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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add a comment |
up vote
2
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favorite
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
New contributor
What formulas do you know for the area of a triangle?
– rogerl
3 hours ago
add a comment |
up vote
2
down vote
favorite
up vote
2
down vote
favorite
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
New contributor
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
New contributor
New contributor
edited 3 hours ago
gimusi
89.4k74495
89.4k74495
New contributor
asked 3 hours ago
Joshuap88
62
62
New contributor
New contributor
What formulas do you know for the area of a triangle?
– rogerl
3 hours ago
add a comment |
What formulas do you know for the area of a triangle?
– rogerl
3 hours ago
What formulas do you know for the area of a triangle?
– rogerl
3 hours ago
What formulas do you know for the area of a triangle?
– rogerl
3 hours ago
add a comment |
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
$$A=frac12 absin theta$$
Yeah, that's probably easier than Heron.
– Ben W
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
@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
2 hours ago
add a comment |
up vote
3
down vote
If you have $x$, then you have all three side lengths. Now just apply Heron's formula.
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
2 hours ago
@BenW With your method how would you use the law of cosine?
– gimusi
2 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
2 hours ago
|
show 1 more comment
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
$$A=frac12 absin theta$$
Yeah, that's probably easier than Heron.
– Ben W
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
@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
2 hours ago
add a comment |
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
$$A=frac12 absin theta$$
Yeah, that's probably easier than Heron.
– Ben W
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
@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
2 hours ago
add a comment |
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
$$A=frac12 absin theta$$
HINT
By law of cosine we have
$$c^2 = a^2 + b^2-2abcos theta$$
then we can find the area by
$$A=frac12 absin theta$$
answered 3 hours ago
gimusi
89.4k74495
89.4k74495
Yeah, that's probably easier than Heron.
– Ben W
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
@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
2 hours ago
add a comment |
Yeah, that's probably easier than Heron.
– Ben W
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
@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
2 hours ago
Yeah, that's probably easier than Heron.
– Ben W
3 hours ago
Yeah, that's probably easier than Heron.
– Ben W
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
@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
2 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
2 hours ago
add a comment |
up vote
3
down vote
If you have $x$, then you have all three side lengths. Now just apply Heron's formula.
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
2 hours ago
@BenW With your method how would you use the law of cosine?
– gimusi
2 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
2 hours ago
|
show 1 more comment
up vote
3
down vote
If you have $x$, then you have all three side lengths. Now just apply Heron's formula.
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
2 hours ago
@BenW With your method how would you use the law of cosine?
– gimusi
2 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
2 hours ago
|
show 1 more comment
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.
If you have $x$, then you have all three side lengths. Now just apply Heron's formula.
answered 3 hours ago
Ben W
1,223510
1,223510
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
2 hours ago
@BenW With your method how would you use the law of cosine?
– gimusi
2 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
2 hours ago
|
show 1 more comment
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
2 hours ago
@BenW With your method how would you use the law of cosine?
– gimusi
2 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
2 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
2 hours ago
@TobyMak $x$ can by surely found easily by hands from the quadratic equation obtined by the law of cosine.
– gimusi
2 hours ago
@BenW With your method how would you use the law of cosine?
– gimusi
2 hours ago
@BenW With your method how would you use the law of cosine?
– gimusi
2 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
2 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
2 hours ago
|
show 1 more comment
Joshuap88 is a new contributor. Be nice, and check out our Code of Conduct.
Joshuap88 is a new contributor. Be nice, and check out our Code of Conduct.
Joshuap88 is a new contributor. Be nice, and check out our Code of Conduct.
Joshuap88 is a new contributor. Be nice, and check out our Code of Conduct.
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What formulas do you know for the area of a triangle?
– rogerl
3 hours ago