I'm having trouble on Law of Cosines and Area of a Triangle. Could Someone help?











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I'm having trouble with Question 10. I have figured out x, which is 2. I'm not sure how I can find the area of the triangle. Please look at the attachment above.






trigonometry







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  • What formulas do you know for the area of a triangle?
    – rogerl
    4 hours ago















up vote
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enter image description here



I'm having trouble with Question 10. I have figured out x, which is 2. I'm not sure how I can find the area of the triangle. Please look at the attachment above.






trigonometry







share|cite|improve this question









New contributor




Joshuap88 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.




















  • What formulas do you know for the area of a triangle?
    – rogerl
    4 hours ago













up vote
3
down vote

favorite
1









up vote
3
down vote

favorite
1






1





enter image description here



I'm having trouble with Question 10. I have figured out x, which is 2. I'm not sure how I can find the area of the triangle. Please look at the attachment above.






trigonometry







share|cite|improve this question









New contributor




Joshuap88 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











enter image description here



I'm having trouble with Question 10. I have figured out x, which is 2. I'm not sure how I can find the area of the triangle. Please look at the attachment above.






trigonometry




trigonometry






share|cite|improve this question









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Joshuap88 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 question









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edited 4 hours ago









gimusi

89.4k74495




89.4k74495






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asked 4 hours ago









Joshuap88

112




112




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Joshuap88 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.






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Check out our Code of Conduct.












  • What formulas do you know for the area of a triangle?
    – rogerl
    4 hours ago


















  • What formulas do you know for the area of a triangle?
    – rogerl
    4 hours ago
















What formulas do you know for the area of a triangle?
– rogerl
4 hours ago




What formulas do you know for the area of a triangle?
– rogerl
4 hours ago










2 Answers
2






active

oldest

votes

















up vote
4
down vote













HINT



By law of cosine we have



$$c^2 = a^2 + b^2-2abcos theta$$



then we can find the area by



$$A=frac12 absin theta$$






share|cite|improve this answer





















  • Yeah, that's probably easier than Heron.
    – Ben W
    4 hours ago










  • @BenW Yes indeed I think that the idea and the spirit for the homework is that. But Heron is always good invoke :)
    – gimusi
    3 hours ago










  • @gimusi It's interesting that we get two solutions: one with $x = -frac{12}{11}$ and one with $x=2$.
    – Toby Mak
    3 hours ago










  • @TobyMak I didn't check that. I'll take a look.
    – gimusi
    3 hours ago






  • 1




    @TobyMak Sorry, I didn't see that clearly but of course we need to esclude -12/11 therefore x=2 is the good one!
    – gimusi
    3 hours ago


















up vote
3
down vote













If you have $x$, then you have all three side lengths. Now just apply Heron's formula.






share|cite|improve this answer





















  • Thank you! This is definitely helpful if I don't have a calculator!
    – Joshuap88
    3 hours ago










  • @Joshuap88 You would still need to find $x$ using a calculator.
    – Toby Mak
    3 hours ago










  • @TobyMak $x$ can by surely found easily by hands from the quadratic equation obtined by the law of cosine.
    – gimusi
    3 hours ago










  • @BenW With your method how would you use the law of cosine?
    – gimusi
    3 hours ago






  • 1




    Didn't you use the law of cosines to find $x$? Now you no longer need it. Whether you use Heron's formula or $A=frac{1}{2}absintheta$, you are done with LoC.
    – Ben W
    3 hours 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
4
down vote













HINT



By law of cosine we have



$$c^2 = a^2 + b^2-2abcos theta$$



then we can find the area by



$$A=frac12 absin theta$$






share|cite|improve this answer





















  • Yeah, that's probably easier than Heron.
    – Ben W
    4 hours ago










  • @BenW Yes indeed I think that the idea and the spirit for the homework is that. But Heron is always good invoke :)
    – gimusi
    3 hours ago










  • @gimusi It's interesting that we get two solutions: one with $x = -frac{12}{11}$ and one with $x=2$.
    – Toby Mak
    3 hours ago










  • @TobyMak I didn't check that. I'll take a look.
    – gimusi
    3 hours ago






  • 1




    @TobyMak Sorry, I didn't see that clearly but of course we need to esclude -12/11 therefore x=2 is the good one!
    – gimusi
    3 hours ago















up vote
4
down vote













HINT



By law of cosine we have



$$c^2 = a^2 + b^2-2abcos theta$$



then we can find the area by



$$A=frac12 absin theta$$






share|cite|improve this answer





















  • Yeah, that's probably easier than Heron.
    – Ben W
    4 hours ago










  • @BenW Yes indeed I think that the idea and the spirit for the homework is that. But Heron is always good invoke :)
    – gimusi
    3 hours ago










  • @gimusi It's interesting that we get two solutions: one with $x = -frac{12}{11}$ and one with $x=2$.
    – Toby Mak
    3 hours ago










  • @TobyMak I didn't check that. I'll take a look.
    – gimusi
    3 hours ago






  • 1




    @TobyMak Sorry, I didn't see that clearly but of course we need to esclude -12/11 therefore x=2 is the good one!
    – gimusi
    3 hours ago













up vote
4
down vote










up vote
4
down vote









HINT



By law of cosine we have



$$c^2 = a^2 + b^2-2abcos theta$$



then we can find the area by



$$A=frac12 absin theta$$






share|cite|improve this answer












HINT



By law of cosine we have



$$c^2 = a^2 + b^2-2abcos theta$$



then we can find the area by



$$A=frac12 absin theta$$







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered 4 hours ago









gimusi

89.4k74495




89.4k74495












  • Yeah, that's probably easier than Heron.
    – Ben W
    4 hours ago










  • @BenW Yes indeed I think that the idea and the spirit for the homework is that. But Heron is always good invoke :)
    – gimusi
    3 hours ago










  • @gimusi It's interesting that we get two solutions: one with $x = -frac{12}{11}$ and one with $x=2$.
    – Toby Mak
    3 hours ago










  • @TobyMak I didn't check that. I'll take a look.
    – gimusi
    3 hours ago






  • 1




    @TobyMak Sorry, I didn't see that clearly but of course we need to esclude -12/11 therefore x=2 is the good one!
    – gimusi
    3 hours ago


















  • Yeah, that's probably easier than Heron.
    – Ben W
    4 hours ago










  • @BenW Yes indeed I think that the idea and the spirit for the homework is that. But Heron is always good invoke :)
    – gimusi
    3 hours ago










  • @gimusi It's interesting that we get two solutions: one with $x = -frac{12}{11}$ and one with $x=2$.
    – Toby Mak
    3 hours ago










  • @TobyMak I didn't check that. I'll take a look.
    – gimusi
    3 hours ago






  • 1




    @TobyMak Sorry, I didn't see that clearly but of course we need to esclude -12/11 therefore x=2 is the good one!
    – gimusi
    3 hours ago
















Yeah, that's probably easier than Heron.
– Ben W
4 hours ago




Yeah, that's probably easier than Heron.
– Ben W
4 hours ago












@BenW Yes indeed I think that the idea and the spirit for the homework is that. But Heron is always good invoke :)
– gimusi
3 hours ago




@BenW Yes indeed I think that the idea and the spirit for the homework is that. But Heron is always good invoke :)
– gimusi
3 hours ago












@gimusi It's interesting that we get two solutions: one with $x = -frac{12}{11}$ and one with $x=2$.
– Toby Mak
3 hours ago




@gimusi It's interesting that we get two solutions: one with $x = -frac{12}{11}$ and one with $x=2$.
– Toby Mak
3 hours ago












@TobyMak I didn't check that. I'll take a look.
– gimusi
3 hours ago




@TobyMak I didn't check that. I'll take a look.
– gimusi
3 hours ago




1




1




@TobyMak Sorry, I didn't see that clearly but of course we need to esclude -12/11 therefore x=2 is the good one!
– gimusi
3 hours ago




@TobyMak Sorry, I didn't see that clearly but of course we need to esclude -12/11 therefore x=2 is the good one!
– gimusi
3 hours ago










up vote
3
down vote













If you have $x$, then you have all three side lengths. Now just apply Heron's formula.






share|cite|improve this answer





















  • Thank you! This is definitely helpful if I don't have a calculator!
    – Joshuap88
    3 hours ago










  • @Joshuap88 You would still need to find $x$ using a calculator.
    – Toby Mak
    3 hours ago










  • @TobyMak $x$ can by surely found easily by hands from the quadratic equation obtined by the law of cosine.
    – gimusi
    3 hours ago










  • @BenW With your method how would you use the law of cosine?
    – gimusi
    3 hours ago






  • 1




    Didn't you use the law of cosines to find $x$? Now you no longer need it. Whether you use Heron's formula or $A=frac{1}{2}absintheta$, you are done with LoC.
    – Ben W
    3 hours ago

















up vote
3
down vote













If you have $x$, then you have all three side lengths. Now just apply Heron's formula.






share|cite|improve this answer





















  • Thank you! This is definitely helpful if I don't have a calculator!
    – Joshuap88
    3 hours ago










  • @Joshuap88 You would still need to find $x$ using a calculator.
    – Toby Mak
    3 hours ago










  • @TobyMak $x$ can by surely found easily by hands from the quadratic equation obtined by the law of cosine.
    – gimusi
    3 hours ago










  • @BenW With your method how would you use the law of cosine?
    – gimusi
    3 hours ago






  • 1




    Didn't you use the law of cosines to find $x$? Now you no longer need it. Whether you use Heron's formula or $A=frac{1}{2}absintheta$, you are done with LoC.
    – Ben W
    3 hours ago















up vote
3
down vote










up vote
3
down vote









If you have $x$, then you have all three side lengths. Now just apply Heron's formula.






share|cite|improve this answer












If you have $x$, then you have all three side lengths. Now just apply Heron's formula.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered 4 hours ago









Ben W

1,224510




1,224510












  • Thank you! This is definitely helpful if I don't have a calculator!
    – Joshuap88
    3 hours ago










  • @Joshuap88 You would still need to find $x$ using a calculator.
    – Toby Mak
    3 hours ago










  • @TobyMak $x$ can by surely found easily by hands from the quadratic equation obtined by the law of cosine.
    – gimusi
    3 hours ago










  • @BenW With your method how would you use the law of cosine?
    – gimusi
    3 hours ago






  • 1




    Didn't you use the law of cosines to find $x$? Now you no longer need it. Whether you use Heron's formula or $A=frac{1}{2}absintheta$, you are done with LoC.
    – Ben W
    3 hours ago




















  • Thank you! This is definitely helpful if I don't have a calculator!
    – Joshuap88
    3 hours ago










  • @Joshuap88 You would still need to find $x$ using a calculator.
    – Toby Mak
    3 hours ago










  • @TobyMak $x$ can by surely found easily by hands from the quadratic equation obtined by the law of cosine.
    – gimusi
    3 hours ago










  • @BenW With your method how would you use the law of cosine?
    – gimusi
    3 hours ago






  • 1




    Didn't you use the law of cosines to find $x$? Now you no longer need it. Whether you use Heron's formula or $A=frac{1}{2}absintheta$, you are done with LoC.
    – Ben W
    3 hours ago


















Thank you! This is definitely helpful if I don't have a calculator!
– Joshuap88
3 hours ago




Thank you! This is definitely helpful if I don't have a calculator!
– Joshuap88
3 hours ago












@Joshuap88 You would still need to find $x$ using a calculator.
– Toby Mak
3 hours ago




@Joshuap88 You would still need to find $x$ using a calculator.
– Toby Mak
3 hours ago












@TobyMak $x$ can by surely found easily by hands from the quadratic equation obtined by the law of cosine.
– gimusi
3 hours ago




@TobyMak $x$ can by surely found easily by hands from the quadratic equation obtined by the law of cosine.
– gimusi
3 hours ago












@BenW With your method how would you use the law of cosine?
– gimusi
3 hours ago




@BenW With your method how would you use the law of cosine?
– gimusi
3 hours ago




1




1




Didn't you use the law of cosines to find $x$? Now you no longer need it. Whether you use Heron's formula or $A=frac{1}{2}absintheta$, you are done with LoC.
– Ben W
3 hours ago






Didn't you use the law of cosines to find $x$? Now you no longer need it. Whether you use Heron's formula or $A=frac{1}{2}absintheta$, you are done with LoC.
– Ben W
3 hours ago












Joshuap88 is a new contributor. Be nice, and check out our Code of Conduct.










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