Math / Trigonometry

Question $\begin{array}{l} \text{Using the figure below, find the exact values of the given trigonometric functions.} \\ (a) \ \sin \theta = \square \\ (b) \ \sin \phi = \square \\ (c) \ \cos \theta = \square \\ (d) \ \cos \phi = \square \\ (e) \ \tan \theta = \square \\ (f) \ \tan \phi = \square \\ \text{The extracted text from the attached image:} \\ 4 \\ \text{O} \\ 16 \\ \end{array}$

Studdy Solution

STEP 1

What is this asking? We need to find the sine, cosine, and tangent of angles $\theta$ and $\phi$ in a right triangle, knowing two of its sides. Watch out! Don't mix up which sides are opposite and adjacent to which angle!
Also, we'll need to find the hypotenuse first.

STEP 2

1. Find the hypotenuse
2. Calculate sin θ, cos θ, and tan θ
3. Calculate sin φ, cos φ, and tan φ

STEP 3

Alright, let's do this!
We've got a right triangle, so we can use the Pythagorean theorem: $a^2 + b^2 = c^2$ , where a and b are the legs and c is the hypotenuse.
In our triangle, $a = 4$ and $b = 16$ .

STEP 4

Let's plug in the values: $4^2 + 16^2 = c^2$ .
This simplifies to $16 + 256 = c^2$ , so $c^2 = 272$ .

STEP 5

Taking the square root of both sides gives us $c = \sqrt{272}$ .
We can simplify this!
Since $272 = 16 \cdot 17$ , we have $c = \sqrt{16 \cdot 17} = \sqrt{16} \cdot \sqrt{17} = 4\sqrt{17}$ .
So, our hypotenuse is $4\sqrt{17}$ !

STEP 6

Remember SOH CAH TOA?
For angle $\theta$ , the opposite side is 16, the adjacent side is 4, and the hypotenuse is $4\sqrt{17}$ .

STEP 7

$\sin \theta = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{16}{4\sqrt{17}} = \frac{4}{\sqrt{17}}$ .
We can rationalize the denominator by multiplying the numerator and denominator by $\sqrt{17}$ to get $\sin \theta = \frac{4\sqrt{17}}{17}$ .

STEP 8

$\cos \theta = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{4}{4\sqrt{17}} = \frac{1}{\sqrt{17}}$ .
Rationalizing gives us $\cos \theta = \frac{\sqrt{17}}{17}$ .

STEP 9

$\tan \theta = \frac{\text{opposite}}{\text{adjacent}} = \frac{16}{4} = 4$ .

STEP 10

Now for angle $\phi$ !
The opposite side is 4, the adjacent side is 16, and the hypotenuse is still $4\sqrt{17}$ .

STEP 11

$\sin \phi = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{4}{4\sqrt{17}} = \frac{1}{\sqrt{17}}$ .
Rationalizing gives us $\sin \phi = \frac{\sqrt{17}}{17}$ .

STEP 12

$\cos \phi = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{16}{4\sqrt{17}} = \frac{4}{\sqrt{17}}$ .
Rationalizing gives us $\cos \phi = \frac{4\sqrt{17}}{17}$ .

STEP 13

$\tan \phi = \frac{\text{opposite}}{\text{adjacent}} = \frac{4}{16} = \frac{1}{4}$ .

STEP 14

(a) $\sin \theta = \frac{4\sqrt{17}}{17}$ (b) $\sin \phi = \frac{\sqrt{17}}{17}$ (c) $\cos \theta = \frac{\sqrt{17}}{17}$ (d) $\cos \phi = \frac{4\sqrt{17}}{17}$ (e) $\tan \theta = 4$ (f) $\tan \phi = \frac{1}{4}$

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Studdy Solution

STEP 1

What is this asking? We need to find the sine, cosine, and tangent of angles $\theta$ and $\phi$ in a right triangle, knowing two of its sides. Watch out! Don't mix up which sides are opposite and adjacent to which angle!
Also, we'll need to find the hypotenuse first.

STEP 2

1. Find the hypotenuse
2. Calculate sin θ, cos θ, and tan θ
3. Calculate sin φ, cos φ, and tan φ

STEP 3

Alright, let's do this!
We've got a right triangle, so we can use the Pythagorean theorem: $a^2 + b^2 = c^2$ , where a and b are the legs and c is the hypotenuse.
In our triangle, $a = 4$ and $b = 16$ .

STEP 4

Let's plug in the values: $4^2 + 16^2 = c^2$ .
This simplifies to $16 + 256 = c^2$ , so $c^2 = 272$ .

STEP 5

Taking the square root of both sides gives us $c = \sqrt{272}$ .
We can simplify this!
Since $272 = 16 \cdot 17$ , we have $c = \sqrt{16 \cdot 17} = \sqrt{16} \cdot \sqrt{17} = 4\sqrt{17}$ .
So, our hypotenuse is $4\sqrt{17}$ !

STEP 6

Remember SOH CAH TOA?
For angle $\theta$ , the opposite side is 16, the adjacent side is 4, and the hypotenuse is $4\sqrt{17}$ .

STEP 7

$\sin \theta = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{16}{4\sqrt{17}} = \frac{4}{\sqrt{17}}$ .
We can rationalize the denominator by multiplying the numerator and denominator by $\sqrt{17}$ to get $\sin \theta = \frac{4\sqrt{17}}{17}$ .

STEP 8

$\cos \theta = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{4}{4\sqrt{17}} = \frac{1}{\sqrt{17}}$ .
Rationalizing gives us $\cos \theta = \frac{\sqrt{17}}{17}$ .

STEP 9

$\tan \theta = \frac{\text{opposite}}{\text{adjacent}} = \frac{16}{4} = 4$ .

STEP 10

Now for angle $\phi$ !
The opposite side is 4, the adjacent side is 16, and the hypotenuse is still $4\sqrt{17}$ .

STEP 11

$\sin \phi = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{4}{4\sqrt{17}} = \frac{1}{\sqrt{17}}$ .
Rationalizing gives us $\sin \phi = \frac{\sqrt{17}}{17}$ .

STEP 12

$\cos \phi = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{16}{4\sqrt{17}} = \frac{4}{\sqrt{17}}$ .
Rationalizing gives us $\cos \phi = \frac{4\sqrt{17}}{17}$ .

STEP 13

$\tan \phi = \frac{\text{opposite}}{\text{adjacent}} = \frac{4}{16} = \frac{1}{4}$ .

STEP 14

(a) $\sin \theta = \frac{4\sqrt{17}}{17}$ (b) $\sin \phi = \frac{\sqrt{17}}{17}$ (c) $\cos \theta = \frac{\sqrt{17}}{17}$ (d) $\cos \phi = \frac{4\sqrt{17}}{17}$ (e) $\tan \theta = 4$ (f) $\tan \phi = \frac{1}{4}$

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Studdy Solution

STEP 1

What is this asking? We need to find the sine, cosine, and tangent of angles θ\thetaθ and ϕ\phiϕ in a right triangle, knowing two of its sides. Watch out! Don't mix up which sides are opposite and adjacent to which angle!Also, we'll need to find the hypotenuse first.

STEP 2

1. Find the hypotenuse2. Calculate sin θ, cos θ, and tan θ3. Calculate sin φ, cos φ, and tan φ

STEP 3

Alright, let's **do this**!We've got a right triangle, so we can use the **Pythagorean theorem**: a2+b2=c2a^2 + b^2 = c^2a2+b2=c2, where *a* and *b* are the legs and *c* is the hypotenuse.In our triangle, a=4a = 4a=4 and b=16b = 16b=16.

STEP 4

Let's plug in the values: 42+162=c24^2 + 16^2 = c^242+162=c2.This simplifies to 16+256=c216 + 256 = c^216+256=c2, so c2=272c^2 = 272c2=272.

STEP 5

STEP 6

Remember **SOH CAH TOA**?For angle θ\thetaθ, the **opposite** side is 16, the **adjacent** side is 4, and the **hypotenuse** is 4174\sqrt{17}417​.

STEP 7

STEP 8

cos⁡θ=adjacenthypotenuse=4417=117\cos \theta = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{4}{4\sqrt{17}} = \frac{1}{\sqrt{17}}cosθ=hypotenuseadjacent​=417​4​=17​1​.Rationalizing gives us cos⁡θ=1717\cos \theta = \frac{\sqrt{17}}{17}cosθ=1717​​.

STEP 9

tan⁡θ=oppositeadjacent=164=4\tan \theta = \frac{\text{opposite}}{\text{adjacent}} = \frac{16}{4} = 4tanθ=adjacentopposite​=416​=4.

STEP 10

Now for angle ϕ\phiϕ!The **opposite** side is 4, the **adjacent** side is 16, and the **hypotenuse** is still 4174\sqrt{17}417​.

STEP 11

sin⁡ϕ=oppositehypotenuse=4417=117\sin \phi = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{4}{4\sqrt{17}} = \frac{1}{\sqrt{17}}sinϕ=hypotenuseopposite​=417​4​=17​1​.Rationalizing gives us sin⁡ϕ=1717\sin \phi = \frac{\sqrt{17}}{17}sinϕ=1717​​.

STEP 12

cos⁡ϕ=adjacenthypotenuse=16417=417\cos \phi = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{16}{4\sqrt{17}} = \frac{4}{\sqrt{17}}cosϕ=hypotenuseadjacent​=417​16​=17​4​.Rationalizing gives us cos⁡ϕ=41717\cos \phi = \frac{4\sqrt{17}}{17}cosϕ=17417​​.

STEP 13

tan⁡ϕ=oppositeadjacent=416=14\tan \phi = \frac{\text{opposite}}{\text{adjacent}} = \frac{4}{16} = \frac{1}{4}tanϕ=adjacentopposite​=164​=41​.

STEP 14

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What is this asking? We need to find the sine, cosine, and tangent of angles $\theta$ and $\phi$ in a right triangle, knowing two of its sides. Watch out! Don't mix up which sides are opposite and adjacent to which angle!
Also, we'll need to find the hypotenuse first.

1. Find the hypotenuse
2. Calculate sin θ, cos θ, and tan θ
3. Calculate sin φ, cos φ, and tan φ

Alright, let's do this!
We've got a right triangle, so we can use the Pythagorean theorem: $a^2 + b^2 = c^2$ , where a and b are the legs and c is the hypotenuse.
In our triangle, $a = 4$ and $b = 16$ .

Let's plug in the values: $4^2 + 16^2 = c^2$ .
This simplifies to $16 + 256 = c^2$ , so $c^2 = 272$ .

Remember SOH CAH TOA?
For angle $\theta$ , the opposite side is 16, the adjacent side is 4, and the hypotenuse is $4\sqrt{17}$ .

$\cos \theta = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{4}{4\sqrt{17}} = \frac{1}{\sqrt{17}}$ .
Rationalizing gives us $\cos \theta = \frac{\sqrt{17}}{17}$ .

$\tan \theta = \frac{\text{opposite}}{\text{adjacent}} = \frac{16}{4} = 4$ .

Now for angle $\phi$ !
The opposite side is 4, the adjacent side is 16, and the hypotenuse is still $4\sqrt{17}$ .

$\sin \phi = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{4}{4\sqrt{17}} = \frac{1}{\sqrt{17}}$ .
Rationalizing gives us $\sin \phi = \frac{\sqrt{17}}{17}$ .

$\cos \phi = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{16}{4\sqrt{17}} = \frac{4}{\sqrt{17}}$ .
Rationalizing gives us $\cos \phi = \frac{4\sqrt{17}}{17}$ .

$\tan \phi = \frac{\text{opposite}}{\text{adjacent}} = \frac{4}{16} = \frac{1}{4}$ .