Precalculus Study Guide: Trigonometry, Functions, and Logarithms

Q1. Find the length of a guy wire. The height of a radio tower is 500 feet, and the ground on one side of the tower slopes upward at an angle of 13 degrees. How long must the wire be if it is attached to the top of the tower and anchored 900 feet from the base?

Background

Topic: Trigonometry - Law of Cosines

This question tests your ability to apply the Law of Cosines to solve for the length of a side in a triangle when two sides and the included angle are known.

Key formula:

$c^2 = a^2 + b^2 - 2ab\cos(C)$

• $a$ and $b$ are the lengths of the sides (500 ft and 900 ft)

• $C$ is the angle between them (13 degrees)

• $c$ is the length of the wire (what you're solving for)

Step-by-Step Guidance

1. Draw a diagram to visualize the triangle formed by the tower, the anchor point, and the wire.

2. Label the sides: $a = 500$ ft (height), $b = 900$ ft (distance from base), and $C = 13^ ext{o}$ (angle between sides).

3. Set up the Law of Cosines formula: $c^2 = 500^2 + 900^2 - 2 \times 500 \times 900 \times \cos(13^ ext{o})$

4. Calculate $\cos(13^ ext{o})$ and substitute all values into the formula, but do not compute the final value yet.

Try solving on your own before revealing the answer!

Final Answer: The wire must be approximately 520.6 ft long.

Using the Law of Cosines and plugging in the values, you get $c \approx 520.6$ ft.

This is based on the correct application of the Law of Cosines for the triangle formed by the tower, anchor, and wire.

Q2. Determine whether the given information results in one triangle, two triangles, or no triangle at all. Solve any triangle(s) that result.

Background

Topic: Trigonometry - Law of Sines and Ambiguous Case (SSA)

This question tests your understanding of the Law of Sines and the ambiguous case, where two sides and a non-included angle are given (SSA).

Key formula:

$\frac{\sin A}{a} = \frac{\sin B}{b}$

• $a = 3$, $b = 7$, $A = 12^ ext{o}$

• Use Law of Sines to find $B$ and check for possible triangles.

Step-by-Step Guidance

1. Write the Law of Sines equation: $\frac{\sin 12^ ext{o}}{3} = \frac{\sin B}{7}$

2. Solve for $\sin B$: $\sin B = 7 \times \frac{\sin 12^ ext{o}}{3}$

3. Check if $\sin B$ is between 0 and 1 to determine if a triangle is possible.

4. If $\sin B$ is valid, find angle $B$ and use $180^ ext{o} - (A + B)$ to find angle $C$.

Try solving on your own before revealing the answer!

Final Answer: One triangle is possible, with $B \approx 28.24^ ext{o}$ and $C \approx 139.76^ ext{o}$.

Law of Sines confirms only one triangle can be formed with the given values.

Q3. From a point 200 feet in front of a public library, the angles of elevation to the base of the flagpole and the top of the flagpole are 30 degrees and 60 degrees, respectively. What is the height of the flagpole?

Background

Topic: Trigonometry - Right Triangle Applications

This question tests your ability to use tangent ratios to solve for the height of an object given angles of elevation and horizontal distance.

Key formula:

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

• Use $\tan(30^ ext{o})$ and $\tan(60^ ext{o})$ with the adjacent side of 200 ft.

Step-by-Step Guidance

1. Set up two equations: $y = 200 \tan(30^ ext{o})$ for the base, and $y + h = 200 \tan(60^ ext{o})$ for the top.

2. Solve for $y$ (height to base) and $y + h$ (height to top).

3. Subtract $y$ from $y + h$ to find the height of the flagpole ($h$).

4. Plug in the values for $\tan(30^ ext{o})$ and $\tan(60^ ext{o})$ but do not compute the final value yet.

Try solving on your own before revealing the answer!

Final Answer: The height of the flagpole is approximately 230.94 ft.

Subtracting the height to the base from the height to the top gives the flagpole's height.

Q4. Write the equation of a sine function that has the given characteristics: Amplitude = 2, Period = $\frac{2\pi}{\lambda}$, Phase Shift = $\pi$

Background

Topic: Trigonometric Functions - Sine Function Transformations

This question tests your understanding of how to write the equation of a sine function given amplitude, period, and phase shift.

Key formula:

$y = A \sin(Bx - C)$

• $A$ = amplitude

• $B$ = $\frac{2\pi}{\text{period}}$

• $C$ = phase shift

Step-by-Step Guidance

1. Identify amplitude ($A = 2$), period ($\frac{2\pi}{\lambda}$), and phase shift ($\pi$).

2. Calculate $B$ using $B = \frac{2\pi}{\text{period}}$.

3. Write the equation in the form $y = 2 \sin(Bx - \pi)$.

4. Substitute the calculated value of $B$ but do not simplify the equation fully yet.

Try solving on your own before revealing the answer!

Final Answer: $y = 2\sin\left(\frac{2\pi}{\lambda}x - \pi\right)$

This equation matches the given amplitude, period, and phase shift.

Q5. Find the exact value of $\cos\left[\sin^{-1}\left(\frac{3}{5}\right)\right]$

Background

Topic: Trigonometric Inverse Functions

This question tests your ability to evaluate composite trigonometric expressions using right triangle relationships.

Key formula:

If $\sin\theta = \frac{3}{5}$, then $\cos\theta = \sqrt{1 - \sin^2\theta}$

• Use Pythagorean identity: $\sin^2\theta + \cos^2\theta = 1$

Step-by-Step Guidance

1. Let $\theta = \sin^{-1}\left(\frac{3}{5}\right)$, so $\sin\theta = \frac{3}{5}$.

2. Draw a right triangle with opposite side 3 and hypotenuse 5.

3. Find the adjacent side using the Pythagorean theorem: $x^2 + 3^2 = 5^2$.

4. Calculate $\cos\theta = \frac{x}{5}$, but do not compute the final value yet.

Try solving on your own before revealing the answer!

Final Answer: $\cos\left[\sin^{-1}\left(\frac{3}{5}\right)\right] = \frac{4}{5}$

Using the triangle, the adjacent side is 4, so $\cos\theta = \frac{4}{5}$.