Trigonometry

At its core, trigonometry (trig for short) is the math of triangles, specifically the relationship between a triangle’s angles and the lengths of its sides. But because of how it is set up, trig becomes much bigger than triangles—it is also the math of circles, waves, and anything that repeats (like sound, light, tides, and spinning objects). Here is the breakdown of what it is, what it does, and how to use it to solve problems.

1. What is Trig? (The Basics)

Trig is built on three main functions: Sine ($\sin$), Cosine ($\cos$), and Tangent ($\tan$). Imagine a right triangle (a triangle with one $90^\circ$ corner). Pick one of the other angles (let’s call it angle $\theta$). The triangle has three sides relative to that angle:

• Hypotenuse ($H$): The longest side, opposite the $90^\circ$ angle.

• Opposite ($O$): The side across from the angle $\theta$.

• Adjacent ($A$): The side next to the angle $\theta$ (not the hypotenuse).

The trig functions are just ratios (fractions) of these sides:

• Sine ($\sin$): $\frac{\text{Opposite}}{\text{Hypotenuse}}$ (SOH)

• Cosine ($\cos$): $\frac{\text{Adjacent}}{\text{Hypotenuse}}$ (CAH)

• Tangent ($\tan$): $\frac{\text{Opposite}}{\text{Adjacent}}$ (TOA)

(Memory trick: SOH-CAH-TOA)

2. What Does Trig Do?

Trig is essentially a translator between angles and distances. It does two main things:

1. It finds missing sides: If you know an angle and one side of a right triangle, you can find any other side.

2. It finds missing angles: If you know the lengths of two sides of a right triangle, you can find the exact angles using inverse trig on a calculator ($\sin^{-1}$, $\cos^{-1}$, $\tan^{-1}$).

Beyond Triangles (The Unit Circle & Waves)

When you take trig off a triangle and put it on a circle (called the "Unit Circle"), the sine and cosine functions create perfect waves. This is why trig is the foundation of physics, music, engineering, and anything that spins or oscillates.

3. How to Solve Problems with Trig (Step-by-Step)

Here is a reliable 4-step system to solve almost any basic trig problem:

1. Step 1: Draw and Label the Triangle

   • Draw the triangle and write down everything you know.

   • Put a mark on the angle you are working with.

   • Label the sides as $O$, $A$, and $H$.

2. Step 2: Decide which rule to use (SOH-CAH-TOA)

   • Look at what you have and what you need to find.

     • Do you have the Opposite and Hypotenuse? Use Sine.

     • Do you have the Adjacent and Hypotenuse? Use Cosine.

     • Do you have the Opposite and Adjacent? Use Tangent.

3. Step 3: Write the equation

   • Plug your numbers into the formula.

   • Example: If you have an angle of $30^\circ$, the Hypotenuse is $10$, and you need to find the Opposite side, you use Sine:
     $\sin(30^\circ) = \frac{O}{10}$

4. Step 4: Solve the math

   • Use algebra to get the missing piece alone.

     • If finding a side: Multiply ($O = 10 \times \sin(30^\circ)$). Type $10 \times \sin(30^\circ)$ into a calculator. You get $5$.

     • If finding an angle: Use the inverse button. If $\sin(\theta) = 0.5$, then $\theta = \sin^{-1}(0.5)$. Type that into a calculator. You get $30^\circ$.

Example Problem

Problem: A $20\,\text{ft}$ ladder is leaning against a wall. It makes a $60^\circ$ angle with the ground. How high up the wall does the ladder reach?

• Step 1: Draw it. The ladder is the Hypotenuse ($H = 20\,\text{ft}$). The height up the wall is the Opposite ($O = ?$) to the $60^\circ$ angle.

• Step 2: We have $H$ and need $O$. We use Sine (SOH).

• Step 3: Write the equation:
  $\sin(60^\circ) = \frac{O}{20}$

• Step 4: Solve for $O$:
  $O = 20 \times \sin(60^\circ)$
  Calculator: $20 \times 0.866 = 17.3\,\text{ft}$. The ladder reaches $17.3\,\text{ft}$ up the wall.

Summary

Trig is the math of angles and triangles. To solve a problem, just draw the triangle, label the sides ($O$, $A$, $H$), choose the right rule using SOH-CAH-TOA, and plug in your numbers!

Practice Problems

Level 1: Finding Missing Sides (Using SOH-CAH-TOA)

For these, solve for $x$. Round to the nearest tenth.

1. A right triangle has an angle of $35^\circ$. The hypotenuse is $12\,\text{cm}$. Find the length of the side opposite the $35^\circ$ angle. (Hint: Which rule uses Opposite and Hypotenuse?)

2. A right triangle has an angle of $50^\circ$. The side adjacent to the angle is $8\,\text{m}$. Find the length of the hypotenuse. (Hint: Which rule uses Adjacent and Hypotenuse?)

3. A ramp makes a $15^\circ$ angle with the ground. The horizontal length of the ramp (adjacent side) is $20\,\text{ft}$. How high is the top of the ramp (opposite side)?

4. A kite string is $100\,\text{yd}$ long. The string makes a $40^\circ$ angle with the ground. How far is the kite horizontally from the person holding the string (adjacent side)?

Level 2: Finding Missing Angles (Using Inverse Trig)

For these, find the angle $\theta$ to the nearest whole degree.

5. A triangle has an opposite side of $5\,\text{cm}$ and a hypotenuse of $9\,\text{cm}$. Find the angle $\theta$. (Hint: Look at the SOH part. You need the inverse of Sine.)

6. A wheelchair ramp is $10\,\text{ft}$ long and rises $2\,\text{ft}$ high. What angle does the ramp make with the ground?

7. A ladder is leaning against a wall. The base of the ladder is $4\,\text{ft}$ away from the wall (adjacent). The ladder touches the wall $15\,\text{ft}$ off the ground (opposite). What angle does the ladder make with the ground?

Level 3: Word Problems & Real Life

Draw a picture for these!

8. The Surveyor Problem: A surveyor stands $50\,\text{m}$ away from the base of a building. She measures the angle from the ground to the top of the building to be $62^\circ$. How tall is the building? (You have the Adjacent side and need the Opposite side).

9. The Airplane Problem: A plane takes off at an angle of $10^\circ$. After traveling $3\,\text{mi}$ in the air, what is its altitude (vertical height above the ground)? (You have the Hypotenuse and need the Opposite side).

10. The Shadow Problem: A $25\,\text{ft}$ tall tree casts a shadow on the ground. The angle from the tip of the shadow back up to the top of the tree is $40^\circ$. How long is the shadow? (The tree is the Opposite side; the shadow is the Adjacent side).

11. The Slippery Slope: A road sign says the road ahead has a $12\%$ grade (meaning it rises $12\,\text{ft}$ for every $100\,\text{ft}$ of horizontal distance). What is the angle of the road?

12. The Multi-Step Challenge (Pythagoras + Trig): A right triangle has a hypotenuse of $13\,\text{cm}$ and a base (adjacent side) of $5\,\text{cm}$.

    • a) Use the Pythagorean theorem ($a^2 + b^2 = c^2$) to find the length of the opposite side.

    • b) Now find the angle $\theta$ using your answer from part (a).

Answers

1. $6.9\,\text{cm}$ (Calculation: $12 \times \sin(35^\circ)$)

2. $12.4\,\text{m}$ (Calculation: $\frac{8}{\cos(50^\circ)}$)

3. $5.4\,\text{ft}$ (Calculation: $20 \times \tan(15^\circ)$)

4. $76.6\,\text{yd}$ (Calculation: $100 \times \cos(40^\circ)$)

5. $\theta \approx 34^\circ$ (Calculation: $\sin^{-1}\left(\frac{5}{9}\right)$)

6. $\theta \approx 12^\circ$ (Calculation: $\sin^{-1}\left(\frac{2}{10}\right)$)

7. $\theta \approx 75^\circ$ (Calculation: $\tan^{-1}\left(\frac{15}{4}\right)$)

8. $94.0\,\text{m}$ (Calculation: $50 \times \tan(62^\circ)$)

9. $0.5\,\text{mi}$ (Calculation: $3 \times \sin(10^\circ)$)

10. $29.8\,\text{ft}$ (Calculation: $\frac{25}{\tan(40^\circ)}$)

11. $\theta \approx 6.8^\circ$ (Calculation: $\tan^{-1}\left(\frac{12}{100}\right)$)

12. 
    

• a) $12\,\text{cm}$ ($5^2 + b^2 = 13^2$)

• b) $\theta \approx 67^\circ$ ($\tan^{-1}\left(\frac{12}{5}\right)$)

Calculator Tips

• If you are finding a side, your calculator should just be using the normal $\sin$, $\cos$, or $\tan$ buttons.

• If you are finding an angle, your calculator must be using the inverse buttons (usually accessed by pressing "2nd" or "Shift" + $\sin$, $\cos$, or $\tan$).