Math / Trigonometry

Question1. Jennifer says that $\sin A=\frac{6}{14}, \cos A=\frac{8}{14}$ , and $\tan A=\frac{6}{8}$ . Do you agree or disagree? If you agree, explain why. If you disagree, explain what she did incorrectly and provide the correct solution.

Studdy Solution

STEP 1

What is this asking? Is Jennifer right about the sine, cosine, and tangent of angle A, given some side lengths? Watch out! Remember SOH CAH TOA!
Don't mix up which sides go with which trig function.
Also, make sure your triangle is a right triangle before using these relationships!

STEP 2

1. Check for Right Triangle
2. Calculate Sine
3. Calculate Cosine
4. Calculate Tangent

STEP 3

Let's see if this is a right triangle using the Pythagorean Theorem.
If $a^2 + b^2 = c^2$ , where $c$ is the longest side, then we've got a right triangle!
We have sides with lengths $6$ , $8$ , and $14$ .
Is $6^2 + 8^2$ equal to $14^2$ ?
Let's find out!

STEP 4

$6^2 + 8^2 = 36 + 64 = 100$ $14^2 = 196$ Since $100$ does not equal $196$ , this is not a right triangle.

STEP 5

Jennifer says $\sin A = \frac{6}{14}$ .
Since sine is defined as the opposite side over the hypotenuse, and $6$ could be the opposite side and $14$ could be the hypotenuse (if it were a right triangle), this might seem right at first glance.
But, since it's not a right triangle, we can't use SOH CAH TOA!

STEP 6

Jennifer's $\cos A = \frac{8}{14}$ has the same issue.
Cosine is the adjacent side over the hypotenuse, so $8$ and $14$ are in the right spots, but we can't use cosine here because it's not a right triangle!

STEP 7

Finally, Jennifer's $\tan A = \frac{6}{8}$ looks like opposite over adjacent, which is correct for tangent.
But again, we can't use tangent because it's not a right triangle!

STEP 8

Jennifer is incorrect!
We can't use SOH CAH TOA because the triangle isn't a right triangle.
Since $6^2 + 8^2 \ne 14^2$ , the Pythagorean Theorem isn't satisfied.

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Question1. Jennifer says that $\sin A=\frac{6}{14}, \cos A=\frac{8}{14}$ , and $\tan A=\frac{6}{8}$ . Do you agree or disagree? If you agree, explain why. If you disagree, explain what she did incorrectly and provide the correct solution.

Studdy Solution

STEP 1

What is this asking? Is Jennifer right about the sine, cosine, and tangent of angle A, given some side lengths? Watch out! Remember SOH CAH TOA!
Don't mix up which sides go with which trig function.
Also, make sure your triangle is a right triangle before using these relationships!

STEP 2

1. Check for Right Triangle
2. Calculate Sine
3. Calculate Cosine
4. Calculate Tangent

STEP 3

Let's see if this is a right triangle using the Pythagorean Theorem.
If $a^2 + b^2 = c^2$ , where $c$ is the longest side, then we've got a right triangle!
We have sides with lengths $6$ , $8$ , and $14$ .
Is $6^2 + 8^2$ equal to $14^2$ ?
Let's find out!

STEP 4

$6^2 + 8^2 = 36 + 64 = 100$ $14^2 = 196$ Since $100$ does not equal $196$ , this is not a right triangle.

STEP 5

STEP 6

Jennifer's $\cos A = \frac{8}{14}$ has the same issue.
Cosine is the adjacent side over the hypotenuse, so $8$ and $14$ are in the right spots, but we can't use cosine here because it's not a right triangle!

STEP 7

Finally, Jennifer's $\tan A = \frac{6}{8}$ looks like opposite over adjacent, which is correct for tangent.
But again, we can't use tangent because it's not a right triangle!

STEP 8

Jennifer is incorrect!
We can't use SOH CAH TOA because the triangle isn't a right triangle.
Since $6^2 + 8^2 \ne 14^2$ , the Pythagorean Theorem isn't satisfied.

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Question1. Jennifer says that sin⁡A=614,cos⁡A=814\sin A=\frac{6}{14}, \cos A=\frac{8}{14}sinA=146​,cosA=148​, and tan⁡A=68\tan A=\frac{6}{8}tanA=86​. Do you agree or disagree? If you agree, explain why. If you disagree, explain what she did incorrectly and provide the correct solution.

Studdy Solution

STEP 1

What is this asking? Is Jennifer right about the sine, cosine, and tangent of angle A, given some side lengths? Watch out! Remember SOH CAH TOA!Don't mix up which sides go with which trig function.Also, make sure your triangle is a right triangle before using these relationships!

STEP 2

1. Check for Right Triangle2. Calculate Sine3. Calculate Cosine4. Calculate Tangent

STEP 3

Let's see if this is a right triangle using the Pythagorean Theorem.If a2+b2=c2a^2 + b^2 = c^2a2+b2=c2, where ccc is the **longest side**, then we've got a right triangle!We have sides with lengths 666, 888, and 141414.Is 62+826^2 + 8^262+82 equal to 14214^2142?Let's find out!

STEP 4

62+82=36+64=1006^2 + 8^2 = 36 + 64 = 10062+82=36+64=100 142=19614^2 = 196142=196 Since 100100100 does not equal 196196196, this is *not* a right triangle.

STEP 5

STEP 6

Jennifer's cos⁡A=814\cos A = \frac{8}{14}cosA=148​ has the same issue.Cosine is the **adjacent side** over the **hypotenuse**, so 888 and 141414 are in the right spots, but we can't use cosine here because it's not a right triangle!

STEP 7

Finally, Jennifer's tan⁡A=68\tan A = \frac{6}{8}tanA=86​ looks like **opposite** over **adjacent**, which is correct for tangent.But again, we can't use tangent because it's not a right triangle!

STEP 8

Jennifer is incorrect!We can't use SOH CAH TOA because the triangle isn't a right triangle.Since 62+82≠1426^2 + 8^2 \ne 14^262+82=142, the Pythagorean Theorem isn't satisfied.

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Studdy solves anything!

Start learning now

Download Studdy AI Tutor now. Learn with ease and get all help you need to be successful at school.

Question1. Jennifer says that $\sin A=\frac{6}{14}, \cos A=\frac{8}{14}$ , and $\tan A=\frac{6}{8}$ . Do you agree or disagree? If you agree, explain why. If you disagree, explain what she did incorrectly and provide the correct solution.

What is this asking? Is Jennifer right about the sine, cosine, and tangent of angle A, given some side lengths? Watch out! Remember SOH CAH TOA!
Don't mix up which sides go with which trig function.
Also, make sure your triangle is a right triangle before using these relationships!

1. Check for Right Triangle
2. Calculate Sine
3. Calculate Cosine
4. Calculate Tangent

Let's see if this is a right triangle using the Pythagorean Theorem.
If $a^2 + b^2 = c^2$ , where $c$ is the longest side, then we've got a right triangle!
We have sides with lengths $6$ , $8$ , and $14$ .
Is $6^2 + 8^2$ equal to $14^2$ ?
Let's find out!

$6^2 + 8^2 = 36 + 64 = 100$ $14^2 = 196$ Since $100$ does not equal $196$ , this is not a right triangle.

Jennifer's $\cos A = \frac{8}{14}$ has the same issue.
Cosine is the adjacent side over the hypotenuse, so $8$ and $14$ are in the right spots, but we can't use cosine here because it's not a right triangle!

Finally, Jennifer's $\tan A = \frac{6}{8}$ looks like opposite over adjacent, which is correct for tangent.
But again, we can't use tangent because it's not a right triangle!

Jennifer is incorrect!
We can't use SOH CAH TOA because the triangle isn't a right triangle.
Since $6^2 + 8^2 \ne 14^2$ , the Pythagorean Theorem isn't satisfied.