chapter 5 individual test - a
name:
show all work & good luck!!
score: /100
1. use your right triangle tools to find the missing values in the triangles below. show all work!
a.
b.

x =
x =
c.

θ =
x =
y=
2. find the missing side lengths of each special right triangle.
a)
b)

a =
b =
b =
c =
3. in the sophomore class at green high school, 95 students have their driver’s license and 105 students have a job. 78 students have both. there are 335 students in the entire class.
a) what is the probability that a student has their driver’s license or has a job? show your calculations.
b) what is the probability a student does not have their driver’s license?

Question

chapter 5 individual test - a
name:
show all work & good luck!!
score:  /100
1. use your right triangle tools to find the missing values in the triangles below. show all work!
a.
b.

x =
x =
c.

θ =
x =
y=
2. find the missing side lengths of each special right triangle.
a)
b)

a =
b =
b =
c =
3. in the sophomore class at green high school, 95 students have their driver’s license and 105 students have a job. 78 students have both. there are 335 students in the entire class.
a) what is the probability that a student has their driver’s license or has a job? show your calculations.
b) what is the probability a student does not have their driver’s license?

Accepted Answer

### Problem 1a
# Explanation:
## Step1: Identify trigonometric ratio
In right triangle $ABC$, $\angle A = 90^\circ$, $\angle C = 35^\circ$, adjacent side $AC = 12$, opposite side $AB = x$. Use tangent: $\tan\theta=\frac{\text{opposite}}{\text{adjacent}}$
$\tan(35^\circ)=\frac{x}{12}$

## Step2: Solve for $x$
Multiply both sides by 12: $x = 12\times\tan(35^\circ)$
Calculate $\tan(35^\circ)\approx0.7002$, so $x\approx12\times0.7002 = 8.4024$ (approx 8.4)

# Answer:
$x\approx 8.4$ (or more precise $12\tan(35^\circ)$)

### Problem 1b
# Explanation:
## Step1: Identify trigonometric ratio
In right triangle, $\angle$ with side 31 is adjacent to $x$? Wait, no: the side 31 is one leg, hypotenuse 47? Wait, no, the triangle has right angle, one leg 31, hypotenuse 47? Wait, no, the side labeled $x$ and 31 are legs? Wait, no, the hypotenuse is 47. Wait, no: right triangle, one leg 31, hypotenuse 47? No, wait, the angle: wait, the triangle has right angle, one leg 31, the other leg $x$, hypotenuse 47? Wait, no, Pythagoras: $x^2 + 31^2 = 47^2$? Wait, no, maybe $x$ is opposite to an angle, but the triangle has right angle, 31 is one leg, hypotenuse 47? Wait, no, let's check: $x^2 + 31^2 = 47^2$ → $x^2 = 47^2 - 31^2 = (47 - 31)(47 + 31)=16\times78 = 1248$ → $x=\sqrt{1248}\approx35.33$. Wait, maybe I misread. Wait, the triangle: right angle, one leg 31, hypotenuse 47? No, maybe the 31 is adjacent, $x$ is opposite, and hypotenuse 47? Wait, $\sin\theta=\frac{x}{47}$, but no, the leg 31: $\cos\theta=\frac{31}{47}$, so $\theta=\cos^{-1}(\frac{31}{47})$, then $x = 47\sin\theta$. Alternatively, Pythagoras: $x=\sqrt{47^2 - 31^2}=\sqrt{2209 - 961}=\sqrt{1248}\approx35.33$

## Step1: Apply Pythagorean theorem
$x = \sqrt{47^2 - 31^2}$

## Step2: Calculate
$47^2 = 2209$, $31^2 = 961$, $2209 - 961 = 1248$, $\sqrt{1248}=\sqrt{16\times78}=4\sqrt{78}\approx35.33$

# Answer:
$x\approx 35.33$ (or $4\sqrt{78}$)

### Problem 1c
# Explanation:
## Step1: Find $\theta$
In right triangle, one angle $44^\circ$, right angle, so $\theta = 90^\circ - 44^\circ = 46^\circ$

## Step2: Find $x$ (opposite to 98? Wait, 98 is one leg, angle $44^\circ$: $\tan(44^\circ)=\frac{x}{98}$? Wait, no: the side 98 is opposite to $\theta$? Wait, the triangle has right angle, side 98 is one leg, angle $44^\circ$ at the other acute angle. So: $\tan(44^\circ)=\frac{x}{98}$? Wait, no: $\tan(44^\circ)=\frac{\text{opposite}}{\text{adjacent}}$. If the angle is $44^\circ$, then the side opposite is $x$, adjacent is 98? Wait, no: the side labeled 98 is adjacent to $\theta$, and opposite to $44^\circ$. Wait, $\theta = 90 - 44 = 46^\circ$. Then, for angle $44^\circ$: $\tan(44^\circ)=\frac{x}{98}$ → $x = 98\tan(44^\circ)$. For $y$: $\cos(44^\circ)=\frac{98}{y}$ → $y=\frac{98}{\cos(44^\circ)}$

## Step3: Calculate $\theta$
$\theta = 90^\circ - 44^\circ = 46^\circ$

## Step4: Calculate $x$
$\tan(44^\circ)\approx0.9657$, so $x\approx98\times0.9657\approx94.64$

## Step5: Calculate $y$
$\cos(44^\circ)\approx0.7193$, so $y\approx\frac{98}{0.7193}\approx136.24$

# Answer:
$\theta = 46^\circ$, $x\approx94.64$, $y\approx136.24$

### Problem 2a
# Explanation:
## Step1: Identify triangle type
It's a 45-45-90 triangle, so legs are equal, hypotenuse $= a\sqrt{2}$ (since $a = b$, hypotenuse $= a\sqrt{2}$). Given hypotenuse $14\sqrt{2}$, so $a\sqrt{2}=14\sqrt{2}$ → $a = 14$, so $b = a = 14$

## Step2: Solve for $a$ and $b$
Since $a = b$ (45-45-90 triangle), and hypotenuse $= a\sqrt{2}$, so $a = \frac{14\sqrt{2}}{\sqrt{2}} = 14$, $b = 14$

# Answer:
$a = 14$, $b = 14$

### Problem 2b
# Explanation:
## Step1: Identify triangle type
30-60-90 triangle, opposite 30° is 5 (shorter leg), longer leg $b$ is $\sqrt{3}$ times shorter leg, hypotenuse $c$ is 2 times shorter leg.

## Step2: Find $b$ and $c$
Shorter leg (opposite 30°) is 5? Wait, no: the right angle, angle 30°, the side opposite 30° is 5? Wait, no: in 30-60-90 triangle, shorter leg is opposite 30°, longer leg opposite 60°, hypotenuse twice shorter leg. Wait, the side labeled 5: is it opposite 30°? No, 30° angle: the side opposite 30° is shorter. Wait, the triangle has right angle, angle 30°, side 5 is opposite 60°? Wait, no: let's see, the side with length 5 is one leg, angle 30°: so $\sin(30^\circ)=\frac{5}{c}$ (since 5 is opposite 30°? Wait, no, 30° angle: opposite side is shorter. Wait, maybe 5 is opposite 60°, so $\sin(60^\circ)=\frac{5}{c}$, $\cos(60^\circ)=\frac{b}{c}$, $\tan(30^\circ)=\frac{5}{b}$. Wait, no: let's correct. In 30-60-90 triangle:
- Shorter leg (opposite 30°): $s$
- Longer leg (opposite 60°): $s\sqrt{3}$
- Hypotenuse: $2s$

Given that one leg is 5: if 5 is the longer leg (opposite 60°), then $s\sqrt{3}=5$ → $s=\frac{5}{\sqrt{3}}$, hypotenuse $2s=\frac{10}{\sqrt{3}}$. But maybe 5 is the shorter leg: then longer leg $b = 5\sqrt{3}$, hypotenuse $c = 10$. Wait, the side labeled 5: let's check the triangle: right angle, angle 30°, side 5 is opposite 30°? No, 30° angle's opposite side is shorter. Wait, the triangle has right angle, angle 30°, side 5 is one leg (let's say opposite 60°), so:

$\tan(30^\circ)=\frac{5}{b}$ → $b = \frac{5}{\tan(30^\circ)} = 5\sqrt{3}\approx8.66$

$\sin(30^\circ)=\frac{5}{c}$ → $c = \frac{5}{\sin(30^\circ)} = 10$

Wait, yes: if 5 is opposite 30°, then hypotenuse $c = 10$, longer leg $b = 5\sqrt{3}$. Wait, the side labeled 5: is it opposite 30°? Then:

- Shorter leg (opposite 30°): 5 → hypotenuse $c = 2\times5 = 10$
- Longer leg (opposite 60°): $b = 5\sqrt{3}\approx8.66$

Yes, that makes sense.

## Step1: Determine sides
Shorter leg (opposite 30°) = 5 → hypotenuse $c = 2\times5 = 10$

Longer leg (opposite 60°) $b = 5\sqrt{3}$

# Answer:
$b = 5\sqrt{3}$ (or ≈8.66), $c = 10$

### Problem 3a
# Explanation:
## Step1: Use principle of inclusion-exclusion
Let $A$ be students with driver's license, $B$ with job. $n(A)=95$, $n(B)=105$, $n(A\cap B)=78$, total $N=335$

Number of students with license or job: $n(A\cup B)=n(A)+n(B)-n(A\cap B)=95+105-78=122$

## Step2: Calculate probability
Probability $P(A\cup B)=\frac{n(A\cup B)}{N}=\frac{122}{335}\approx0.3642$ (or simplify $\frac{122}{335}$)

# Answer:
$\frac{122}{335}$ (or ≈0.364)

### Problem 3b
# Explanation:
## Step1: Number of students without license
Total students: 335, students with license: 95. So without license: $335 - 95 = 240$

## Step2: Calculate probability
Probability $P(\text{no license})=\frac{240}{335}=\frac{48}{67}\approx0.7164$

# Answer:
$\frac{48}{67}$ (or ≈0.716)

Sovi.AI - AI Math Tutor

QUESTION IMAGE

Question

Explanation:

Problem 1a

Step1: Identify trigonometric ratio

Step2: Solve for \(x\)

Step1: Identify trigonometric ratio

Step1: Apply Pythagorean theorem

Step2: Calculate

Step1: Find \(\theta\)

Step3: Calculate \(\theta\)

Step4: Calculate \(x\)

Step5: Calculate \(y\)

Answer:

Problem 1b