Right Triangle Trigonometry: Definitions and Applications

Introduction to Right Triangle Trigonometry

This video introduces right triangle trigonometry, covering the definitions of the six trigonometric functions, methods for finding their values, and related identities. The primary goal is to establish a foundational understanding for future applications in real-world scenarios.

Real-World Applications of Right Triangle Geometry

Triangle geometry plays a crucial role in various engineering and architectural applications, particularly in structural design:

• Suspension Bridges: The guy wires in suspension bridges are arranged in triangular forms.
• Truss Bridges: The beams of a truss bridge are welded together to form triangles.

Triangles are preferentially used in construction because they excel at supporting forces, such as the weight of automobiles and the bridge structure itself. This inherent stability makes them ideal for bearing loads.

Fundamentals of Right Triangles

Before delving into trigonometric functions, it's essential to recall the basic properties of right triangles:

• Definition: A triangle is classified as a right triangle if it contains one right angle, which measures exactly $90^ ext{o}$.
• Sides of a Right Triangle:
  • Hypotenuse: This is the side directly opposite the right angle and is always the longest side of the triangle. Its length is denoted as $c$ units.
  • Legs: The remaining two sides are called legs. Their lengths are denoted as $a$ units and $b$ units.
• Units of Measure: The side lengths $(a, b, c)$ represent measures of length and can be expressed in various units such as feet, inches, meters, miles, or kilometers, depending on the specific application.
• Pythagorean Theorem: This fundamental theorem states that the square of the length of the hypotenuse ($c$) is equal to the sum of the squares of the lengths of the two legs ($a$ and $b$).
  • Formula: $c^2 = a^2 + b^2$

Constructing a Right Triangle from an Angle

To define trigonometric functions, we often construct a right triangle using an angle $heta$:

• Angle Placement: Imagine an angle $heta$ (theta) drawn in the coordinate plane. Its initial side lies along the positive x-axis, and its terminal side extends into Quadrant I.
• Acute Angle: Since the terminal side is in Quadrant I, $heta$ is an acute angle, meaning its measure is between $0^ ext{o}$ and $90^ ext{o}$, or between $0$ and $rac{\pi}{2}$ radians.
• Triangle Formation:
  • The hypotenuse of the right triangle lies along the terminal side of $heta$ and has a length of $c$ units.
  • One leg of length $a$ units lies on the initial side of $heta$. This leg is adjacent to $heta$.
  • The other leg of length $b$ units is drawn perpendicular to the initial side, forming the right angle. This leg is opposite to $heta$.

Ratios of Sides and Similar Triangles

From the three sides ($a, b, c$) of a right triangle, six distinct ratios can be formed:

1. $b/c$
2. $a/c$
3. $b/a$
4. $c/b$
5. $c/a$
6. $a/b$

These ratios are not arbitrary; they possess a crucial property related to similar triangles:

• Definition of Similar Triangles: Two triangles are similar if their corresponding angles have equal measures.
• Property of Similar Triangles: When triangles are similar, the ratios of their corresponding sides are equivalent, regardless of their absolute size.
• Demonstration: Consider two right triangles that share an acute angle $heta$. Since both possess a $90^ ext{o}$ angle and share $heta$, their third angles must also be equal. Therefore, the triangles are similar.
  • If one triangle has sides $a', b', c'$ and a larger similar triangle has sides $a, b, c$, then:
    • rac{b}{c} = rac{b'}{c'}
    • rac{a}{c} = rac{a'}{c'}
    • rac{b}{a} = rac{b'}{a'}
    • rac{c}{b} = rac{c'}{b'}
    • rac{c}{a} = rac{c'}{a'}
    • rac{a}{b} = rac{a'}{b'}
• Key Implication: These ratios depend only on the measure of the angle $heta$ and not on the specific size of the right triangle. This property allows us to define the trigonometric functions.

Definition of the Six Trigonometric Functions

Based on the ratios of the sides of a right triangle relative to an acute angle $heta$, the six trigonometric functions are defined:

Let:

• Opposite (b): The length of the leg opposite angle $heta$.
• Adjacent (a): The length of the leg adjacent to angle $heta$.
• Hypotenuse (c): The length of the hypotenuse.

1. Sine (sin) of $heta$:

   • Spelling: S-I-N-E
   • Abbreviation: $ext{sin} heta$
   • Ratio: rac{ ext{Opposite}}{ ext{Hypotenuse}} = rac{b}{c}

2. Cosine (cos) of $heta$:

   • Spelling: C-O-S-I-N-E
   • Abbreviation: $ext{cos} heta$
   • Ratio: rac{ ext{Adjacent}}{ ext{Hypotenuse}} = rac{a}{c}

3. Tangent (tan) of $heta$:

   • Spelling: T-A-N-G-E-N-T
   • Abbreviation: $ext{tan} heta$
   • Ratio: rac{ ext{Opposite}}{ ext{Adjacent}} = rac{b}{a}

4. Cosecant (csc) of $heta$:

   • Spelling: C-O-S-E-C-A-N-T
   • Abbreviation: $ext{csc} heta$
   • Ratio: rac{ ext{Hypotenuse}}{ ext{Opposite}} = rac{c}{b}

5. Secant (sec) of $heta$:

   • Spelling: S-E-C-A-N-T
   • Abbreviation: $ext{sec} heta$
   • Ratio: rac{ ext{Hypotenuse}}{ ext{Adjacent}} = rac{c}{a}

6. Cotangent (cot) of $heta$:

   • Spelling: C-O-T-A-N-G-E-N-T
   • Abbreviation: $ext{cot} heta$
   • Ratio: rac{ ext{Adjacent}}{ ext{Opposite}} = rac{a}{b}

Important Note: For an acute angle $heta$ (between $0^ ext{o}$ and $90^ ext{o}$), the values of all six trigonometric functions are always positive.

Example: Finding Trigonometric Function Values

Problem: Given a right triangle where the side opposite $heta$ is $2$ units and the side adjacent to $heta$ is $\sqrt{3}$ units, find the values of all six trigonometric functions of $heta$.

Solution Steps:

1. Identify Known Sides:

   • Opposite leg ($b$) $= 2$
   • Adjacent leg ($a$) $= \sqrt{3}$
   • The hypotenuse is opposite the right angle, which is the longest side.

2. Find the Hypotenuse (c) using the Pythagorean Theorem:

   • $a^2 + b^2 = c^2$
   • $(\sqrt{3})^2 + (2)^2 = c^2$
   • $3 + 4 = c^2$
   • $7 = c^2$
   • $c = \sqrt{7}$ (We take the positive root since length must be positive).

3. Calculate Each Trigonometric Function:

   • $\text{sin } \theta = \frac{\text{Opposite}}{\text{Hypotenuse}} = \frac{2}{\sqrt{7}}$ (Rationalize the denominator by multiplying by $rac{\sqrt{7}}{\sqrt{7}}$) $= \frac{2\sqrt{7}}{7}$
   • $\text{cos } \theta = \frac{\text{Adjacent}}{\text{Hypotenuse}} = \frac{\sqrt{3}}{\sqrt{7}}$ (Rationalize) $= \frac{\sqrt{3 \cdot 7}}{7} = \frac{\sqrt{21}}{7}$
   • $\text{tan } \theta = \frac{\text{Opposite}}{\text{Adjacent}} = \frac{2}{\sqrt{3}}$ (Rationalize) $= \frac{2\sqrt{3}}{3}$
   • $\text{csc } \theta = \frac{\text{Hypotenuse}}{\text{Opposite}} = \frac{\sqrt{7}}{2}$
   • $\text{sec } \theta = \frac{\text{Hypotenuse}}{\text{Adjacent}} = \frac{\sqrt{7}}{\sqrt{3}}$ (Rationalize) $= \frac{\sqrt{7 \cdot 3}}{3} = \frac{\sqrt{21}}{3}$
   • $\text{cot } \theta = \frac{\text{Adjacent}}{\text{Opposite}} = \frac{\sqrt{3}}{2}$

Summary of Steps for Finding Trigonometric Values

When given two sides of a right triangle and needing to find the values of the trigonometric functions, follow these four steps:

1. Draw a Right Triangle: Sketch the triangle, clearly indicating the acute angle $heta$. This helps in visualizing which sides are opposite, adjacent, or the hypotenuse relative to $heta$.
2. Label Known Sides: Write down the given lengths for the two known sides on your diagram.
3. Find the Third Side: Utilize the Pythagorean theorem ($a^2 + b^2 = c^2$) to calculate the length of the unknown third side.
4. Apply Ratio Definitions: Use the definitions of sine, cosine, tangent, cosecant, secant, and cotangent (Opposite/Hypotenuse, Adjacent/Hypotenuse, etc.) to determine the value of each function for angle $heta$. Remember to rationalize denominators where necessary.

Application Example: Friction Force on a Ramp

Problem: A box weighing $20$ pounds slides down a ramp that is $10$ feet long and $3$ feet high. The force due to friction is calculated using the formula $F = W \text{sin } \theta$, where $W$ is the weight of the box in pounds and $heta$ is the angle of inclination of the ramp. Find the friction force (in pounds), rounded to the nearest tenth.

Solution Steps:

1. Draw and Label the Right Triangle:

   • The length of the ramp is the hypotenuse ($c$) $= 10$ feet.
   • The height of the ramp is the vertical leg ($b$) $= 3$ feet.
   • The horizontal leg represents the ground ($a$) $= b_{\text{horizontal}}$. We'll identify $heta$ as the angle between the horizontal ground and the ramp (hypotenuse). Thus, the vertical leg (3 ft) is opposite $heta$.

2. Find the Length of the Missing Side (Horizontal Leg):

   • Using the Pythagorean theorem: $a^2 + b^2 = c^2$
   • $(b_{\text{horizontal}})^2 + (3)^2 = (10)^2$
   • $(b_{\text{horizontal}})^2 + 9 = 100$
   • $(b_{\text{horizontal}})^2 = 100 - 9$
   • $(b_{\text{horizontal}})^2 = 91$
   • $b_{\text{horizontal}} = \sqrt{91}$ feet (We only need the positive root for physical length).

3. Calculate $\text{sin } \theta$:

   • For the angle $heta$ (at the base of the ramp), the side opposite it is the height (3 feet), and the hypotenuse is the ramp length (10 feet).
   • $\text{sin } \theta = \frac{\text{Opposite}}{\text{Hypotenuse}} = \frac{3}{10}$

4. Calculate the Friction Force (F):

   • The given formula is $F = W \text{sin } \theta$
   • Substitute the weight ($W = 20$ lbs) and the calculated $\text{sin } \theta$
   • $F = 20 \text{ lbs} \times \frac{3}{10}$
   • $F = \frac{60}{10}$
   • $F = 6$ pounds

The friction force is $6$ pounds.