Math / Trigonometry

QuestionA right triangle has side lengths 3, 4, and 5 as shown below. Use these lengths to find $\cos{B}$ , $\tan{B}$ , and $\sin{B}$ .
$\cos{B} =$ $\tan{B} =$ $\sin{B} =$

Studdy Solution

STEP 1

What is this asking? We need to find the cosine, tangent, and sine of angle B in a right triangle with sides 3, 4, and 5. Watch out! Make sure to identify the correct sides relative to angle B: adjacent, opposite, and hypotenuse.

STEP 2

1. Identify the sides
2. Calculate cos B
3. Calculate tan B
4. Calculate sin B

STEP 3

Looking at angle B, we can see that the side adjacent to it is BC, which has a length of $3$ .
The side opposite to angle B is AC, with a length of $4$ .
Lastly, the hypotenuse, which is always opposite the right angle, is AB with a length of $5$ .

STEP 4

Remember SOH CAH TOA!
Cosine is adjacent over hypotenuse.
So, $\cos{B} = \frac{\text{adjacent}}{\text{hypotenuse}}$ .

STEP 5

We've already identified the adjacent side as having a length of $3$ and the hypotenuse as having a length of $5$ .
Therefore, $\cos{B} = \frac{3}{5}$ .

STEP 6

Again, using SOH CAH TOA, tangent is opposite over adjacent.
So, $\tan{B} = \frac{\text{opposite}}{\text{adjacent}}$ .

STEP 7

We've identified the opposite side as having length $4$ and the adjacent side as having length $3$ .
Thus, $\tan{B} = \frac{4}{3}$ .

STEP 8

One last time with SOH CAH TOA, sine is opposite over hypotenuse.
So, $\sin{B} = \frac{\text{opposite}}{\text{hypotenuse}}$ .

STEP 9

We've identified the opposite side as having length $4$ and the hypotenuse as having length $5$ .
Therefore, $\sin{B} = \frac{4}{5}$ .

STEP 10

$\cos{B} = \frac{3}{5}$ , $\tan{B} = \frac{4}{3}$ , and $\sin{B} = \frac{4}{5}$ .

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QuestionA right triangle has side lengths 3, 4, and 5 as shown below. Use these lengths to find $\cos{B}$ , $\tan{B}$ , and $\sin{B}$ .
$\cos{B} =$ $\tan{B} =$ $\sin{B} =$

Studdy Solution

STEP 1

What is this asking? We need to find the cosine, tangent, and sine of angle B in a right triangle with sides 3, 4, and 5. Watch out! Make sure to identify the correct sides relative to angle B: adjacent, opposite, and hypotenuse.

STEP 2

1. Identify the sides
2. Calculate cos B
3. Calculate tan B
4. Calculate sin B

STEP 3

Looking at angle B, we can see that the side adjacent to it is BC, which has a length of $3$ .
The side opposite to angle B is AC, with a length of $4$ .
Lastly, the hypotenuse, which is always opposite the right angle, is AB with a length of $5$ .

STEP 4

Remember SOH CAH TOA!
Cosine is adjacent over hypotenuse.
So, $\cos{B} = \frac{\text{adjacent}}{\text{hypotenuse}}$ .

STEP 5

We've already identified the adjacent side as having a length of $3$ and the hypotenuse as having a length of $5$ .
Therefore, $\cos{B} = \frac{3}{5}$ .

STEP 6

Again, using SOH CAH TOA, tangent is opposite over adjacent.
So, $\tan{B} = \frac{\text{opposite}}{\text{adjacent}}$ .

STEP 7

We've identified the opposite side as having length $4$ and the adjacent side as having length $3$ .
Thus, $\tan{B} = \frac{4}{3}$ .

STEP 8

One last time with SOH CAH TOA, sine is opposite over hypotenuse.
So, $\sin{B} = \frac{\text{opposite}}{\text{hypotenuse}}$ .

STEP 9

We've identified the opposite side as having length $4$ and the hypotenuse as having length $5$ .
Therefore, $\sin{B} = \frac{4}{5}$ .

STEP 10

$\cos{B} = \frac{3}{5}$ , $\tan{B} = \frac{4}{3}$ , and $\sin{B} = \frac{4}{5}$ .

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QuestionA right triangle has side lengths 3, 4, and 5 as shown below. Use these lengths to find cos⁡B\cos{B}cosB, tan⁡B\tan{B}tanB, and sin⁡B\sin{B}sinB.cos⁡B=\cos{B} = cosB= tan⁡B=\tan{B} = tanB= sin⁡B=\sin{B} = sinB=

Studdy Solution

STEP 1

What is this asking? We need to find the cosine, tangent, and sine of angle B in a right triangle with sides 3, 4, and 5. Watch out! Make sure to identify the correct sides relative to angle B: **adjacent**, **opposite**, and **hypotenuse**.

STEP 2

1. Identify the sides2. Calculate cos B3. Calculate tan B4. Calculate sin B

STEP 3

Looking at angle B, we can see that the side **adjacent** to it is BC, which has a length of 333.The side **opposite** to angle B is AC, with a length of 444.Lastly, the **hypotenuse**, which is always opposite the right angle, is AB with a length of 555.

STEP 4

Remember **SOH CAH TOA**!Cosine is **adjacent** over **hypotenuse**.So, cos⁡B=adjacenthypotenuse\cos{B} = \frac{\text{adjacent}}{\text{hypotenuse}}cosB=hypotenuseadjacent​.

STEP 5

We've already identified the **adjacent** side as having a length of 333 and the **hypotenuse** as having a length of 555.Therefore, cos⁡B=35\cos{B} = \frac{3}{5}cosB=53​.

STEP 6

Again, using **SOH CAH TOA**, tangent is **opposite** over **adjacent**.So, tan⁡B=oppositeadjacent\tan{B} = \frac{\text{opposite}}{\text{adjacent}}tanB=adjacentopposite​.

STEP 7

We've identified the **opposite** side as having length 444 and the **adjacent** side as having length 333.Thus, tan⁡B=43\tan{B} = \frac{4}{3}tanB=34​.

STEP 8

One last time with **SOH CAH TOA**, sine is **opposite** over **hypotenuse**.So, sin⁡B=oppositehypotenuse\sin{B} = \frac{\text{opposite}}{\text{hypotenuse}}sinB=hypotenuseopposite​.

STEP 9

We've identified the **opposite** side as having length 444 and the **hypotenuse** as having length 555.Therefore, sin⁡B=45\sin{B} = \frac{4}{5}sinB=54​.

STEP 10

cos⁡B=35\cos{B} = \frac{3}{5}cosB=53​, tan⁡B=43\tan{B} = \frac{4}{3}tanB=34​, and sin⁡B=45\sin{B} = \frac{4}{5}sinB=54​.

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QuestionA right triangle has side lengths 3, 4, and 5 as shown below. Use these lengths to find $\cos{B}$ , $\tan{B}$ , and $\sin{B}$ .
$\cos{B} =$ $\tan{B} =$ $\sin{B} =$

What is this asking? We need to find the cosine, tangent, and sine of angle B in a right triangle with sides 3, 4, and 5. Watch out! Make sure to identify the correct sides relative to angle B: adjacent, opposite, and hypotenuse.

1. Identify the sides
2. Calculate cos B
3. Calculate tan B
4. Calculate sin B

Looking at angle B, we can see that the side adjacent to it is BC, which has a length of $3$ .
The side opposite to angle B is AC, with a length of $4$ .
Lastly, the hypotenuse, which is always opposite the right angle, is AB with a length of $5$ .

Remember SOH CAH TOA!
Cosine is adjacent over hypotenuse.
So, $\cos{B} = \frac{\text{adjacent}}{\text{hypotenuse}}$ .

We've already identified the adjacent side as having a length of $3$ and the hypotenuse as having a length of $5$ .
Therefore, $\cos{B} = \frac{3}{5}$ .

Again, using SOH CAH TOA, tangent is opposite over adjacent.
So, $\tan{B} = \frac{\text{opposite}}{\text{adjacent}}$ .

We've identified the opposite side as having length $4$ and the adjacent side as having length $3$ .
Thus, $\tan{B} = \frac{4}{3}$ .

One last time with SOH CAH TOA, sine is opposite over hypotenuse.
So, $\sin{B} = \frac{\text{opposite}}{\text{hypotenuse}}$ .

We've identified the opposite side as having length $4$ and the hypotenuse as having length $5$ .
Therefore, $\sin{B} = \frac{4}{5}$ .

$\cos{B} = \frac{3}{5}$ , $\tan{B} = \frac{4}{3}$ , and $\sin{B} = \frac{4}{5}$ .