A triangle has sides $3x+1$, $x+2$, $x+3$ and an angle with a known cosine; find $x$ and area [closed]
1
$begingroup$
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
edited Dec 4 '18 at 4:09
Blue
47.9k870153
asked Dec 3 '18 at 23:26
Joshuap88Joshuap88
142
$endgroup$
closed as off-topic by Jyrki Lahtonen, Saad, José Carlos Santos, DRF, Chinnapparaj R Dec 4 '18 at 11:13
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 improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Jyrki Lahtonen, Saad, José Carlos Santos, DRF, Chinnapparaj R
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
1
$begingroup$
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
edited Dec 4 '18 at 4:09
Blue
47.9k870153
asked Dec 3 '18 at 23:26
Joshuap88Joshuap88
142
$endgroup$
closed as off-topic by Jyrki Lahtonen, Saad, José Carlos Santos, DRF, Chinnapparaj R Dec 4 '18 at 11:13
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 improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Jyrki Lahtonen, Saad, José Carlos Santos, DRF, Chinnapparaj R
If this question can be reworded to fit the rules in the help center, please edit the question.
$begingroup$
What did you try? Do you have some formulas for finding the areas of triangles?
$endgroup$
– DRF
Dec 4 '18 at 8:51
add a comment |
1
1
1
1
$begingroup$
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
edited Dec 4 '18 at 4:09
Blue
47.9k870153
asked Dec 3 '18 at 23:26
Joshuap88Joshuap88
142
$endgroup$
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
edited Dec 4 '18 at 4:09
Blue
47.9k870153
asked Dec 3 '18 at 23:26
Joshuap88Joshuap88
142
edited Dec 4 '18 at 4:09
Blue
47.9k870153
asked Dec 3 '18 at 23:26
Joshuap88Joshuap88
142
edited Dec 4 '18 at 4:09
Blue
47.9k870153
edited Dec 4 '18 at 4:09
Blue
47.9k870153
edited Dec 4 '18 at 4:09
Blue
47.9k870153
47.9k870153
asked Dec 3 '18 at 23:26
Joshuap88Joshuap88
142
asked Dec 3 '18 at 23:26
Joshuap88Joshuap88
142
asked Dec 3 '18 at 23:26
Joshuap88Joshuap88
142
142
closed as off-topic by Jyrki Lahtonen, Saad, José Carlos Santos, DRF, Chinnapparaj R Dec 4 '18 at 11:13
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 improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Jyrki Lahtonen, Saad, José Carlos Santos, DRF, Chinnapparaj R
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by Jyrki Lahtonen, Saad, José Carlos Santos, DRF, Chinnapparaj R Dec 4 '18 at 11:13
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 improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Jyrki Lahtonen, Saad, José Carlos Santos, DRF, Chinnapparaj R
If this question can be reworded to fit the rules in the help center, please edit the question.
$begingroup$
What did you try? Do you have some formulas for finding the areas of triangles?
$endgroup$
– DRF
Dec 4 '18 at 8:51
add a comment |
$begingroup$
What did you try? Do you have some formulas for finding the areas of triangles?
$endgroup$
– DRF
Dec 4 '18 at 8:51
$begingroup$
What did you try? Do you have some formulas for finding the areas of triangles?
$endgroup$
– DRF
Dec 4 '18 at 8:51
$begingroup$
What did you try? Do you have some formulas for finding the areas of triangles?
$endgroup$
– DRF
Dec 4 '18 at 8:51
add a comment |
2 Answers
2
active
oldest
votes
6
$begingroup$
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 Dec 3 '18 at 23:31
gimusigimusi
92.8k84494
$endgroup$
$begingroup$
Yeah, that's probably easier than Heron.
$endgroup$
– Ben W
Dec 3 '18 at 23:34
$begingroup$
@BenW Yes indeed I think that the idea and the spirit for the homework is that. But Heron is always good invoke :)
$endgroup$
– gimusi
Dec 3 '18 at 23:41
$begingroup$
@gimusi It's interesting that we get two solutions: one with $x = -frac{12}{11}$ and one with $x=2$.
$endgroup$
– Toby Mak
Dec 3 '18 at 23:44
$begingroup$
@TobyMak I didn't check that. I'll take a look.
$endgroup$
– gimusi
Dec 3 '18 at 23:51
1
$begingroup$
@TobyMak Sorry, I didn't see that clearly but of course we need to esclude -12/11 therefore x=2 is the good one!
$endgroup$
– gimusi
Dec 4 '18 at 0:08
add a comment |
5
$begingroup$
If you have $x$, then you have all three side lengths. Now just apply Heron's formula.
answered Dec 3 '18 at 23:29
Ben WBen W
2,234615
$endgroup$
$begingroup$
Thank you! This is definitely helpful if I don't have a calculator!
$endgroup$
– Joshuap88
Dec 3 '18 at 23:38
$begingroup$
@Joshuap88 You would still need to find $x$ using a calculator.
$endgroup$
– Toby Mak
Dec 3 '18 at 23:53
$begingroup$
@TobyMak $x$ can by surely found easily by hands from the quadratic equation obtined by the law of cosine.
$endgroup$
– gimusi
Dec 4 '18 at 0:19
$begingroup$
@BenW With your method how would you use the law of cosine?
$endgroup$
– gimusi
Dec 4 '18 at 0:20
1
$begingroup$
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.
$endgroup$
– Ben W
Dec 4 '18 at 0:24
|
show 1 more comment
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
6
$begingroup$
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 Dec 3 '18 at 23:31
gimusigimusi
92.8k84494
$endgroup$
$begingroup$
Yeah, that's probably easier than Heron.
$endgroup$
– Ben W
Dec 3 '18 at 23:34
$begingroup$
@BenW Yes indeed I think that the idea and the spirit for the homework is that. But Heron is always good invoke :)
$endgroup$
– gimusi
Dec 3 '18 at 23:41
$begingroup$
@gimusi It's interesting that we get two solutions: one with $x = -frac{12}{11}$ and one with $x=2$.
$endgroup$
– Toby Mak
Dec 3 '18 at 23:44
$begingroup$
@TobyMak I didn't check that. I'll take a look.
$endgroup$
– gimusi
Dec 3 '18 at 23:51
1
$begingroup$
@TobyMak Sorry, I didn't see that clearly but of course we need to esclude -12/11 therefore x=2 is the good one!
$endgroup$
– gimusi
Dec 4 '18 at 0:08
add a comment |
6
$begingroup$
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 Dec 3 '18 at 23:31
gimusigimusi
92.8k84494
$endgroup$
$begingroup$
Yeah, that's probably easier than Heron.
$endgroup$
– Ben W
Dec 3 '18 at 23:34
$begingroup$
@BenW Yes indeed I think that the idea and the spirit for the homework is that. But Heron is always good invoke :)
$endgroup$
– gimusi
Dec 3 '18 at 23:41
$begingroup$
@gimusi It's interesting that we get two solutions: one with $x = -frac{12}{11}$ and one with $x=2$.
$endgroup$
– Toby Mak
Dec 3 '18 at 23:44
$begingroup$
@TobyMak I didn't check that. I'll take a look.
$endgroup$
– gimusi
Dec 3 '18 at 23:51
1
$begingroup$
@TobyMak Sorry, I didn't see that clearly but of course we need to esclude -12/11 therefore x=2 is the good one!
$endgroup$
– gimusi
Dec 4 '18 at 0:08
add a comment |
6
6
6
$begingroup$
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 Dec 3 '18 at 23:31
gimusigimusi
92.8k84494
$endgroup$
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 Dec 3 '18 at 23:31
gimusigimusi
92.8k84494
answered Dec 3 '18 at 23:31
gimusigimusi
92.8k84494
answered Dec 3 '18 at 23:31
gimusigimusi
92.8k84494
answered Dec 3 '18 at 23:31
gimusigimusi
92.8k84494
92.8k84494
$begingroup$
Yeah, that's probably easier than Heron.
$endgroup$
– Ben W
Dec 3 '18 at 23:34
$begingroup$
@BenW Yes indeed I think that the idea and the spirit for the homework is that. But Heron is always good invoke :)
$endgroup$
– gimusi
Dec 3 '18 at 23:41
$begingroup$
@gimusi It's interesting that we get two solutions: one with $x = -frac{12}{11}$ and one with $x=2$.
$endgroup$
– Toby Mak
Dec 3 '18 at 23:44
$begingroup$
@TobyMak I didn't check that. I'll take a look.
$endgroup$
– gimusi
Dec 3 '18 at 23:51
1
$begingroup$
@TobyMak Sorry, I didn't see that clearly but of course we need to esclude -12/11 therefore x=2 is the good one!
$endgroup$
– gimusi
Dec 4 '18 at 0:08
add a comment |
$begingroup$
Yeah, that's probably easier than Heron.
$endgroup$
– Ben W
Dec 3 '18 at 23:34
$begingroup$
@BenW Yes indeed I think that the idea and the spirit for the homework is that. But Heron is always good invoke :)
$endgroup$
– gimusi
Dec 3 '18 at 23:41
$begingroup$
@gimusi It's interesting that we get two solutions: one with $x = -frac{12}{11}$ and one with $x=2$.
$endgroup$
– Toby Mak
Dec 3 '18 at 23:44
$begingroup$
@TobyMak I didn't check that. I'll take a look.
$endgroup$
– gimusi
Dec 3 '18 at 23:51
1
$begingroup$
@TobyMak Sorry, I didn't see that clearly but of course we need to esclude -12/11 therefore x=2 is the good one!
$endgroup$
– gimusi
Dec 4 '18 at 0:08
$begingroup$
Yeah, that's probably easier than Heron.
$endgroup$
– Ben W
Dec 3 '18 at 23:34
$begingroup$
Yeah, that's probably easier than Heron.
$endgroup$
– Ben W
Dec 3 '18 at 23:34
$begingroup$
@BenW Yes indeed I think that the idea and the spirit for the homework is that. But Heron is always good invoke :)
$endgroup$
– gimusi
Dec 3 '18 at 23:41
$begingroup$
@BenW Yes indeed I think that the idea and the spirit for the homework is that. But Heron is always good invoke :)
$endgroup$
– gimusi
Dec 3 '18 at 23:41
$begingroup$
@gimusi It's interesting that we get two solutions: one with $x = -frac{12}{11}$ and one with $x=2$.
$endgroup$
– Toby Mak
Dec 3 '18 at 23:44
$begingroup$
@gimusi It's interesting that we get two solutions: one with $x = -frac{12}{11}$ and one with $x=2$.
$endgroup$
– Toby Mak
Dec 3 '18 at 23:44
$begingroup$
@TobyMak I didn't check that. I'll take a look.
$endgroup$
– gimusi
Dec 3 '18 at 23:51
$begingroup$
@TobyMak I didn't check that. I'll take a look.
$endgroup$
– gimusi
Dec 3 '18 at 23:51
1
1
$begingroup$
@TobyMak Sorry, I didn't see that clearly but of course we need to esclude -12/11 therefore x=2 is the good one!
$endgroup$
– gimusi
Dec 4 '18 at 0:08
$begingroup$
@TobyMak Sorry, I didn't see that clearly but of course we need to esclude -12/11 therefore x=2 is the good one!
$endgroup$
– gimusi
Dec 4 '18 at 0:08
add a comment |
5
$begingroup$
If you have $x$, then you have all three side lengths. Now just apply Heron's formula.
answered Dec 3 '18 at 23:29
Ben WBen W
2,234615
$endgroup$
$begingroup$
Thank you! This is definitely helpful if I don't have a calculator!
$endgroup$
– Joshuap88
Dec 3 '18 at 23:38
$begingroup$
@Joshuap88 You would still need to find $x$ using a calculator.
$endgroup$
– Toby Mak
Dec 3 '18 at 23:53
$begingroup$
@TobyMak $x$ can by surely found easily by hands from the quadratic equation obtined by the law of cosine.
$endgroup$
– gimusi
Dec 4 '18 at 0:19
$begingroup$
@BenW With your method how would you use the law of cosine?
$endgroup$
– gimusi
Dec 4 '18 at 0:20
1
$begingroup$
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.
$endgroup$
– Ben W
Dec 4 '18 at 0:24
|
show 1 more comment
5
$begingroup$
If you have $x$, then you have all three side lengths. Now just apply Heron's formula.
answered Dec 3 '18 at 23:29
Ben WBen W
2,234615
$endgroup$
$begingroup$
Thank you! This is definitely helpful if I don't have a calculator!
$endgroup$
– Joshuap88
Dec 3 '18 at 23:38
$begingroup$
@Joshuap88 You would still need to find $x$ using a calculator.
$endgroup$
– Toby Mak
Dec 3 '18 at 23:53
$begingroup$
@TobyMak $x$ can by surely found easily by hands from the quadratic equation obtined by the law of cosine.
$endgroup$
– gimusi
Dec 4 '18 at 0:19
$begingroup$
@BenW With your method how would you use the law of cosine?
$endgroup$
– gimusi
Dec 4 '18 at 0:20
1
$begingroup$
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.
$endgroup$
– Ben W
Dec 4 '18 at 0:24
|
show 1 more comment
5
5
5
$begingroup$
If you have $x$, then you have all three side lengths. Now just apply Heron's formula.
answered Dec 3 '18 at 23:29
Ben WBen W
2,234615
$endgroup$
If you have $x$, then you have all three side lengths. Now just apply Heron's formula.
answered Dec 3 '18 at 23:29
Ben WBen W
2,234615
answered Dec 3 '18 at 23:29
Ben WBen W
2,234615
answered Dec 3 '18 at 23:29
Ben WBen W
2,234615
answered Dec 3 '18 at 23:29
Ben WBen W
2,234615
2,234615
$begingroup$
Thank you! This is definitely helpful if I don't have a calculator!
$endgroup$
– Joshuap88
Dec 3 '18 at 23:38
$begingroup$
@Joshuap88 You would still need to find $x$ using a calculator.
$endgroup$
– Toby Mak
Dec 3 '18 at 23:53
$begingroup$
@TobyMak $x$ can by surely found easily by hands from the quadratic equation obtined by the law of cosine.
$endgroup$
– gimusi
Dec 4 '18 at 0:19
$begingroup$
@BenW With your method how would you use the law of cosine?
$endgroup$
– gimusi
Dec 4 '18 at 0:20
1
$begingroup$
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.
$endgroup$
– Ben W
Dec 4 '18 at 0:24
|
show 1 more comment
$begingroup$
Thank you! This is definitely helpful if I don't have a calculator!
$endgroup$
– Joshuap88
Dec 3 '18 at 23:38
$begingroup$
@Joshuap88 You would still need to find $x$ using a calculator.
$endgroup$
– Toby Mak
Dec 3 '18 at 23:53
$begingroup$
@TobyMak $x$ can by surely found easily by hands from the quadratic equation obtined by the law of cosine.
$endgroup$
– gimusi
Dec 4 '18 at 0:19
$begingroup$
@BenW With your method how would you use the law of cosine?
$endgroup$
– gimusi
Dec 4 '18 at 0:20
1
$begingroup$
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.
$endgroup$
– Ben W
Dec 4 '18 at 0:24
$begingroup$
Thank you! This is definitely helpful if I don't have a calculator!
$endgroup$
– Joshuap88
Dec 3 '18 at 23:38
$begingroup$
Thank you! This is definitely helpful if I don't have a calculator!
$endgroup$
– Joshuap88
Dec 3 '18 at 23:38
$begingroup$
@Joshuap88 You would still need to find $x$ using a calculator.
$endgroup$
– Toby Mak
Dec 3 '18 at 23:53
$begingroup$
@Joshuap88 You would still need to find $x$ using a calculator.
$endgroup$
– Toby Mak
Dec 3 '18 at 23:53
$begingroup$
@TobyMak $x$ can by surely found easily by hands from the quadratic equation obtined by the law of cosine.
$endgroup$
– gimusi
Dec 4 '18 at 0:19
$begingroup$
@TobyMak $x$ can by surely found easily by hands from the quadratic equation obtined by the law of cosine.
$endgroup$
– gimusi
Dec 4 '18 at 0:19
$begingroup$
@BenW With your method how would you use the law of cosine?
$endgroup$
– gimusi
Dec 4 '18 at 0:20
$begingroup$
@BenW With your method how would you use the law of cosine?
$endgroup$
– gimusi
Dec 4 '18 at 0:20
1
1
$begingroup$
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.
$endgroup$
– Ben W
Dec 4 '18 at 0:24
$begingroup$
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.
$endgroup$
– Ben W
Dec 4 '18 at 0:24
|
show 1 more comment
$begingroup$
What did you try? Do you have some formulas for finding the areas of triangles?
$endgroup$
– DRF
Dec 4 '18 at 8:51