out, because sometimes it is learned to students that $arcsine = frac1sin$. The thing is, $sin$ on itself cannot be cancelled out, as it is a trigonometric function of an angle, not $s . i . n$ as variables.






share|cite|improve this answer








New contributor




deadlyvirus is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

















  • Ah, I took to long to post a comment. At least @gimusi got to it!
    – deadlyvirus
    1 hour ago










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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes








up vote
3
down vote



accepted










The following step is completely wrong



$$sintheta = frac36 sin120^circ45 iff colorredtheta = frac36(120^circ)45 = 95^circ$$



we have that $sin120^circ=sqrt 3/2$ and then



$$sintheta = frac36sin120^circ45 iff sin theta = frac36sqrt 32cdot 45 iff theta = arcsin left(frac2sqrt 39right)$$






share|cite|improve this answer






















  • I'm sorry, my class hasn't covered arcsin yet. Could you please show how to get the answer in degrees?
    – LuminousNutria
    1 hour ago










  • Note that $arcsin x$ is the inverse function of $sin x$, we need that to find the value for $theta$ here. You should have itnon your calculator and you should also be able to obtain the result in degree directly or convrting by radians.
    – gimusi
    1 hour ago










  • Note also that in that case we can use $arcsin x$ to obtain the result because we are sure that the angle is acute otherwise, when the angle is obtuse, we need to take $pi - arcsin x$.
    – gimusi
    1 hour ago














up vote
3
down vote



accepted










The following step is completely wrong



$$sintheta = frac36 sin120^circ45 iff colorredtheta = frac36(120^circ)45 = 95^circ$$



we have that $sin120^circ=sqrt 3/2$ and then



$$sintheta = frac36sin120^circ45 iff sin theta = frac36sqrt 32cdot 45 iff theta = arcsin left(frac2sqrt 39right)$$






share|cite|improve this answer






















  • I'm sorry, my class hasn't covered arcsin yet. Could you please show how to get the answer in degrees?
    – LuminousNutria
    1 hour ago










  • Note that $arcsin x$ is the inverse function of $sin x$, we need that to find the value for $theta$ here. You should have itnon your calculator and you should also be able to obtain the result in degree directly or convrting by radians.
    – gimusi
    1 hour ago










  • Note also that in that case we can use $arcsin x$ to obtain the result because we are sure that the angle is acute otherwise, when the angle is obtuse, we need to take $pi - arcsin x$.
    – gimusi
    1 hour ago












up vote
3
down vote



accepted







up vote
3
down vote



accepted






The following step is completely wrong



$$sintheta = frac36 sin120^circ45 iff colorredtheta = frac36(120^circ)45 = 95^circ$$



we have that $sin120^circ=sqrt 3/2$ and then



$$sintheta = frac36sin120^circ45 iff sin theta = frac36sqrt 32cdot 45 iff theta = arcsin left(frac2sqrt 39right)$$






share|cite|improve this answer














The following step is completely wrong



$$sintheta = frac36 sin120^circ45 iff colorredtheta = frac36(120^circ)45 = 95^circ$$



we have that $sin120^circ=sqrt 3/2$ and then



$$sintheta = frac36sin120^circ45 iff sin theta = frac36sqrt 32cdot 45 iff theta = arcsin left(frac2sqrt 39right)$$







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited 1 hour ago

























answered 1 hour ago









gimusi

78.1k73889




78.1k73889











  • I'm sorry, my class hasn't covered arcsin yet. Could you please show how to get the answer in degrees?
    – LuminousNutria
    1 hour ago










  • Note that $arcsin x$ is the inverse function of $sin x$, we need that to find the value for $theta$ here. You should have itnon your calculator and you should also be able to obtain the result in degree directly or convrting by radians.
    – gimusi
    1 hour ago










  • Note also that in that case we can use $arcsin x$ to obtain the result because we are sure that the angle is acute otherwise, when the angle is obtuse, we need to take $pi - arcsin x$.
    – gimusi
    1 hour ago
















  • I'm sorry, my class hasn't covered arcsin yet. Could you please show how to get the answer in degrees?
    – LuminousNutria
    1 hour ago










  • Note that $arcsin x$ is the inverse function of $sin x$, we need that to find the value for $theta$ here. You should have itnon your calculator and you should also be able to obtain the result in degree directly or convrting by radians.
    – gimusi
    1 hour ago










  • Note also that in that case we can use $arcsin x$ to obtain the result because we are sure that the angle is acute otherwise, when the angle is obtuse, we need to take $pi - arcsin x$.
    – gimusi
    1 hour ago















I'm sorry, my class hasn't covered arcsin yet. Could you please show how to get the answer in degrees?
– LuminousNutria
1 hour ago




I'm sorry, my class hasn't covered arcsin yet. Could you please show how to get the answer in degrees?
– LuminousNutria
1 hour ago












Note that $arcsin x$ is the inverse function of $sin x$, we need that to find the value for $theta$ here. You should have itnon your calculator and you should also be able to obtain the result in degree directly or convrting by radians.
– gimusi
1 hour ago




Note that $arcsin x$ is the inverse function of $sin x$, we need that to find the value for $theta$ here. You should have itnon your calculator and you should also be able to obtain the result in degree directly or convrting by radians.
– gimusi
1 hour ago












Note also that in that case we can use $arcsin x$ to obtain the result because we are sure that the angle is acute otherwise, when the angle is obtuse, we need to take $pi - arcsin x$.
– gimusi
1 hour ago




Note also that in that case we can use $arcsin x$ to obtain the result because we are sure that the angle is acute otherwise, when the angle is obtuse, we need to take $pi - arcsin x$.
– gimusi
1 hour ago










up vote
1
down vote













Most of your steps are correct, but it is from the second last step you made a fundamental mistake.



Using arcsine ($sine^-1$ on calculator):



$$sintheta = frac36sin120^circ45 $$
$$theta = arcsine(frac36sin120^circ45) $$
$$theta = 43.8537^circ... $$
$$theta approx 44^circ $$



Just as @Dr. Sonnhard Graubner pointed out. The mistake you made was that you interpreted arsine and sin$theta$ canceling out, because sometimes it is learned to students that $arcsine = frac1sin$. The thing is, $sin$ on itself cannot be cancelled out, as it is a trigonometric function of an angle, not $s . i . n$ as variables.






share|cite|improve this answer








New contributor




deadlyvirus is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

















  • Ah, I took to long to post a comment. At least @gimusi got to it!
    – deadlyvirus
    1 hour ago














up vote
1
down vote













Most of your steps are correct, but it is from the second last step you made a fundamental mistake.



Using arcsine ($sine^-1$ on calculator):



$$sintheta = frac36sin120^circ45 $$
$$theta = arcsine(frac36sin120^circ45) $$
$$theta = 43.8537^circ... $$
$$theta approx 44^circ $$



Just as @Dr. Sonnhard Graubner pointed out. The mistake you made was that you interpreted arsine and sin$theta$ canceling out, because sometimes it is learned to students that $arcsine = frac1sin$. The thing is, $sin$ on itself cannot be cancelled out, as it is a trigonometric function of an angle, not $s . i . n$ as variables.






share|cite|improve this answer








New contributor




deadlyvirus is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

















  • Ah, I took to long to post a comment. At least @gimusi got to it!
    – deadlyvirus
    1 hour ago












up vote
1
down vote










up vote
1
down vote









Most of your steps are correct, but it is from the second last step you made a fundamental mistake.



Using arcsine ($sine^-1$ on calculator):



$$sintheta = frac36sin120^circ45 $$
$$theta = arcsine(frac36sin120^circ45) $$
$$theta = 43.8537^circ... $$
$$theta approx 44^circ $$



Just as @Dr. Sonnhard Graubner pointed out. The mistake you made was that you interpreted arsine and sin$theta$ canceling out, because sometimes it is learned to students that $arcsine = frac1sin$. The thing is, $sin$ on itself cannot be cancelled out, as it is a trigonometric function of an angle, not $s . i . n$ as variables.






share|cite|improve this answer








New contributor




deadlyvirus is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









Most of your steps are correct, but it is from the second last step you made a fundamental mistake.



Using arcsine ($sine^-1$ on calculator):



$$sintheta = frac36sin120^circ45 $$
$$theta = arcsine(frac36sin120^circ45) $$
$$theta = 43.8537^circ... $$
$$theta approx 44^circ $$



Just as @Dr. Sonnhard Graubner pointed out. The mistake you made was that you interpreted arsine and sin$theta$ canceling out, because sometimes it is learned to students that $arcsine = frac1sin$. The thing is, $sin$ on itself cannot be cancelled out, as it is a trigonometric function of an angle, not $s . i . n$ as variables.







share|cite|improve this answer








New contributor




deadlyvirus is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









share|cite|improve this answer



share|cite|improve this answer






New contributor




deadlyvirus is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









answered 1 hour ago









deadlyvirus

364




364




New contributor




deadlyvirus is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.





New contributor





deadlyvirus is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.






deadlyvirus is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











  • Ah, I took to long to post a comment. At least @gimusi got to it!
    – deadlyvirus
    1 hour ago
















  • Ah, I took to long to post a comment. At least @gimusi got to it!
    – deadlyvirus
    1 hour ago















Ah, I took to long to post a comment. At least @gimusi got to it!
– deadlyvirus
1 hour ago




Ah, I took to long to post a comment. At least @gimusi got to it!
– deadlyvirus
1 hour ago

















 

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