Solving Right Triangles and Applications of Trigonometry

Solving Right Triangles

Finding Missing Side Lengths

Solving right triangles is a fundamental topic in precalculus, involving the determination of unknown side lengths or angles using trigonometric ratios and the Pythagorean Theorem.

• Pythagorean Theorem: For a right triangle with legs a and b, and hypotenuse c: $a^2 + b^2 = c^2$

• Trigonometric Ratios:

  • Sine: $\sin(\theta) = \frac{\text{opposite}}{\text{hypotenuse}}$

  • Cosine: $\cos(\theta) = \frac{\text{adjacent}}{\text{hypotenuse}}$

  • Tangent: $\tan(\theta) = \frac{\text{opposite}}{\text{adjacent}}$

• SOH-CAH-TOA: Mnemonic for remembering the trigonometric ratios.

Example: Given a right triangle with one side and one angle, use the appropriate trigonometric ratio to solve for the missing side. Step-by-step:

1. Identify the given side and angle.

2. Choose the correct trigonometric ratio.

3. Solve for the unknown side.

4. If necessary, use the Pythagorean Theorem to find the third side.

Finding Missing Angles

If two side lengths are given, the missing angles can be found using inverse trigonometric functions.

• Inverse Trigonometric Functions:

  • $\theta = \sin^{-1}\left(\frac{\text{opposite}}{\text{hypotenuse}}\right)$

  • $\theta = \cos^{-1}\left(\frac{\text{adjacent}}{\text{hypotenuse}}\right)$

  • $\theta = \tan^{-1}\left(\frac{\text{opposite}}{\text{adjacent}}\right)$

• Sum of angles in a triangle: $90^\circ + x + y = 180^\circ$ (for right triangles, one angle is always $90^\circ$)

Example: Given a triangle with sides 3, 4, and 5, find the missing angles using inverse trigonometric functions. Practice: Given a right triangle with sides 3 and 4, calculate all missing angles in degrees (rounded to 3 decimal places). Solution:

• $x = \tan^{-1}\left(\frac{3}{4}\right) \approx 36.870^\circ$

• $y = \tan^{-1}\left(\frac{4}{3}\right) \approx 53.130^\circ$

Applications of Trigonometry

Word Problems Involving Right Triangles

Trigonometry is widely used to solve real-world problems involving heights, distances, and angles of elevation or depression.

• Angle of Elevation: The angle formed by the line of sight above the horizontal when looking up at an object.

• Angle of Depression: The angle formed by the line of sight below the horizontal when looking down at an object.

Example: The Grand Lighthouse problem: Given the height of the lighthouse and the distance from the shore, use trigonometric ratios to determine the distance to a boat based on the angle of elevation. Solution:

• Set up a right triangle with the known height and distance.

• Use $\tan(\theta) = \frac{\text{opposite}}{\text{adjacent}}$ to solve for the unknown distance.

Inclined Paths and Elevation Changes

Problems involving hiking paths or ramps often require finding the angle of inclination or the length of the path using trigonometric ratios.

Example: Hiking Path problem: Given the elevation change and the horizontal distance, determine the angle of inclination. Solution:

• Use $\tan(\theta) = \frac{\text{vertical change}}{\text{horizontal distance}}$

• Solve for $\theta$ using the inverse tangent function.

Trigonometric Cofunctions

Cofunctions of Complementary Angles

Cofunctions are pairs of trigonometric functions that are equal when their angles are complementary (sum to $90^\circ$).

• Sine and Cosine: $\sin(\theta) = \cos(90^\circ - \theta)$

• Tangent and Cotangent: $\tan(\theta) = \cot(90^\circ - \theta)$

• Secant and Cosecant: $\sec(\theta) = \csc(90^\circ - \theta)$

Example: Given a right triangle with a $31^\circ$ angle and hypotenuse of 8.0, calculate the side opposite the $31^\circ$ angle and the side adjacent to the $31^\circ$ angle. Solution:

• Opposite: $y = 8.0 \times \sin(31^\circ) \approx 4.129$

• Adjacent: $x = 8.0 \times \cos(31^\circ) \approx 6.860$

Summary Table: Trigonometric Ratios

Function	Ratio	Inverse
Sine	$\sin(\theta) = \frac{\text{opposite}}{\text{hypotenuse}}$	$\theta = \sin^{-1}\left(\frac{\text{opposite}}{\text{hypotenuse}}\right)$
Cosine	$\cos(\theta) = \frac{\text{adjacent}}{\text{hypotenuse}}$	$\theta = \cos^{-1}\left(\frac{\text{adjacent}}{\text{hypotenuse}}\right)$
Tangent	$\tan(\theta) = \frac{\text{opposite}}{\text{adjacent}}$	$\theta = \tan^{-1}\left(\frac{\text{opposite}}{\text{adjacent}}\right)$

Additional info: These notes cover key precalculus concepts from Chapter 6 (The Trigonometric Functions) and Chapter 8 (Applications of Trigonometry), including solving right triangles, trigonometric ratios, cofunctions, and real-world applications.