express the sine, cosine and tangent of angles a and b as a fraction and as a decimal rounded to the nearest ten - thousandths.
1. $sin a = \frac{o}{h}$,
2. $cos a = \frac{a}{h}$,
3. $ an a = \frac{o}{a}$,
4. $cos b = $,
5. $ an b = $,
6. $sin b = $,
7. find the measure of $\angle a$ and $\angle b$ to the nearest degree.
$m\angle a = $ $^\circ$ and $m\angle b = $ $^\circ$
for each of the following, write the equation to find the missing value. then, write the equation you will enter in your calculator. round answer(s) to nearest tenth for sides, and nearest degree for angles.
8. $x \approx$ ; $y \approx$
9. $x \approx$ ; $y \approx$
10. $x \approx$ ; $y \approx$
11. $x \approx$ ; $y \approx$

Question

express the sine, cosine and tangent of angles a and b as a fraction and as a decimal rounded to the nearest ten - thousandths.
1. $sin a = \frac{o}{h}$,
2. $cos a = \frac{a}{h}$,
3. $	an a = \frac{o}{a}$,
4. $cos b = $,
5. $	an b = $,
6. $sin b = $,
7. find the measure of $\angle a$ and $\angle b$ to the nearest degree.
$m\angle a = $ $^\circ$ and $m\angle b = $ $^\circ$
for each of the following, write the equation to find the missing value. then, write the equation you will enter in your calculator. round answer(s) to nearest tenth for sides, and nearest degree for angles.
8. $x \approx$ ; $y \approx$
9. $x \approx$ ; $y \approx$
10. $x \approx$ ; $y \approx$
11. $x \approx$ ; $y \approx$

Accepted Answer

### 1. $\boldsymbol{\sin A}$
# Explanation:
## Step1: Recall SOHCAHTOA (sin = opposite/hypotenuse)
In $	riangle ABC$, $\angle C = 90^\circ$, opposite side to $A$ is $BC = 15$, hypotenuse $AB = 17$. So $\sin A=\frac{15}{17}$.
## Step2: Calculate decimal
$\frac{15}{17}\approx 0.8824$
# Answer: $\frac{15}{17}$, $0.8824$

### 2. $\boldsymbol{\cos A}$
# Explanation:
## Step1: Recall SOHCAHTOA (cos = adjacent/hypotenuse)
Adjacent side to $A$ is $AC = 8$, hypotenuse $AB = 17$. So $\cos A=\frac{8}{17}$.
## Step2: Calculate decimal
$\frac{8}{17}\approx 0.4706$
# Answer: $\frac{8}{17}$, $0.4706$

### 3. $\boldsymbol{	an A}$
# Explanation:
## Step1: Recall SOHCAHTOA (tan = opposite/adjacent)
Opposite side to $A$ is $BC = 15$, adjacent side $AC = 8$. So $	an A=\frac{15}{8}$.
## Step2: Calculate decimal
$\frac{15}{8}=1.875$
# Answer: $\frac{15}{8}$, $1.875$

### 4. $\boldsymbol{\cos B}$
# Explanation:
## Step1: Recall SOHCAHTOA (cos = adjacent/hypotenuse)
Adjacent side to $B$ is $BC = 15$, hypotenuse $AB = 17$. So $\cos B=\frac{15}{17}$.
## Step2: Calculate decimal
$\frac{15}{17}\approx 0.8824$
# Answer: $\frac{15}{17}$, $0.8824$

### 5. $\boldsymbol{	an B}$
# Explanation:
## Step1: Recall SOHCAHTOA (tan = opposite/adjacent)
Opposite side to $B$ is $AC = 8$, adjacent side $BC = 15$. So $	an B=\frac{8}{15}$.
## Step2: Calculate decimal
$\frac{8}{15}\approx 0.5333$
# Answer: $\frac{8}{15}$, $0.5333$

### 6. $\boldsymbol{\sin B}$
# Explanation:
## Step1: Recall SOHCAHTOA (sin = opposite/hypotenuse)
Opposite side to $B$ is $AC = 8$, hypotenuse $AB = 17$. So $\sin B=\frac{8}{17}$.
## Step2: Calculate decimal
$\frac{8}{17}\approx 0.4706$
# Answer: $\frac{8}{17}$, $0.4706$

### 7. $\boldsymbol{m\angle A}$ and $\boldsymbol{m\angle B}$
# Explanation:
## Step1: Find $m\angle A$
Use $\sin A=\frac{15}{17}\approx 0.8824$. $\angle A=\arcsin(0.8824)\approx 62^\circ$ (since $\sin 62^\circ\approx 0.8829$).
## Step2: Find $m\angle B$
In a right triangle, $\angle A+\angle B = 90^\circ$, so $\angle B = 90^\circ - 62^\circ = 28^\circ$ (or use $\cos B=\frac{15}{17}$, $\angle B=\arccos(0.8824)\approx 28^\circ$).
# Answer: $m\angle A = 62^\circ$ and $m\angle B = 28^\circ$

### 8. $\boldsymbol{x}$ and $\boldsymbol{y}$ (triangle with $36^\circ$)
# Explanation:
## Step1: Find $y$ (adjacent side, $\cos 36^\circ=\frac{y}{8}$)
$y = 8\cos 36^\circ$. $\cos 36^\circ\approx 0.8090$, so $y\approx 8	imes 0.8090 = 6.5$ (rounded to nearest tenth).
## Step2: Already found $x = 4.7$ (from $\sin 36^\circ=\frac{x}{8}$, $x = 8\sin 36^\circ\approx 4.7$)
# Answer: $x\approx 4.7$; $y\approx 6.5$

### 9. $\boldsymbol{x}$ and $\boldsymbol{y}$ (right triangle with sides 4, 8)
# Explanation:
## Step1: Find $x$ (angle, $\cos x=\frac{4}{8}=0.5$)
$x=\arccos(0.5)=60^\circ$ (since $\cos 60^\circ = 0.5$).
## Step2: Find $y$ (angle, $y = 90^\circ - 60^\circ = 30^\circ$)
# Answer: $x\approx 60^\circ$; $y\approx 30^\circ$

### 10. $\boldsymbol{x}$ and $\boldsymbol{y}$ (right triangle with side 5, angle $50^\circ$)
# Explanation:
## Step1: Find $x$ (opposite side, $	an 50^\circ=\frac{x}{5}$)
$x = 5	an 50^\circ$. $	an 50^\circ\approx 1.1918$, so $x\approx 5	imes 1.1918 = 5.959\approx 6.0$ (rounded to nearest tenth).
## Step2: Find $y$ (adjacent side, $\cos 50^\circ=\frac{5}{y}$)
$y=\frac{5}{\cos 50^\circ}$. $\cos 50^\circ\approx 0.6428$, so $y\approx\frac{5}{0.6428}\approx 7.8$ (rounded to nearest tenth).
# Answer: $x\approx 6.0$; $y\approx 7.8$

### 11. $\boldsymbol{x}$ and $\boldsymbol{y}$ (right triangle with angle $64^\circ$, hypotenuse 10)
# Explanation:
## Step1: Find $x$ (opposite side, $\sin 64^\circ=\frac{x}{10}$)
$x = 10\sin 64^\circ$. $\sin 64^\circ\approx 0.8988$, so $x\approx 10	imes 0.8988 = 8.988\approx 9.0$ (rounded to nearest tenth).
## Step2: Find $y$ (adjacent side, $\cos 64^\circ=\frac{10}{y}$? Wait, no: $\cos 64^\circ=\frac{x}{y}$? Wait, right triangle: hypotenuse is not 10, wait the triangle has right angle, angle $64^\circ$, side $x$ (opposite to $64^\circ$), side $y$ (adjacent). Wait, hypotenuse? Wait, the triangle: right angle, angle $64^\circ$, one leg $x$, hypotenuse? No, the diagram: right angle, angle $64^\circ$, side $x$ (opposite), side $y$ (adjacent), and the other side? Wait, no, the problem says "10" is a side. Wait, correct: $	an 64^\circ=\frac{y}{x}$? No, wait, let's re - express: In right triangle, angle $64^\circ$, opposite side $y$, adjacent side $x$, and hypotenuse? Wait, no, the diagram shows right angle, angle $64^\circ$, side $x$ (one leg), side $y$ (another leg), and the hypotenuse? Wait, no, the problem says "10" is a side. Wait, maybe $\sin 64^\circ=\frac{y}{10}$? No, the user's diagram: right triangle, angle $64^\circ$, side $x$ (opposite to $64^\circ$), side $y$ (adjacent), and the hypotenuse is 10? Wait, no, the correct approach: If the right triangle has angle $64^\circ$, and the side adjacent to $64^\circ$ is $x$, opposite is $y$, and hypotenuse? Wait, no, the problem says "10" is a side. Wait, let's use $\cos 64^\circ=\frac{x}{10}$? No, $\sin 64^\circ=\frac{y}{10}$, $\cos 64^\circ=\frac{x}{10}$? Wait, no, the correct way: Let's assume the hypotenuse is 10. Then $x = 10\cos 64^\circ$, $y = 10\sin 64^\circ$. Wait, $\cos 64^\circ\approx 0.4384$, so $x = 10	imes 0.4384 = 4.4$ (rounded to nearest tenth). $\sin 64^\circ\approx 0.8988$, so $y = 10	imes 0.8988 = 9.0$ (rounded to nearest tenth). Wait, but maybe the 10 is the hypotenuse. So:
## Step1: Find $x$ (adjacent, $\cos 64^\circ=\frac{x}{10}$)
$x = 10\cos 64^\circ\approx 10	imes 0.4384 = 4.4$ (rounded to nearest tenth).
## Step2: Find $y$ (opposite, $\sin 64^\circ=\frac{y}{10}$)
$y = 10\sin 64^\circ\approx 10	imes 0.8988 = 9.0$ (rounded to nearest tenth).
Wait, but the user's previous problem: Let's correct. If the right triangle has angle $64^\circ$, and the side opposite to $64^\circ$ is $y$, adjacent is $x$, and hypotenuse is 10, then:
$x = 10\cos 64^\circ\approx 4.4$, $y = 10\sin 64^\circ\approx 9.0$
# Answer: $x\approx 4.4$; $y\approx 9.0$ (Note: There may be a mis - interpretation of the diagram, but based on standard right - triangle trigonometry, this is the solution)

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QUESTION IMAGE

Question

Explanation:

1. $\boldsymbol{\sin A}$

Step1: Recall SOHCAHTOA (sin = opposite/hypotenuse)

Step2: Calculate decimal

Step1: Recall SOHCAHTOA (cos = adjacent/hypotenuse)

Step2: Calculate decimal

Step1: Recall SOHCAHTOA (tan = opposite/adjacent)

Step2: Calculate decimal

Answer:

2. $\boldsymbol{\cos A}$