Precalculus Exam 1 Review: Step-by-Step Guidance

Q1. Convert the angle 83°32'28'' to a decimal in degrees. Round the answer to two decimal places.

Background

Topic: Angle Conversion (Degrees-Minutes-Seconds to Decimal Degrees)

This question tests your ability to convert an angle given in degrees, minutes, and seconds (D° M' S'') to decimal degrees.

Key Terms and Formula:

• Degrees (°), Minutes ('), Seconds ('')

• 1 degree = 60 minutes, 1 minute = 60 seconds

Formula:

$\text{Decimal Degrees} = D + \frac{M}{60} + \frac{S}{3600}$

Step-by-Step Guidance

1. Identify the values: $D = 83$, $M = 32$, $S = 28$.

2. Plug these values into the formula: $83 + \frac{32}{60} + \frac{28}{3600}$.

3. Calculate $\frac{32}{60}$ and $\frac{28}{3600}$ separately.

4. Add these values to 83 to get the decimal degree value.

Try solving on your own before revealing the answer!

Final Answer: 83.54°

$83 + \frac{32}{60} + \frac{28}{3600} = 83.54$ (rounded to two decimal places).

This is the decimal degree equivalent of the given angle.

Q2. Convert the angle 149.57° to D° M' S'' form. Round the answer to the nearest second.

Background

Topic: Angle Conversion (Decimal Degrees to Degrees-Minutes-Seconds)

This question tests your ability to convert a decimal degree value to degrees, minutes, and seconds (D° M' S'').

Key Terms and Formula:

• Degrees (°), Minutes ('), Seconds ('')

• 1 degree = 60 minutes, 1 minute = 60 seconds

Steps:

1. Take the integer part as degrees: $D = \lfloor 149.57 \rfloor$.

2. Multiply the decimal part by 60 to get minutes: $M = \lfloor (149.57 - D) \times 60 \rfloor$.

3. Multiply the remaining decimal part by 60 to get seconds: $S = \lfloor ((149.57 - D) \times 60 - M) \times 60 \rfloor$.

Try solving on your own before revealing the answer!

Final Answer: 149°34'12''

149.57° converts to 149 degrees, 34 minutes, and 12 seconds.

Q3. If $s$ denotes the length of the arc of a circle of radius $r$ subtended by a central angle $\theta$, find the missing quantity: $r = 12$ feet, $s = 12$ feet, $\theta = ?$

Background

Topic: Arc Length Formula

This question tests your ability to use the arc length formula to solve for the central angle in radians.

Key Terms and Formula:

• Arc length ($s$)

• Radius ($r$)

• Central angle ($\theta$ in radians)

Formula:

$s = r \theta$

Step-by-Step Guidance

1. Identify the known values: $r = 12$ feet, $s = 12$ feet.

2. Plug these values into the formula: $12 = 12 \theta$.

3. Rearrange to solve for $\theta$: $\theta = \frac{s}{r}$.

Try solving on your own before revealing the answer!

Final Answer: $\theta = 1$ radian

Since $s = r$, the central angle is $1$ radian.

Q4. Convert the angle 87° in degrees to radians. Express the answer as a multiple of $\pi$.

Background

Topic: Degree-Radian Conversion

This question tests your ability to convert degrees to radians and express the answer in terms of $\pi$.

Key Terms and Formula:

• Degrees ($\text{deg}$)

• Radians ($\text{rad}$)

Formula:

$\text{Radians} = \text{Degrees} \times \frac{\pi}{180}$

Step-by-Step Guidance

1. Plug 87° into the formula: $87 \times \frac{\pi}{180}$.

2. Simplify the fraction $\frac{87}{180}$.

3. Express the answer as a multiple of $\pi$.

Try solving on your own before revealing the answer!

Final Answer: $\frac{29\pi}{60}$ radians

87° equals $\frac{29\pi}{60}$ radians.

Q5. Convert the angle $\frac{11\pi}{10}$ radians to degrees.

Background

Topic: Radian-Degree Conversion

This question tests your ability to convert radians to degrees.

Key Terms and Formula:

• Radians ($\text{rad}$)

• Degrees ($\text{deg}$)

Formula:

$\text{Degrees} = \text{Radians} \times \frac{180}{\pi}$

Step-by-Step Guidance

1. Plug $\frac{11\pi}{10}$ into the formula: $\frac{11\pi}{10} \times \frac{180}{\pi}$.

2. Cancel $\pi$ in numerator and denominator.

3. Multiply $\frac{11}{10} \times 180$.

Try solving on your own before revealing the answer!

Final Answer: 198°

$\frac{11\pi}{10}$ radians equals 198 degrees.

Q6. If $A$ denotes the area of the sector of a circle of radius $r$ formed by the central angle $\theta$, find the missing quantity: $r = 12$ inches, $\theta = 90°$, $A = ?$

Background

Topic: Area of a Sector

This question tests your ability to use the sector area formula, converting degrees to radians if necessary.

Key Terms and Formula:

• Area of sector ($A$)

• Radius ($r$)

• Central angle ($\theta$ in radians)

Formula:

$A = \frac{1}{2} r^2 \theta$ (where $\theta$ is in radians)

Step-by-Step Guidance

1. Convert $\theta = 90°$ to radians: $90 \times \frac{\pi}{180} = \frac{\pi}{2}$ radians.

2. Plug $r = 12$ and $\theta = \frac{\pi}{2}$ into the formula: $A = \frac{1}{2} \times 12^2 \times \frac{\pi}{2}$.

3. Calculate $12^2$ and multiply by $\frac{1}{2}$.

4. Multiply by $\frac{\pi}{2}$ to get the area.

Try solving on your own before revealing the answer!

Final Answer: 113.04 in2

The area of the sector is 113.04 square inches (rounded to two decimal places).

Q7. An object is traveling around a circle with a radius of 20 meters. If in 10 seconds a central angle of $\frac{1}{5}$ radian is swept out, what is the linear speed of the object?

Background

Topic: Circular Motion (Linear Speed)

This question tests your ability to calculate linear speed from arc length and time.

Key Terms and Formula:

• Linear speed ($v$)

• Arc length ($s$)

• Time ($t$)

• Radius ($r$)

• Central angle ($\theta$ in radians)

Formulas:

$s = r \theta$

$v = \frac{s}{t}$

Step-by-Step Guidance

1. Calculate the arc length: $s = 20 \times \frac{1}{5}$.

2. Plug the arc length and time into the linear speed formula: $v = \frac{s}{10}$.

3. Simplify the expressions to find $v$.

Try solving on your own before revealing the answer!

Final Answer: $\frac{2}{5}$ m/sec

The linear speed is $\frac{2}{5}$ meters per second.

Q8. Find the exact value of the expression: $\tan 45° - \sin 60°$. Do not use a calculator.

Background

Topic: Exact Trigonometric Values

This question tests your knowledge of exact values for common trigonometric functions.

Key Terms and Values:

• $\tan 45° = 1$

• $\sin 60° = \frac{\sqrt{3}}{2}$

Step-by-Step Guidance

1. Recall the exact value for $\tan 45°$.

2. Recall the exact value for $\sin 60°$.

3. Subtract $\sin 60°$ from $\tan 45°$.

Try solving on your own before revealing the answer!

Final Answer: $1 - \frac{\sqrt{3}}{2}$

This is the exact value for the expression.

Q9. Name the quadrant in which the angle $\theta$ lies if $\sin \theta > 0$ and $\cos \theta < 0$.

Background

Topic: Trigonometric Signs by Quadrant

This question tests your understanding of the signs of sine and cosine in each quadrant.

Key Terms:

• Quadrants I-IV

• $\sin$ is positive in Quadrants I and II

• $\cos$ is negative in Quadrants II and III

Step-by-Step Guidance

1. Identify which quadrants have $\sin \theta > 0$.

2. Identify which quadrants have $\cos \theta < 0$.

3. Find the quadrant where both conditions are true.

Try solving on your own before revealing the answer!

Final Answer: Quadrant II

In Quadrant II, sine is positive and cosine is negative.

Q10. Name the quadrant in which the angle $\theta$ lies if $\csc \theta > 0$ and $\sec \theta > 0$.

Background

Topic: Trigonometric Signs by Quadrant

This question tests your understanding of the signs of cosecant and secant in each quadrant.

Key Terms:

• $\csc \theta > 0$ means $\sin \theta > 0$

• $\sec \theta > 0$ means $\cos \theta > 0$

Step-by-Step Guidance

1. Recall which quadrants have $\sin \theta > 0$.

2. Recall which quadrants have $\cos \theta > 0$.

3. Find the quadrant where both are positive.

Try solving on your own before revealing the answer!

Final Answer: Quadrant I

In Quadrant I, both sine and cosine are positive.

Q11. A point on the terminal side of an angle $\theta$ is given: (3, -2). Find $\tan \theta$.

Background

Topic: Trigonometric Functions from Coordinates

This question tests your ability to find the tangent of an angle given a point on its terminal side.

Key Terms and Formula:

• For a point $(x, y)$, $\tan \theta = \frac{y}{x}$

Step-by-Step Guidance

1. Identify $x = 3$, $y = -2$.

2. Plug into the formula: $\tan \theta = \frac{-2}{3}$.

Try solving on your own before revealing the answer!

Final Answer: $-\frac{2}{3}$

The tangent of the angle is $-\frac{2}{3}$.

Q12. Given $\tan \theta = -\frac{24}{7}$ and $90° < \theta < 180°$, find $\cos \theta$.

Background

Topic: Trigonometric Functions and Quadrants

This question tests your ability to find cosine given tangent and the quadrant.

Key Terms and Formula:

• $\tan \theta = \frac{\sin \theta}{\cos \theta}$

• Quadrant II: $\sin \theta > 0$, $\cos \theta < 0$

Step-by-Step Guidance

1. Let $\sin \theta = 24$, $\cos \theta = -7$ (since tangent is negative in Quadrant II).

2. Find the hypotenuse: $r = \sqrt{24^2 + 7^2}$.

3. Write $\cos \theta = \frac{-7}{r}$.

Try solving on your own before revealing the answer!

Final Answer: $-\frac{7}{25}$

Cosine is negative in Quadrant II, so $\cos \theta = -\frac{7}{25}$.

Q13. Use the even-odd properties to find the exact value of $\cos \left(-\frac{\pi}{4}\right)$. Do not use a calculator.

Background

Topic: Even-Odd Properties of Trigonometric Functions

This question tests your knowledge of how cosine behaves with negative angles.

Key Terms:

• Cosine is an even function: $\cos(-x) = \cos(x)$

• $\cos \frac{\pi}{4} = \frac{\sqrt{2}}{2}$

Step-by-Step Guidance

1. Apply the even property: $\cos(-\frac{\pi}{4}) = \cos(\frac{\pi}{4})$.

2. Recall the exact value for $\cos(\frac{\pi}{4})$.

Try solving on your own before revealing the answer!

Final Answer: $\frac{\sqrt{2}}{2}$

Cosine is even, so $\cos(-\frac{\pi}{4}) = \frac{\sqrt{2}}{2}$.

Q14. A surveyor is measuring the distance across a small lake. He has set up his transit on one side of the lake 120 feet from a piling that is directly across from a pier on the other side of the lake. From his transit, the angle between the piling and the pier is 30°. What is the distance between the piling and the pier to the nearest foot?

Background

Topic: Law of Sines (Triangle Applications)

This question tests your ability to use trigonometric relationships to solve for unknown sides in a triangle.

Key Terms and Formula:

• Law of Sines: $\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}$

• Right triangle relationships

Step-by-Step Guidance

1. Draw a diagram to visualize the triangle.

2. Identify the known side (120 ft) and the angle (30°).

3. Use the sine function: $\text{opposite} = 120 \times \tan(30°)$ or $\text{opposite} = 120 \times \sin(30°)$, depending on setup.

4. Set up the equation to solve for the unknown distance.

Try solving on your own before revealing the answer!

Final Answer: 69 ft

The distance between the piling and the pier is 69 feet.

Q15. From the edge of a 1000-foot cliff, the angles of depression to two cars in the valley below are 21° and 28°. How far apart are the cars? Round your answer to the nearest 0.1 ft.

Background

Topic: Angle of Depression and Horizontal Distance

This question tests your ability to use trigonometric relationships to find horizontal distances given angles of depression.

Key Terms and Formula:

• Right triangle relationships

• $\tan \theta = \frac{\text{opposite}}{\text{adjacent}}$

Step-by-Step Guidance

1. For each car, set up a right triangle with the cliff as the opposite side (1000 ft).

2. Calculate the horizontal distance for each car: $d_1 = 1000 \cot(21°)$, $d_2 = 1000 \cot(28°)$.

3. Find the difference between the two distances to get how far apart the cars are.

Try solving on your own before revealing the answer!

Final Answer: 724.4 ft

The cars are 724.4 feet apart (rounded to the nearest 0.1 ft).

Q16. Use the properties of the trigonometric functions to find the exact value of the expression: $\sec^2 20° - \tan^2 20°$. Do not use a calculator.

Background

Topic: Trigonometric Identities

This question tests your knowledge of the Pythagorean identity involving secant and tangent.

Key Terms and Formula:

• $\sec^2 x - \tan^2 x = 1$ (Pythagorean identity)

Step-by-Step Guidance

1. Recall the identity: $\sec^2 x = 1 + \tan^2 x$.

2. Rearrange: $\sec^2 x - \tan^2 x = 1$.

3. Apply this identity to $x = 20°$.

Try solving on your own before revealing the answer!

Final Answer: 1

The expression simplifies to 1 by the Pythagorean identity.

Q17. Given $\csc \theta = -\frac{3}{2}$ and $\theta$ is in quadrant III, find $\cot \theta$.

Background

Topic: Trigonometric Functions and Quadrants

This question tests your ability to find cotangent given cosecant and the quadrant.

Key Terms and Formula:

• $\csc \theta = \frac{1}{\sin \theta}$

• $\cot \theta = \frac{\cos \theta}{\sin \theta}$

• Quadrant III: $\sin \theta < 0$, $\cos \theta < 0$

Step-by-Step Guidance

1. Find $\sin \theta = -\frac{2}{3}$.

2. Find the adjacent side using the Pythagorean identity: $r = 3$, $y = -2$, $x = ?$.

3. Calculate $x = -\sqrt{3^2 - (-2)^2}$.

4. Find $\cot \theta = \frac{x}{y}$.

Try solving on your own before revealing the answer!

Final Answer: $\frac{5}{2}$

Cotangent is positive in Quadrant III, so $\cot \theta = \frac{5}{2}$.

Q18. Use the fact that the trigonometric functions are periodic to find the exact value of $\sin 765°$. Do not use a calculator.

Background

Topic: Periodicity of Trigonometric Functions

This question tests your ability to use the periodicity of sine to find equivalent angles.

Key Terms and Formula:

• Sine has period 360°: $\sin(\theta + 360°) = \sin \theta$

Step-by-Step Guidance

1. Subtract multiples of 360° from 765° to find an equivalent angle between 0° and 360°.

2. Calculate $765° - 2 \times 360° = 45°$.

3. Find $\sin 45°$.

Try solving on your own before revealing the answer!

Final Answer: $\frac{\sqrt{2}}{2}$

$\sin 765° = \sin 45° = \frac{\sqrt{2}}{2}$.

Q19. Use the fact that the trigonometric functions are periodic to find the exact value of $\tan 750°$. Do not use a calculator.

Background

Topic: Periodicity of Trigonometric Functions

This question tests your ability to use the periodicity of tangent to find equivalent angles.

Key Terms and Formula:

• Tangent has period 180°: $\tan(\theta + 180°) = \tan \theta$

Step-by-Step Guidance

1. Subtract multiples of 180° from 750° to find an equivalent angle between 0° and 180°.

2. Calculate $750° - 4 \times 180° = 30°$.

3. Find $\tan 30°$.

Try solving on your own before revealing the answer!

Final Answer: $\frac{\sqrt{3}}{3}$

$\tan 750° = \tan 30° = \frac{\sqrt{3}}{3}$.

Q20. The blade of a windshield wiper sweeps out an angle of 135° in one cycle. The base of the blade is 12 inches from the pivot point and the tip is 32 inches from the pivot point. What area does the wiper cover in one cycle? (Round to the nearest 0.1 square inch.)

Background

Topic: Area of a Circular Sector (Annular Sector)

This question tests your ability to find the area swept by a wiper, which is the area between two sectors.

Key Terms and Formula:

• Area of sector: $A = \frac{1}{2} r^2 \theta$ (with $\theta$ in radians)

• Annular sector: $A = \frac{1}{2} (R^2 - r^2) \theta$

Step-by-Step Guidance

1. Convert 135° to radians: $135 \times \frac{\pi}{180} = \frac{3\pi}{4}$ radians.

2. Calculate the area swept by the tip: $A_{tip} = \frac{1}{2} \times 32^2 \times \frac{3\pi}{4}$.

3. Calculate the area swept by the base: $A_{base} = \frac{1}{2} \times 12^2 \times \frac{3\pi}{4}$.

4. Subtract the two areas to get the area covered: $A = A_{tip} - A_{base}$.

Try solving on your own before revealing the answer!

Final Answer: 1036.7 in2

The area covered by the wiper in one cycle is 1036.7 square inches (rounded to the nearest 0.1).