Applications and Vectors
Apply trigonometry to real-world problems and vector quantities
You have spent time learning about sine, cosine, and tangent. You can find missing sides and angles in triangles. You understand the unit circle. But here is the question that might be nagging at you: when am I ever going to use this?
The answer is: all the time, in ways you might not expect. Every time a pilot navigates through the sky, every time an engineer calculates whether a bridge can handle the load, every time a physicist predicts where a ball will land, trigonometry is doing the heavy lifting. And at the heart of many of these applications is a concept called a vector.
So, if the word “vector” seems scary to you, then know that you are not alone. Many students hear the word and imagine something complicated and abstract. But you need to realize something: you already know how to work with vectors. You may not believe it, but, really, you do. If you have ever given directions like “go 3 blocks north, then 2 blocks east,” you have described vectors. If you have ever felt the wind push you sideways while you are walking forward, you have experienced vector addition. This lesson is about taking that intuition and making it precise.
Core Concepts
Bearings and Navigation
Before GPS, sailors and pilots needed a precise way to describe direction. They developed the system of bearings, which is still used today in aviation, marine navigation, and surveying.
A bearing tells you which direction to travel, measured as an angle from a reference direction. There are two common ways to express bearings:
Compass Bearings: These use cardinal directions (N, S, E, W) as references. You start at either North or South, then specify the angle toward East or West.
- N 30° E means: start facing North, rotate 30° toward the East
- S 45° W means: start facing South, rotate 45° toward the West
- N 0° E is the same as due North
- N 90° E is the same as due East
True Bearings (Three-Figure Bearings): These measure the angle clockwise from North, always using three digits.
- 000° is due North
- 090° is due East
- 180° is due South
- 270° is due West
- 045° is Northeast (N 45° E in compass bearing)
True bearings are common in aviation because they are unambiguous: every direction corresponds to exactly one number between 000° and 360°.
The key to solving navigation problems is drawing a clear diagram. Always:
- Draw a vertical line representing North-South
- Mark the bearing angle from North
- Use right triangles to find north-south and east-west components
Angle of Elevation and Depression
When you look up at something above you, the angle your line of sight makes with the horizontal is called the angle of elevation. When you look down at something below you, that angle is called the angle of depression.
Here is the crucial insight: the angle of elevation from point A to point B equals the angle of depression from point B to point A. They are alternate interior angles formed by a horizontal line and the line of sight, with a transversal (your line of sight) crossing them.
These angles appear constantly in real-world problems:
- Surveyors measuring the height of a building
- Pilots calculating descent angles
- Engineers designing wheelchair ramps
- Astronomers measuring the position of stars above the horizon
The setup is always the same: you have a right triangle where one angle is the angle of elevation or depression, one side is a known distance (often horizontal or vertical), and you need to find another side using trigonometric ratios.
Introduction to Vectors
A scalar is a quantity that has only magnitude (size). Temperature, mass, and time are scalars. When you say “it took 3 hours,” you have completely described the time.
A vector is a quantity that has both magnitude and direction. When you say “the wind is blowing at 20 mph,” you have not given complete information. From which direction? North? Southwest? The direction matters. Velocity, force, and displacement are all vectors.
Vectors are typically represented in two ways:
Geometric representation: An arrow where the length represents magnitude and the direction of the arrow represents, well, direction. We often name vectors with bold letters like v or with an arrow above them like $\vec{v}$.
Component form: A vector can be written as $\mathbf{v} = \langle a, b \rangle$, where $a$ is the horizontal component and $b$ is the vertical component. Think of this as: “go $a$ units in the x-direction and $b$ units in the y-direction.”
The vector $\langle 3, 4 \rangle$ means: move 3 units right and 4 units up.
Magnitude and Direction
The magnitude of a vector $\mathbf{v} = \langle a, b \rangle$ is its length, written as $|\mathbf{v}|$ or $|\mathbf{v}|$. By the Pythagorean Theorem:
$$|\mathbf{v}| = \sqrt{a^2 + b^2}$$
For example, the vector $\langle 3, 4 \rangle$ has magnitude $\sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5$.
The direction of a vector is the angle it makes with the positive x-axis, measured counterclockwise. If $\theta$ is this angle, then:
$$\tan \theta = \frac{b}{a}$$
So $\theta = \tan^{-1}\left(\frac{b}{a}\right)$, but you need to be careful about which quadrant the vector is in. The arctangent function only gives angles in the first and fourth quadrants, so you may need to add 180° if the vector points into the second or third quadrant.
Converting Between Forms
Often you will need to convert between the two representations.
From magnitude and direction to component form:
If a vector has magnitude $r$ and makes angle $\theta$ with the positive x-axis:
$$a = r \cos \theta$$ $$b = r \sin \theta$$
So $\mathbf{v} = \langle r \cos \theta, r \sin \theta \rangle$.
This should feel familiar: it is exactly how you found the coordinates of a point on the unit circle, just scaled by $r$.
From component form to magnitude and direction:
Given $\mathbf{v} = \langle a, b \rangle$:
$$r = |\mathbf{v}| = \sqrt{a^2 + b^2}$$ $$\theta = \tan^{-1}\left(\frac{b}{a}\right) \text{ (adjusted for quadrant)}$$
Adding Vectors
This is where vectors become powerful. To add two vectors, you simply add their corresponding components:
$$\langle a, b \rangle + \langle c, d \rangle = \langle a + c, b + d \rangle$$
Geometrically, this is like placing the tail of the second vector at the head of the first. The sum (called the resultant) goes from the tail of the first to the head of the second. This is sometimes called the “tip-to-tail” method.
Alternatively, if both vectors start at the same point, the resultant is the diagonal of the parallelogram they form. This is the “parallelogram method.”
Both methods give the same answer, just visualized differently.
Subtracting vectors works similarly:
$$\langle a, b \rangle - \langle c, d \rangle = \langle a - c, b - d \rangle$$
Scalar multiplication scales a vector:
$$k \langle a, b \rangle = \langle ka, kb \rangle$$
If $k > 0$, the vector points in the same direction but is longer or shorter. If $k < 0$, the vector reverses direction.
Applications: Force, Velocity, and Displacement
Displacement is the change in position from one point to another. If you walk 5 meters east, your displacement is the vector $\langle 5, 0 \rangle$ (assuming east is positive x). If you then walk 3 meters north, that is $\langle 0, 3 \rangle$. Your total displacement is $\langle 5, 0 \rangle + \langle 0, 3 \rangle = \langle 5, 3 \rangle$.
Velocity is speed with a direction. A car going 60 mph north has a different velocity than a car going 60 mph east, even though they have the same speed. Velocities add as vectors: if you are on a boat going 10 mph east in a river flowing 3 mph south, your actual velocity relative to the ground is $\langle 10, -3 \rangle$ (if east is positive x and north is positive y).
Force is a push or pull with both magnitude and direction. When multiple forces act on an object, you add them as vectors to find the net force or resultant force. This is fundamental to all of physics and engineering.
Simple Harmonic Motion
Many things in nature oscillate: a pendulum swings back and forth, a spring bounces up and down, a guitar string vibrates, the tides rise and fall. When the motion follows a sine or cosine pattern, we call it simple harmonic motion (SHM).
The position of an object in SHM can be described by:
$$y = A \cos(\omega t)$$ or $$y = A \sin(\omega t)$$
where:
- $A$ is the amplitude: the maximum displacement from equilibrium (how far it swings)
- $\omega$ (omega) is the angular frequency: how fast it oscillates, in radians per unit time
- $t$ is time
The period $T$ is the time for one complete cycle:
$$T = \frac{2\pi}{\omega}$$
The frequency $f$ is how many cycles per unit time:
$$f = \frac{1}{T} = \frac{\omega}{2\pi}$$
Which function you use (sine or cosine) depends on the initial conditions:
- Use $\cos$ if the object starts at maximum displacement (like a pendulum released from the side)
- Use $\sin$ if the object starts at the equilibrium position moving in the positive direction
You can also add a phase shift $\phi$ to account for different starting positions:
$$y = A \cos(\omega t + \phi)$$
This is why trigonometry is so important in physics: the same mathematical functions that describe triangles also describe oscillations, waves, and countless other phenomena.
Notation and Terminology
| Term | Meaning | Example |
|---|---|---|
| Bearing | Direction measured from north | N 30° E, or 030° |
| Angle of elevation | Angle upward from horizontal | Looking up at a plane |
| Angle of depression | Angle downward from horizontal | Looking down from a cliff |
| Vector | Quantity with magnitude and direction | Force, velocity |
| Scalar | Quantity with magnitude only | Speed, temperature |
| Component form | $\mathbf{v} = \langle a, b \rangle$ | $\langle 3, 4 \rangle$ |
| Magnitude | Length of a vector: $ | \mathbf{v} |
| Resultant | Sum of two or more vectors | Net force |
| Amplitude | Maximum displacement in SHM | $A$ in $y = A\cos(\omega t)$ |
| Period | Time for one complete oscillation | $T = \frac{2\pi}{\omega}$ |
| Angular frequency | Rate of oscillation in radians per time | $\omega$ in $y = A\cos(\omega t)$ |
Examples
A plane flies at bearing 040° for 200 miles. Find the north and east components of its displacement.
Solution:
First, understand what bearing 040° means: starting from north, rotate 40° clockwise (toward east).
Draw a diagram with North pointing up. The plane’s path makes a 40° angle with the north direction.
To find the components, we need to set up a right triangle. The 200-mile displacement is the hypotenuse. The angle between the displacement and the north (vertical) axis is 40°.
Using trigonometry:
- The north component is adjacent to the 40° angle: $\text{North} = 200 \cos 40°$
- The east component is opposite to the 40° angle: $\text{East} = 200 \sin 40°$
Calculate: $$\text{North} = 200 \cos 40° \approx 200 \times 0.766 = 153.2 \text{ miles}$$ $$\text{East} = 200 \sin 40° \approx 200 \times 0.643 = 128.6 \text{ miles}$$
Answer: The plane travels approximately 153.2 miles north and 128.6 miles east.
Note: When the bearing is measured from north, use cosine for the north component and sine for the east component. This is the opposite of what you might expect from standard position angles, where cosine gives the x-component. The reason: bearing angles are measured from the vertical (north), not the horizontal (east).
Find the magnitude and direction of the vector $\mathbf{v} = \langle 3, 4 \rangle$.
Solution:
Magnitude: Use the Pythagorean Theorem: $$|\mathbf{v}| = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5$$
Direction: Find the angle with the positive x-axis: $$\tan \theta = \frac{4}{3}$$ $$\theta = \tan^{-1}\left(\frac{4}{3}\right) \approx 53.13°$$
Since both components are positive, the vector is in the first quadrant, so no adjustment is needed.
Answer: The vector has magnitude 5 and direction approximately 53.13° from the positive x-axis.
Check: We can verify by converting back to components:
- $5 \cos 53.13° \approx 5 \times 0.6 = 3$ ✓
- $5 \sin 53.13° \approx 5 \times 0.8 = 4$ ✓
A ship sails N 25° W for 50 km, then N 65° E for 80 km. Find the ship’s displacement from its starting point (magnitude and direction).
Solution:
We need to find the components of each leg, add them, then find the magnitude and direction of the total.
First leg: N 25° W for 50 km
“N 25° W” means starting from north, rotate 25° toward west. West is the negative x-direction.
- North component: $50 \cos 25° \approx 50 \times 0.906 = 45.32$ km
- East component: $-50 \sin 25° \approx -50 \times 0.423 = -21.13$ km (negative because west)
Second leg: N 65° E for 80 km
“N 65° E” means starting from north, rotate 65° toward east.
- North component: $80 \cos 65° \approx 80 \times 0.423 = 33.81$ km
- East component: $80 \sin 65° \approx 80 \times 0.906 = 72.50$ km
Total displacement:
Add the components:
- Total North: $45.32 + 33.81 = 79.13$ km
- Total East: $-21.13 + 72.50 = 51.37$ km
Magnitude: $$d = \sqrt{79.13^2 + 51.37^2} = \sqrt{6261.56 + 2638.87} = \sqrt{8900.43} \approx 94.34 \text{ km}$$
Direction:
The ship is north and east of the starting point, so the bearing will be N __ E.
$$\tan \theta = \frac{51.37}{79.13} \approx 0.649$$ $$\theta = \tan^{-1}(0.649) \approx 33.0°$$
Answer: The ship is approximately 94.3 km from its starting point, at a bearing of N 33° E (or 033°).
Two forces act on an object: one of 100 N and another of 150 N. The angle between them is 60°. Find the magnitude and direction of the resultant force.
Solution:
Place the first force (100 N) along the positive x-axis. The second force (150 N) makes a 60° angle with the first.
Component form of each force:
Force 1: $\mathbf{F}_1 = \langle 100, 0 \rangle$
Force 2: $$\mathbf{F}_2 = \langle 150 \cos 60°, 150 \sin 60° \rangle = \langle 150 \times 0.5, 150 \times 0.866 \rangle = \langle 75, 129.90 \rangle$$
Resultant force:
$$\mathbf{F}_R = \mathbf{F}_1 + \mathbf{F}_2 = \langle 100 + 75, 0 + 129.90 \rangle = \langle 175, 129.90 \rangle$$
Magnitude: $$|\mathbf{F}_R| = \sqrt{175^2 + 129.90^2} = \sqrt{30625 + 16874.01} = \sqrt{47499.01} \approx 217.9 \text{ N}$$
Direction (angle from the 100 N force): $$\tan \theta = \frac{129.90}{175} \approx 0.742$$ $$\theta = \tan^{-1}(0.742) \approx 36.6°$$
Answer: The resultant force has magnitude approximately 217.9 N and makes an angle of approximately 36.6° with the 100 N force.
Note: Notice that the resultant (217.9 N) is less than the sum of the two forces (250 N). This is because the forces are not pointing in the same direction. The resultant would only equal 250 N if the forces were parallel and pointing the same way.
An object oscillates in simple harmonic motion with amplitude 5 cm and period 2 seconds. The object starts at its maximum displacement. Write the equation of motion and find the position at $t = 0.5$ seconds.
Solution:
Identify the parameters:
- Amplitude: $A = 5$ cm
- Period: $T = 2$ seconds
- Initial position: maximum displacement
Find the angular frequency:
$$\omega = \frac{2\pi}{T} = \frac{2\pi}{2} = \pi \text{ rad/s}$$
Choose the appropriate function:
Since the object starts at maximum displacement (at $t = 0$, position $= A$), we use cosine:
$$y = A \cos(\omega t)$$
Why cosine? Because $\cos(0) = 1$, so at $t = 0$: $y = A \cos(0) = A \times 1 = A$.
If we used sine, we would have $y = A \sin(0) = 0$ at $t = 0$, which would mean starting at the equilibrium position, not at maximum displacement.
The equation of motion:
$$y = 5 \cos(\pi t)$$
Position at $t = 0.5$ seconds:
$$y = 5 \cos(\pi \times 0.5) = 5 \cos\left(\frac{\pi}{2}\right) = 5 \times 0 = 0 \text{ cm}$$
Answer: The equation of motion is $y = 5\cos(\pi t)$, and the position at $t = 0.5$ seconds is 0 cm (at the equilibrium position).
Understanding the motion: Starting at maximum displacement (5 cm), the object moves through equilibrium (0 cm) at $t = 0.5$ s, reaches minimum displacement (-5 cm) at $t = 1$ s, returns through equilibrium at $t = 1.5$ s, and is back at maximum displacement at $t = 2$ s. This is one complete cycle.
An airplane is heading 300° (true bearing) at an airspeed of 500 mph. It encounters a wind blowing from 045° at 60 mph. Find the plane’s actual course (ground track) and ground speed.
Solution:
This problem requires careful attention to what “heading,” “wind from,” and “course” mean.
- Heading 300°: The direction the plane’s nose points. This is where the plane would go with no wind.
- Wind from 045°: The wind is coming from the 045° direction, which means it blows toward 225° (045° + 180°).
- Course (ground track): The actual direction the plane travels over the ground.
- Ground speed: The actual speed relative to the ground.
The plane’s velocity through the air plus the wind velocity equals the ground velocity.
Set up a coordinate system: Let east be positive x, north be positive y.
Plane’s airspeed vector (heading 300°):
300° from north, clockwise. To convert to standard position (from east, counterclockwise): $90° - 300° = -210°$, or equivalently $150°$.
Actually, let us work directly with bearing angles. For a bearing of $\beta$:
- East component: $\sin \beta$
- North component: $\cos \beta$
Plane velocity: $\mathbf{v}_p = \langle 500 \sin 300°, 500 \cos 300° \rangle$
Calculate: $$\sin 300° = -\sin 60° = -0.866$$ $$\cos 300° = \cos 60° = 0.5$$
$$\mathbf{v}_p = \langle 500 \times (-0.866), 500 \times 0.5 \rangle = \langle -433, 250 \rangle \text{ mph}$$
Wind velocity (from 045°, so blowing toward 225°):
$$\mathbf{v}_w = \langle 60 \sin 225°, 60 \cos 225° \rangle$$
Calculate: $$\sin 225° = -\frac{\sqrt{2}}{2} \approx -0.707$$ $$\cos 225° = -\frac{\sqrt{2}}{2} \approx -0.707$$
$$\mathbf{v}_w = \langle 60 \times (-0.707), 60 \times (-0.707) \rangle = \langle -42.4, -42.4 \rangle \text{ mph}$$
Ground velocity:
$$\mathbf{v}_g = \mathbf{v}_p + \mathbf{v}_w = \langle -433 + (-42.4), 250 + (-42.4) \rangle = \langle -475.4, 207.6 \rangle \text{ mph}$$
Ground speed (magnitude):
$$|\mathbf{v}_g| = \sqrt{(-475.4)^2 + (207.6)^2} = \sqrt{226005 + 43098} = \sqrt{269103} \approx 518.8 \text{ mph}$$
Course (direction):
The ground velocity has negative east (west) and positive north components, so the plane is actually traveling to the northwest.
$$\tan \theta = \frac{-475.4}{207.6} \approx -2.290$$
This gives the angle from north. Since we are going west of north: $$\theta = \tan^{-1}(2.290) \approx 66.4°$$
The bearing is N 66.4° W, or as a true bearing: $360° - 66.4° = 293.6°$, approximately 294°.
Answer: The airplane’s ground speed is approximately 518.8 mph, and its actual course is approximately 294° (or N 66° W).
What this means: Even though the pilot is pointing the plane at 300°, the wind pushes it off course. The plane ends up traveling at 294°, about 6° to the left of the intended heading. The ground speed (518.8 mph) is actually higher than the airspeed (500 mph) in this case because the wind has a component in the direction of travel.
Key Properties and Rules
Vector Operations
Addition: $\langle a, b \rangle + \langle c, d \rangle = \langle a + c, b + d \rangle$
Subtraction: $\langle a, b \rangle - \langle c, d \rangle = \langle a - c, b - d \rangle$
Scalar multiplication: $k\langle a, b \rangle = \langle ka, kb \rangle$
Magnitude: $|\langle a, b \rangle| = \sqrt{a^2 + b^2}$
Converting Between Forms
| Have | Want | Use |
|---|---|---|
| Magnitude $r$, angle $\theta$ | Components | $a = r\cos\theta$, $b = r\sin\theta$ |
| Components $a$, $b$ | Magnitude | $r = \sqrt{a^2 + b^2}$ |
| Components $a$, $b$ | Direction | $\theta = \tan^{-1}(b/a)$ (adjust for quadrant) |
Bearing Conversions
For a bearing of $\beta$ (measured clockwise from north):
- East component: distance $\times \sin \beta$
- North component: distance $\times \cos \beta$
Simple Harmonic Motion
$$y = A\cos(\omega t) \quad \text{or} \quad y = A\sin(\omega t)$$
| Quantity | Symbol | Formula |
|---|---|---|
| Amplitude | $A$ | Maximum displacement |
| Angular frequency | $\omega$ | $\omega = \frac{2\pi}{T} = 2\pi f$ |
| Period | $T$ | $T = \frac{2\pi}{\omega}$ |
| Frequency | $f$ | $f = \frac{1}{T} = \frac{\omega}{2\pi}$ |
Initial conditions:
- Object starts at maximum displacement: use $y = A\cos(\omega t)$
- Object starts at equilibrium, moving positive: use $y = A\sin(\omega t)$
Common Mistakes to Avoid
- Confusing bearing angles with standard position angles: Bearings are measured clockwise from north; standard position angles are counterclockwise from east.
- Forgetting that “wind from” means the opposite direction: Wind “from the north” blows toward the south.
- Not adjusting arctangent for quadrant: $\tan^{-1}$ only gives angles in Q1 and Q4. If the vector is in Q2 or Q3, add 180°.
- Confusing speed and velocity: Speed is a scalar (just magnitude); velocity is a vector (magnitude and direction).
- Mixing up period and frequency: Period is time per cycle (seconds); frequency is cycles per time (Hz). They are reciprocals.
Real-World Applications
Aviation Navigation
Pilots constantly deal with the difference between heading (where the nose points) and track (where the plane actually goes). Crosswinds push planes off course, and pilots must calculate the “wind correction angle” to maintain their intended track. This is exactly the vector addition problem from Example 6.
Flight computers and autopilots do these trigonometric calculations automatically, but pilots are trained to do them by hand as a backup. The E6B “whiz wheel,” a circular slide rule that has been standard pilot equipment for decades, is essentially a mechanical vector calculator.
Ocean Currents and Sailing
Sailors face the same challenge as pilots, but with ocean currents instead of wind. A boat that can travel at 8 knots through the water might only make 6 knots over the ground if it is fighting a 2-knot current, or it might make 10 knots with a following current.
Modern navigation systems show both speed through water (from the boat’s instruments) and speed over ground (from GPS), and the difference tells you about the current. Competitive sailors and racing teams use detailed current charts and vector calculations to find the fastest route.
Physics: Projectile Motion and Forces
When you throw a ball at an angle, its initial velocity is a vector. Physics uses trigonometry to break this into horizontal and vertical components, which then evolve independently (the horizontal component stays constant while the vertical component changes due to gravity).
Similarly, forces in physics always combine as vectors. An object on a ramp experiences gravitational force pulling straight down, but we break this into components parallel and perpendicular to the ramp to analyze the motion. Bridges, buildings, and machines are all designed using vector analysis of forces.
Engineering: Structural Analysis
Civil engineers must ensure that structures can handle all the forces acting on them. Wind loads, weight loads, and tension in cables all act as vectors. The joints in a bridge truss experience forces from multiple directions, and engineers use vector addition to find the total force each joint must withstand.
Even something as simple as a hanging sign involves vectors: the tension in each supporting cable must add up to a vector that exactly balances the weight of the sign.
Pendulums and Springs
The grandfather clock in a museum, the shock absorbers in your car, the vibrating atoms in a solid, and the electrical oscillations in a radio circuit all exhibit simple harmonic motion. The same equations describe them all.
In music, a vibrating guitar string undergoes SHM (approximately), and the frequency of vibration determines the pitch you hear. The amplitude determines the volume. Every time you listen to music, you are experiencing the trigonometry of simple harmonic motion translated into sound waves.
Self-Test Problems
Problem 1: A hiker walks at bearing 065° for 3 km. How far east and how far north does the hiker travel?
Show Answer
For bearing 065° (65° clockwise from north):
East component: $3 \sin 65° \approx 3 \times 0.906 = 2.72$ km
North component: $3 \cos 65° \approx 3 \times 0.423 = 1.27$ km
The hiker travels approximately 2.72 km east and 1.27 km north.
Problem 2: Find the magnitude and direction of the vector $\mathbf{v} = \langle -5, 12 \rangle$.
Show Answer
Magnitude: $|\mathbf{v}| = \sqrt{(-5)^2 + 12^2} = \sqrt{25 + 144} = \sqrt{169} = 13$
Direction: $\tan \theta = \frac{12}{-5} = -2.4$
$\tan^{-1}(-2.4) \approx -67.4°$
But the vector has a negative x-component and positive y-component, so it is in the second quadrant. We need to add 180°:
$\theta = -67.4° + 180° = 112.6°$
The vector has magnitude 13 and direction approximately 112.6° from the positive x-axis.
Problem 3: A boat travels N 40° E for 25 km, then due east for 15 km. Find the total displacement (magnitude and bearing).
Show Answer
First leg (N 40° E for 25 km):
- North: $25 \cos 40° \approx 19.15$ km
- East: $25 \sin 40° \approx 16.07$ km
Second leg (due east for 15 km):
- North: 0 km
- East: 15 km
Total:
- North: $19.15 + 0 = 19.15$ km
- East: $16.07 + 15 = 31.07$ km
Magnitude: $\sqrt{19.15^2 + 31.07^2} = \sqrt{366.7 + 965.3} = \sqrt{1332} \approx 36.5$ km
Direction: $\tan \theta = \frac{31.07}{19.15} \approx 1.62$, so $\theta \approx 58.3°$ east of north.
The total displacement is approximately 36.5 km at bearing N 58° E (or 058°).
Problem 4: Two forces of 80 N and 60 N act on an object at right angles to each other. Find the magnitude and direction of the resultant force.
Show Answer
Place the 80 N force along the positive x-axis and the 60 N force along the positive y-axis.
Resultant: $\mathbf{F}_R = \langle 80, 60 \rangle$
Magnitude: $|\mathbf{F}_R| = \sqrt{80^2 + 60^2} = \sqrt{6400 + 3600} = \sqrt{10000} = 100$ N
Direction: $\tan \theta = \frac{60}{80} = 0.75$, so $\theta = \tan^{-1}(0.75) \approx 36.9°$
The resultant force has magnitude 100 N at an angle of approximately 36.9° from the 80 N force.
(Notice this is a 3-4-5 right triangle scaled by 20!)
Problem 5: An object in simple harmonic motion has amplitude 8 cm and completes 4 cycles per second. If the object starts at the equilibrium position moving in the positive direction, write the equation of motion.
Show Answer
Given:
- Amplitude: $A = 8$ cm
- Frequency: $f = 4$ Hz (cycles per second)
- Initial condition: starts at equilibrium, moving positive (use sine)
Angular frequency: $\omega = 2\pi f = 2\pi \times 4 = 8\pi$ rad/s
Equation: $y = A\sin(\omega t)$
$$y = 8\sin(8\pi t)$$
where $y$ is in cm and $t$ is in seconds.
Problem 6: From the top of a 200-foot cliff, the angle of depression to a boat is 25°. How far is the boat from the base of the cliff?
Show Answer
The angle of depression from the cliff top equals the angle of elevation from the boat (alternate interior angles).
In the right triangle:
- The vertical side (cliff height) is 200 feet (opposite the 25° angle)
- The horizontal distance is what we want (adjacent to the 25° angle)
Using tangent: $$\tan 25° = \frac{200}{d}$$
$$d = \frac{200}{\tan 25°} = \frac{200}{0.466} \approx 429 \text{ feet}$$
The boat is approximately 429 feet from the base of the cliff.
Problem 7: A weight on a spring oscillates with the equation $y = 6\cos(2\pi t)$, where $y$ is in centimeters and $t$ is in seconds. Find the amplitude, period, and the position at $t = 0.25$ seconds.
Show Answer
From the equation $y = 6\cos(2\pi t)$:
Amplitude: $A = 6$ cm
Angular frequency: $\omega = 2\pi$ rad/s
Period: $T = \frac{2\pi}{\omega} = \frac{2\pi}{2\pi} = 1$ second
Position at $t = 0.25$: $$y = 6\cos(2\pi \times 0.25) = 6\cos\left(\frac{\pi}{2}\right) = 6 \times 0 = 0 \text{ cm}$$
The amplitude is 6 cm, the period is 1 second, and the position at $t = 0.25$ s is 0 cm (at equilibrium).
Summary
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Bearings describe direction using angles from north. Compass bearings use N/S and E/W (like N 30° E), while true bearings use three-digit clockwise angles from north (like 030°). For a bearing $\beta$, the north component is distance $\times \cos\beta$ and the east component is distance $\times \sin\beta$.
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Angles of elevation and depression are measured from the horizontal. They are equal when measured from opposite ends of the same line of sight, and they set up right triangle problems that can be solved with basic trigonometry.
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Vectors have both magnitude and direction. They can be represented geometrically (as arrows) or in component form: $\mathbf{v} = \langle a, b \rangle$.
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The magnitude of $\langle a, b \rangle$ is $\sqrt{a^2 + b^2}$. The direction is found using $\tan^{-1}(b/a)$, adjusted for the correct quadrant.
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To convert from magnitude $r$ and angle $\theta$ to components: $a = r\cos\theta$ and $b = r\sin\theta$.
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Vectors add by adding their components: $\langle a, b \rangle + \langle c, d \rangle = \langle a + c, b + d \rangle$. This models how velocities combine, how forces add, and how displacements accumulate.
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Simple harmonic motion is described by $y = A\cos(\omega t)$ or $y = A\sin(\omega t)$, where $A$ is the amplitude, $\omega$ is the angular frequency, and the period is $T = 2\pi/\omega$.
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These concepts have countless real-world applications: navigation (bearing problems and wind correction), physics (projectile motion, forces, oscillations), engineering (structural analysis), and more.
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The key to solving these problems is always the same: draw a diagram, identify the right triangle or vector relationship, and apply trigonometry. Whether you are calculating a pilot’s ground track or predicting where a pendulum will be in half a second, you are using the same fundamental tools.
You have now seen how the trigonometry you learned about angles and triangles extends to describe motion, forces, and navigation in the real world. Vectors are the language that physics and engineering use to describe quantities with direction, and simple harmonic motion is the pattern that describes everything from heartbeats to radio waves. These applications are why trigonometry is not just an abstract mathematical exercise but an essential tool for understanding how the world works.