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Finding the side and angle of a triangle.
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I'm working on some summer problems so that I can be more prepared when I go into my class in the fall. I found a website full of problems of the content we will be learning but it doesn't have the answers. I need a little guidance on how to do this problem.
The following diagram shows the triangle $ABC$. 
a. Find $AC$.
b. Find $angle BCA$.
For a, I believe I would do the Pythagorean Theorem to find the side. $a^2 + b^2 = c^2$. Is this correct?
For b, to find this angle would I use the sides? As in using soh-cah-toa? So, I could do the sine of $6$ over the hypotenuse, which I would find after part a.
Edit: After reading comments, I used the Law of Cosines for part a and got b = 12.5 as my answer. However, I am not sure I completed it correctly.
I also used the Law of Sines to do part b, I got C = 28.16° as my answer. Can someone please tell me if I completed these two correctly?
trigonometry triangle angle
asked Aug 6 at 18:03
Ella
999
add a comment |Â
up vote
0
down vote
favorite
I'm working on some summer problems so that I can be more prepared when I go into my class in the fall. I found a website full of problems of the content we will be learning but it doesn't have the answers. I need a little guidance on how to do this problem.
The following diagram shows the triangle $ABC$.
a. Find $AC$.
b. Find $angle BCA$.
For a, I believe I would do the Pythagorean Theorem to find the side. $a^2 + b^2 = c^2$. Is this correct?
For b, to find this angle would I use the sides? As in using soh-cah-toa? So, I could do the sine of $6$ over the hypotenuse, which I would find after part a.
Edit: After reading comments, I used the Law of Cosines for part a and got b = 12.5 as my answer. However, I am not sure I completed it correctly.
I also used the Law of Sines to do part b, I got C = 28.16° as my answer. Can someone please tell me if I completed these two correctly?
trigonometry triangle angle
asked Aug 6 at 18:03
Ella
999
3
no. part (a) is the Law of Cosines. After that, the Law of Sines gives you the other angles
– Will Jagy
Aug 6 at 18:07
. . . such as this. Pythagoras applies to right triangles. Plenty of other sources are available.
– Weather Vane
Aug 6 at 18:11
Please read this MathJax tutorial, which explains how to typeset mathematics on this site. You cannot apply the Pythagorean Theorem here since you do not have a right triangle.
– N. F. Taussig
Aug 6 at 18:12
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I'm working on some summer problems so that I can be more prepared when I go into my class in the fall. I found a website full of problems of the content we will be learning but it doesn't have the answers. I need a little guidance on how to do this problem.
The following diagram shows the triangle $ABC$.
a. Find $AC$.
b. Find $angle BCA$.
For a, I believe I would do the Pythagorean Theorem to find the side. $a^2 + b^2 = c^2$. Is this correct?
For b, to find this angle would I use the sides? As in using soh-cah-toa? So, I could do the sine of $6$ over the hypotenuse, which I would find after part a.
Edit: After reading comments, I used the Law of Cosines for part a and got b = 12.5 as my answer. However, I am not sure I completed it correctly.
I also used the Law of Sines to do part b, I got C = 28.16° as my answer. Can someone please tell me if I completed these two correctly?
trigonometry triangle angle
asked Aug 6 at 18:03
Ella
999
I'm working on some summer problems so that I can be more prepared when I go into my class in the fall. I found a website full of problems of the content we will be learning but it doesn't have the answers. I need a little guidance on how to do this problem.
The following diagram shows the triangle $ABC$.
a. Find $AC$.
b. Find $angle BCA$.
For a, I believe I would do the Pythagorean Theorem to find the side. $a^2 + b^2 = c^2$. Is this correct?
For b, to find this angle would I use the sides? As in using soh-cah-toa? So, I could do the sine of $6$ over the hypotenuse, which I would find after part a.
Edit: After reading comments, I used the Law of Cosines for part a and got b = 12.5 as my answer. However, I am not sure I completed it correctly.
I also used the Law of Sines to do part b, I got C = 28.16° as my answer. Can someone please tell me if I completed these two correctly?
trigonometry triangle angle
asked Aug 6 at 18:03
Ella
999
edited Aug 6 at 19:42
asked Aug 6 at 18:03
Ella
999
asked Aug 6 at 18:03
Ella
999
asked Aug 6 at 18:03
Ella
999
999
3
no. part (a) is the Law of Cosines. After that, the Law of Sines gives you the other angles
– Will Jagy
Aug 6 at 18:07
. . . such as this. Pythagoras applies to right triangles. Plenty of other sources are available.
– Weather Vane
Aug 6 at 18:11
Please read this MathJax tutorial, which explains how to typeset mathematics on this site. You cannot apply the Pythagorean Theorem here since you do not have a right triangle.
– N. F. Taussig
Aug 6 at 18:12
add a comment |Â
3
no. part (a) is the Law of Cosines. After that, the Law of Sines gives you the other angles
– Will Jagy
Aug 6 at 18:07
. . . such as this. Pythagoras applies to right triangles. Plenty of other sources are available.
– Weather Vane
Aug 6 at 18:11
Please read this MathJax tutorial, which explains how to typeset mathematics on this site. You cannot apply the Pythagorean Theorem here since you do not have a right triangle.
– N. F. Taussig
Aug 6 at 18:12
3
3
no. part (a) is the Law of Cosines. After that, the Law of Sines gives you the other angles
– Will Jagy
Aug 6 at 18:07
no. part (a) is the Law of Cosines. After that, the Law of Sines gives you the other angles
– Will Jagy
Aug 6 at 18:07
. . . such as this. Pythagoras applies to right triangles. Plenty of other sources are available.
– Weather Vane
Aug 6 at 18:11
. . . such as this. Pythagoras applies to right triangles. Plenty of other sources are available.
– Weather Vane
Aug 6 at 18:11
Please read this MathJax tutorial, which explains how to typeset mathematics on this site. You cannot apply the Pythagorean Theorem here since you do not have a right triangle.
– N. F. Taussig
Aug 6 at 18:12
Please read this MathJax tutorial, which explains how to typeset mathematics on this site. You cannot apply the Pythagorean Theorem here since you do not have a right triangle.
– N. F. Taussig
Aug 6 at 18:12
add a comment |Â
4 Answers
4
active
oldest
votes
up vote
1
down vote
You could use law of cosines.
To find AC we can use the formula $cos B=dfraca^2-b^2+c^22ac$ since we know $angle B$
From the diagram we know that $AC=b$
Now, $$cos100=dfrac100-b^2+36120$$
By solving the above you will get the value of $AC$ which is also $b$.
In the similar way use the formula $cos C=dfraca^2+b^2-c^22ab$ to find the $angle BCA$
answered Aug 6 at 18:31
Key Flex
4,471525
add a comment |Â
up vote
0
down vote
$text We can also do this. We know through sine law: $
$displaystyle fracACsin(angleABC) = fracABsin(angleBCA)$
$displaystyle angleBCA = sin^-1left(fracAB * sin(angleABC)ACright)$
answered Aug 6 at 19:15
mvr950
19618
I got C=28.16°. Does this seem correct?
– Ella
Aug 6 at 19:52
1
$C=28.15°$ would be slightly better, because it is not affected by the rounding to a three digit number of $b$.
– random
Aug 6 at 20:21
Here's the cosine law: google.ca/…
– mvr950
Aug 6 at 20:28
add a comment |Â
up vote
0
down vote
$displaystyle sqrt(36 + 100 - ((2 * 6 * 10)* cos(100)))$
$=displaystyle 12.5235$
$displaystyle sin^-1 left(frac6 * sin(100)12.5235 right)$
$=displaystyle 28.16°$
You can check the answers here:
enter link description here
enter link description here
answered Aug 6 at 20:27
mvr950
19618
add a comment |Â
up vote
0
down vote
For $a$ we can't use Pythagorean Theorem since ABC is not a right triangle but we need the Law of cosines.
For that see the related Does the law of cosines contradict Pythagoras's theorem?
For point b once we have AC by the Law of sines we have
$$fracsin 100AC=fracsin (angle BCA)AB$$
answered Aug 6 at 20:40
gimusi
65.5k73684
add a comment |Â
4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
You could use law of cosines.
To find AC we can use the formula $cos B=dfraca^2-b^2+c^22ac$ since we know $angle B$
From the diagram we know that $AC=b$
Now, $$cos100=dfrac100-b^2+36120$$
By solving the above you will get the value of $AC$ which is also $b$.
In the similar way use the formula $cos C=dfraca^2+b^2-c^22ab$ to find the $angle BCA$
answered Aug 6 at 18:31
Key Flex
4,471525
add a comment |Â
up vote
1
down vote
You could use law of cosines.
To find AC we can use the formula $cos B=dfraca^2-b^2+c^22ac$ since we know $angle B$
From the diagram we know that $AC=b$
Now, $$cos100=dfrac100-b^2+36120$$
By solving the above you will get the value of $AC$ which is also $b$.
In the similar way use the formula $cos C=dfraca^2+b^2-c^22ab$ to find the $angle BCA$
answered Aug 6 at 18:31
Key Flex
4,471525
add a comment |Â
up vote
1
down vote
up vote
1
down vote
You could use law of cosines.
To find AC we can use the formula $cos B=dfraca^2-b^2+c^22ac$ since we know $angle B$
From the diagram we know that $AC=b$
Now, $$cos100=dfrac100-b^2+36120$$
By solving the above you will get the value of $AC$ which is also $b$.
In the similar way use the formula $cos C=dfraca^2+b^2-c^22ab$ to find the $angle BCA$
answered Aug 6 at 18:31
Key Flex
4,471525
You could use law of cosines.
To find AC we can use the formula $cos B=dfraca^2-b^2+c^22ac$ since we know $angle B$
From the diagram we know that $AC=b$
Now, $$cos100=dfrac100-b^2+36120$$
By solving the above you will get the value of $AC$ which is also $b$.
In the similar way use the formula $cos C=dfraca^2+b^2-c^22ab$ to find the $angle BCA$
answered Aug 6 at 18:31
Key Flex
4,471525
answered Aug 6 at 18:31
Key Flex
4,471525
answered Aug 6 at 18:31
Key Flex
4,471525
answered Aug 6 at 18:31
Key Flex
4,471525
4,471525
add a comment |Â
add a comment |Â
up vote
0
down vote
$text We can also do this. We know through sine law: $
$displaystyle fracACsin(angleABC) = fracABsin(angleBCA)$
$displaystyle angleBCA = sin^-1left(fracAB * sin(angleABC)ACright)$
answered Aug 6 at 19:15
mvr950
19618
I got C=28.16°. Does this seem correct?
– Ella
Aug 6 at 19:52
1
$C=28.15°$ would be slightly better, because it is not affected by the rounding to a three digit number of $b$.
– random
Aug 6 at 20:21
Here's the cosine law: google.ca/…
– mvr950
Aug 6 at 20:28
add a comment |Â
up vote
0
down vote
$text We can also do this. We know through sine law: $
$displaystyle fracACsin(angleABC) = fracABsin(angleBCA)$
$displaystyle angleBCA = sin^-1left(fracAB * sin(angleABC)ACright)$
answered Aug 6 at 19:15
mvr950
19618
I got C=28.16°. Does this seem correct?
– Ella
Aug 6 at 19:52
1
$C=28.15°$ would be slightly better, because it is not affected by the rounding to a three digit number of $b$.
– random
Aug 6 at 20:21
Here's the cosine law: google.ca/…
– mvr950
Aug 6 at 20:28
add a comment |Â
up vote
0
down vote
up vote
0
down vote
$text We can also do this. We know through sine law: $
$displaystyle fracACsin(angleABC) = fracABsin(angleBCA)$
$displaystyle angleBCA = sin^-1left(fracAB * sin(angleABC)ACright)$
answered Aug 6 at 19:15
mvr950
19618
$text We can also do this. We know through sine law: $
$displaystyle fracACsin(angleABC) = fracABsin(angleBCA)$
$displaystyle angleBCA = sin^-1left(fracAB * sin(angleABC)ACright)$
answered Aug 6 at 19:15
mvr950
19618
answered Aug 6 at 19:15
mvr950
19618
answered Aug 6 at 19:15
mvr950
19618
answered Aug 6 at 19:15
mvr950
19618
19618
I got C=28.16°. Does this seem correct?
– Ella
Aug 6 at 19:52
1
$C=28.15°$ would be slightly better, because it is not affected by the rounding to a three digit number of $b$.
– random
Aug 6 at 20:21
Here's the cosine law: google.ca/…
– mvr950
Aug 6 at 20:28
add a comment |Â
I got C=28.16°. Does this seem correct?
– Ella
Aug 6 at 19:52
1
$C=28.15°$ would be slightly better, because it is not affected by the rounding to a three digit number of $b$.
– random
Aug 6 at 20:21
Here's the cosine law: google.ca/…
– mvr950
Aug 6 at 20:28
I got C=28.16°. Does this seem correct?
– Ella
Aug 6 at 19:52
I got C=28.16°. Does this seem correct?
– Ella
Aug 6 at 19:52
1
1
$C=28.15°$ would be slightly better, because it is not affected by the rounding to a three digit number of $b$.
– random
Aug 6 at 20:21
$C=28.15°$ would be slightly better, because it is not affected by the rounding to a three digit number of $b$.
– random
Aug 6 at 20:21
Here's the cosine law: google.ca/…
– mvr950
Aug 6 at 20:28
Here's the cosine law: google.ca/…
– mvr950
Aug 6 at 20:28
add a comment |Â
up vote
0
down vote
$displaystyle sqrt(36 + 100 - ((2 * 6 * 10)* cos(100)))$
$=displaystyle 12.5235$
$displaystyle sin^-1 left(frac6 * sin(100)12.5235 right)$
$=displaystyle 28.16°$
You can check the answers here:
enter link description here
enter link description here
answered Aug 6 at 20:27
mvr950
19618
add a comment |Â
up vote
0
down vote
$displaystyle sqrt(36 + 100 - ((2 * 6 * 10)* cos(100)))$
$=displaystyle 12.5235$
$displaystyle sin^-1 left(frac6 * sin(100)12.5235 right)$
$=displaystyle 28.16°$
You can check the answers here:
enter link description here
enter link description here
answered Aug 6 at 20:27
mvr950
19618
add a comment |Â
up vote
0
down vote
up vote
0
down vote
$displaystyle sqrt(36 + 100 - ((2 * 6 * 10)* cos(100)))$
$=displaystyle 12.5235$
$displaystyle sin^-1 left(frac6 * sin(100)12.5235 right)$
$=displaystyle 28.16°$
You can check the answers here:
enter link description here
enter link description here
answered Aug 6 at 20:27
mvr950
19618
$displaystyle sqrt(36 + 100 - ((2 * 6 * 10)* cos(100)))$
$=displaystyle 12.5235$
$displaystyle sin^-1 left(frac6 * sin(100)12.5235 right)$
$=displaystyle 28.16°$
You can check the answers here:
enter link description here
enter link description here
answered Aug 6 at 20:27
mvr950
19618
edited Aug 6 at 20:35
answered Aug 6 at 20:27
mvr950
19618
answered Aug 6 at 20:27
mvr950
19618
answered Aug 6 at 20:27
mvr950
19618
19618
add a comment |Â
add a comment |Â
up vote
0
down vote
For $a$ we can't use Pythagorean Theorem since ABC is not a right triangle but we need the Law of cosines.
For that see the related Does the law of cosines contradict Pythagoras's theorem?
For point b once we have AC by the Law of sines we have
$$fracsin 100AC=fracsin (angle BCA)AB$$
answered Aug 6 at 20:40
gimusi
65.5k73684
add a comment |Â
up vote
0
down vote
For $a$ we can't use Pythagorean Theorem since ABC is not a right triangle but we need the Law of cosines.
For that see the related Does the law of cosines contradict Pythagoras's theorem?
For point b once we have AC by the Law of sines we have
$$fracsin 100AC=fracsin (angle BCA)AB$$
answered Aug 6 at 20:40
gimusi
65.5k73684
add a comment |Â
up vote
0
down vote
up vote
0
down vote
For $a$ we can't use Pythagorean Theorem since ABC is not a right triangle but we need the Law of cosines.
For that see the related Does the law of cosines contradict Pythagoras's theorem?
For point b once we have AC by the Law of sines we have
$$fracsin 100AC=fracsin (angle BCA)AB$$
answered Aug 6 at 20:40
gimusi
65.5k73684
For $a$ we can't use Pythagorean Theorem since ABC is not a right triangle but we need the Law of cosines.
For that see the related Does the law of cosines contradict Pythagoras's theorem?
For point b once we have AC by the Law of sines we have
$$fracsin 100AC=fracsin (angle BCA)AB$$
answered Aug 6 at 20:40
gimusi
65.5k73684
answered Aug 6 at 20:40
gimusi
65.5k73684
answered Aug 6 at 20:40
gimusi
65.5k73684
answered Aug 6 at 20:40
gimusi
65.5k73684
65.5k73684
add a comment |Â
add a comment |Â
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3
no. part (a) is the Law of Cosines. After that, the Law of Sines gives you the other angles
– Will Jagy
Aug 6 at 18:07
. . . such as this. Pythagoras applies to right triangles. Plenty of other sources are available.
– Weather Vane
Aug 6 at 18:11
Please read this MathJax tutorial, which explains how to typeset mathematics on this site. You cannot apply the Pythagorean Theorem here since you do not have a right triangle.
– N. F. Taussig
Aug 6 at 18:12