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Mom's Basement University

Applications of Trigonometric Equations

What You’ll Learn

Section titled “What You’ll Learn”

In this lesson you’ll learn how to set up and solve trigonometric equations in practical, real-world contexts.

The Concept

Section titled “The Concept”

Many real-world problems can be modeled using trigonometric equations. The process usually involves:

  1. Drawing a clear diagram and labeling known and unknown quantities
  2. Choosing the appropriate trigonometric ratio or identity
  3. Setting up the equation
  4. Solving for the unknown variable
  5. Interpreting the answer in context and checking if it makes sense

Common applications include angles of elevation and depression, navigation, physics (waves and motion), and engineering problems.

Worked Example

Section titled “Worked Example”

Example 1: Building height

From a point 80 feet away from the base of a building, the angle of elevation to the top is 52°. How tall is the building?

The height is opposite the 52° angle, and the 80 feet is adjacent. That’s TOA: tan = opposite / adjacent.

tan⁡(52°)=h80\tan(52°) = \frac{h}{80} h=80×tan⁡(52°)≈80×1.2799≈102.4 feeth = 80 \times \tan(52°) \approx 80 \times 1.2799 \approx 102.4 \text{ feet}

Example 2: Ladder distance

A ladder 18 feet long leans against a wall, forming a 68° angle with the ground. How far is the base of the ladder from the wall?

The distance from the wall is adjacent, and the ladder is the hypotenuse. That’s CAH: cos = adjacent / hypotenuse.

cos⁡(68°)=d18\cos(68°) = \frac{d}{18} d=18×cos⁡(68°)≈18×0.3746≈6.74 feetd = 18 \times \cos(68°) \approx 18 \times 0.3746 \approx 6.74 \text{ feet}

Example 3: Finding an angle

A ramp must rise 4 feet over a horizontal distance of 48 feet. What angle does the ramp make with the ground?

We know the opposite (4 ft) and adjacent (48 ft) and need the angle. That’s TOA solved backwards with inverse tangent.

tan⁡(θ)=448≈0.0833\tan(\theta) = \frac{4}{48} \approx 0.0833 θ=tan⁡−1(0.0833)≈4.76°\theta = \tan^{-1}(0.0833) \approx 4.76°

This is within the ADA maximum ramp slope of about 4.76° (a 1:12 ratio), so the ramp meets code.

Example 4: Periodic model

The temperature in a city varies between 30°F in January and 90°F in July. Model the temperature as a function of the month and find when the temperature reaches 60°F.

The average temperature is (30 + 90) / 2 = 60°F. The amplitude is (90 - 30) / 2 = 30°F. The period is 12 months. Using a cosine model (since January is the minimum):

T(t)=−30cos⁡(2π12t)+60T(t) = -30\cos\left(\frac{2\pi}{12}t\right) + 60

Set T(t) = 60 and solve:

−30cos⁡(π6t)+60=60-30\cos\left(\frac{\pi}{6}t\right) + 60 = 60 cos⁡(π6t)=0\cos\left(\frac{\pi}{6}t\right) = 0 π6t=π2orπ6t=3π2\frac{\pi}{6}t = \frac{\pi}{2} \quad \text{or} \quad \frac{\pi}{6}t = \frac{3\pi}{2} t=3ort=9t = 3 \quad \text{or} \quad t = 9

The temperature hits 60°F in month 3 (March) and month 9 (September). That makes sense.

Real-World Application

Section titled “Real-World Application”

Trigonometric equations are used in:

  • Construction and architecture (roof slopes, ramp angles, structural supports)
  • Navigation and surveying (measuring inaccessible distances)
  • Physics (projectile motion, wave analysis)
  • Engineering (designing roads, bridges, and mechanical systems)
  • Climate science (modeling seasonal temperature patterns)
  • Safety (proper ladder placement, wheelchair ramp slopes)

Quiz

Section titled “Quiz”
From 100 feet away, the angle of elevation to the top of a tree is 48°. How tall is the tree?
A 25-foot ladder makes a 70° angle with the ground. How far is the base from the wall?
The most important first step in a trig application problem is:
A ramp rises 3 feet over 36 feet horizontally. The angle is approximately:
A temperature model $T(t) = 20\cos\left(\frac{\pi}{6}t\right) + 70$ has an average temperature of: