3. list the rules for each trig function.
$\sin\theta=\frac{\text{opposite}}{\text{hypotenuse}}$
$\cos\theta=\frac{\text{adjacent}}{\text{hypotenuse}}$
$\tan\theta=\frac{\text{opposite}}{\text{adjacent}}$
4. in the right triangle mnp, $m\angle n = 90^\circ$, and the $\cos m = \frac{5}{12}$, what is the value for the $\sin p$? (hint: draw & label the triangle)
5. look at the figure below. if $f * \sin 75^\circ = e$, then $\frac{e}{f}$ is equivalent to which ratio?
6. identify the following trigonometric ratio that is correct
a. $\sin e = \frac{a}{c}$
b. $\cos b = \frac{e}{f}$
c. $\tan a = \frac{e}{d}$
d. $\cos d = \frac{b}{c}$

Question

3. list the rules for each trig function.
$\sin\theta=\frac{\text{opposite}}{\text{hypotenuse}}$
$\cos\theta=\frac{\text{adjacent}}{\text{hypotenuse}}$
$\tan\theta=\frac{\text{opposite}}{\text{adjacent}}$
4. in the right triangle mnp, $m\angle n = 90^\circ$, and the $\cos m = \frac{5}{12}$, what is the value for the $\sin p$? (hint: draw & label the triangle)
5. look at the figure below. if $f * \sin 75^\circ = e$, then $\frac{e}{f}$ is equivalent to which ratio?
6. identify the following trigonometric ratio that is correct
 a. $\sin e = \frac{a}{c}$
 b. $\cos b = \frac{e}{f}$
 c. $\tan a = \frac{e}{d}$
 d. $\cos d = \frac{b}{c}$

Accepted Answer

### Question 4 Solution:
# Explanation:
## Step1: Analyze the right triangle
In right triangle $ MNP $ with $ \angle N = 90^\circ $, $ \cos M=\frac{5}{12} $. By the definition of cosine, $ \cos M=\frac{\text{adjacent to } M}{\text{hypotenuse}} $. So, adjacent side to $ M $ (which is $ MN $) is $ 5 $, and hypotenuse $ MP $ is $ 12 $.

## Step2: Identify angles and sides for $ \sin P $
In a right triangle, $ \angle M + \angle P=90^\circ $. For $ \sin P $, $ \sin P=\frac{\text{opposite to } P}{\text{hypotenuse}} $. The side opposite to $ P $ is $ MN $ (since $ \angle P $ and $ \angle M $ are complementary, the opposite side of $ P $ is the adjacent side of $ M $).

## Step3: Calculate $ \sin P $
Since opposite side to $ P $ is $ 5 $ and hypotenuse is $ 12 $, $ \sin P = \frac{5}{12} $. Wait, no, wait. Wait, let's re - check. Wait, in right triangle $ MNP $, $ \angle N = 90^\circ $, so sides: $ MN $ (adjacent to $ M $), $ NP $ (opposite to $ M $), $ MP $ (hypotenuse). $ \cos M=\frac{MN}{MP}=\frac{5}{12} $, so $ MN = 5 $, $ MP = 12 $. Now, $ \angle P $: the side opposite to $ P $ is $ MN $, and hypotenuse is $ MP $. Wait, no, $ \angle P $: in triangle $ MNP $, angles sum to $ 180^\circ $, $ \angle N = 90^\circ $, so $ \angle M+\angle P = 90^\circ $. So $ \sin P=\cos M $ (because $ \sin(90^\circ - \theta)=\cos\theta $). Since $ \cos M=\frac{5}{12} $, then $ \sin P=\frac{5}{12} $? Wait, no, wait, maybe I mixed up the sides. Wait, let's draw the triangle: vertices $ M $, $ N $, $ P $, right - angled at $ N $. So $ MN $ and $ NP $ are legs, $ MP $ is hypotenuse. $ \angle M $ is at vertex $ M $, so the sides: adjacent to $ M $ is $ MN $, opposite to $ M $ is $ NP $, hypotenuse $ MP $. $ \cos M=\frac{MN}{MP}=\frac{5}{12} $, so $ MN = 5 $, $ MP = 12 $. Now, $ \angle P $ is at vertex $ P $, so the side opposite to $ P $ is $ MN $, and hypotenuse is $ MP $. So $ \sin P=\frac{\text{opposite to } P}{\text{hypotenuse}}=\frac{MN}{MP}=\frac{5}{12} $. Wait, but that seems too easy. Wait, no, wait, maybe I made a mistake. Wait, $ \sin P=\frac{NP}{MP} $? No, no. Wait, $ \angle P $: the side opposite to $ P $ is $ MN $, because $ \angle P $ is at $ P $, so the side opposite is $ MN $ (from $ P $ to $ M $'s adjacent side). Wait, let's use the co - function identity: $ \sin P=\sin(90^\circ - M)=\cos M $. Since $ \cos M = \frac{5}{12} $, then $ \sin P=\frac{5}{12} $.

# Answer:
$\frac{5}{12}$

### Question 5 Solution:
# Explanation:
## Step1: Start with the given equation
We are given $ f\times\sin75^{\circ}=e $. We need to find $ \frac{e}{f} $.

## Step2: Solve for $ \frac{e}{f} $
Divide both sides of the equation $ f\times\sin75^{\circ}=e $ by $ f $ (assuming $ f
eq0 $). We get $ \frac{e}{f}=\sin75^{\circ} $.

# Answer:
$\sin75^{\circ}$

### Question 6 Solution:
# Explanation:
Let's analyze each option:

- Option a: In triangle $ ABC $, right - angled at $ C $, $ \sin E $ (wait, no, triangle $ ABC $: right - angled at $ C $, angle at $ B $ is $ 70^\circ $. $ \sin B=\frac{\text{opposite to } B}{\text{hypotenuse}}=\frac{AC}{AB}=\frac{b}{c} $, not $ \frac{a}{c} $. So option a is wrong.

- Option b: In triangle $ EFD $, right - angled at $ F $, angle at $ E $ is $ 70^\circ $. In triangle $ ABC $, angle at $ B $ is $ 70^\circ $. $ \cos B=\frac{\text{adjacent to } B}{\text{hypotenuse}}=\frac{BC}{AB}=\frac{a}{c} $, not $ \frac{e}{f} $. Wait, no, let's look at triangle $ EFD $: $ \cos E=\frac{\text{adjacent to } E}{\text{hypotenuse}}=\frac{EF}{ED}=\frac{e}{f} $. And since $ \angle B = \angle E = 70^\circ $, $ \cos B=\cos E=\frac{e}{f} $. So option b is correct.

- Option c: In triangle $ ABC $, $ \tan A=\frac{\text{opposite to } A}{\text{adjacent to } A}=\frac{BC}{AC}=\frac{a}{b} $, not $ \frac{e}{d} $. So option c is wrong.

- Option d: In triangle $ ABC $, $ \cos D $: wait, $ D $ is in triangle $ EFD $. $ \cos D=\frac{\text{adjacent to } D}{\text{hypotenuse}}=\frac{FD}{ED}=\frac{e}{f} $, not $ \frac{b}{c} $. So option d is wrong.

# Answer:
b. $ \cos B=\frac{e}{f} $

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QUESTION IMAGE

Question

Explanation:

Question 4 Solution:

Step1: Analyze the right triangle

Step2: Identify angles and sides for \( \sin P \)

Step3: Calculate \( \sin P \)

Step1: Start with the given equation

Step2: Solve for \( \frac{e}{f} \)

Answer:

Question 5 Solution: