# Geometry/Chapter 19

## Solving Right Triangles[edit | edit source]

In order to solve a right triangle using Trig, a simple acronym is used: SOH-CAH-TOA; SOH, standing for sine (opposite/hypotneuse), CAH, standing for cosine (adjacent/hypotneuse), and TOA, standing for tangent (opposite/adjacent).

This work is mainly done using a calculator, as there are no simple, analytical, formulas for the sine, cosine, and tangent functions.

To solve for angles of a right triangle, you would use the same method of SOH-CAH-TOA, except you use (opp/hyp), (adj/hyp) and (opp/adj)

### Example Problems[edit | edit source]

Example One: Finding the Missing Parts of a Right Triangle

Find the missing angle and the sides in a right triangle with an acute angle of 38 degrees and an hypotenuse of 15 meters.

Solution:

- The other acute angle can be found by realizing that the sum of the angles of a triangle is always 180 degrees. Therefore,
- so . Using SOHCAHTOA, we see that
- = opposite side length / Hypotenuse. Thus, meters. To find the adjacent side
- we use the cosine function and the formula meters.

## Pythagorean Relationships[edit | edit source]

This is one step beyond the basic trig functions or ideas. It involves three equations that can be manipulated to suit the needs of the given problem. The three equations are as follows:

These can be easily manipulated to figure out a long range of complex problems. For this, we will use the example of 1 - sin²θ. When we look at the above equations, we see that it is very similar to the first one (sin²θ + cos²θ = 1). In fact, sin²θ has just been subtracted. Using this principle, we can thus solve the problem as cos²θ = 1-sin²θ. Easy, right?

### Problems for Practice[edit | edit source]

Solve these equations using the Pythagorean relationships:

- tan²θ + 1 =
- sec²θ - 1 =
- csc²θ - cot²θ =
- sin²θ + cos²θ =
- sec²θ - tan²θ =
- sin²θ - 1 =

## Verifying More Complex Identities[edit | edit source]

We now move on to the more complex functions of the trigonometric world. To solve or prove these functions or equations, we have to use all of the math skills we learned before. We will need to use factoring, the Pythagorean relationships table, the inverse trig functions (sec θ = 1/cos θ, csc θ = 1/sin θ, cot θ = 1/tan θ ) and anything else we need to solve the problem.

This is very advanced math. To start this lesson off, lets look at an example:

Prove (make the right side look like the left without touching the left)

- sin θ /(sin θ + cos θ) = tan θ /(1 + tan θ)

We can see that tan θ is being divided by 1 + tan θ. To start the problem off, let’s get common denominators:

- tan θ /(cos θ/cos θ + sin θ/cos θ)

We can now combine the two equations on the bottom, so it looks like this:

- tan θ /((cos θ + sin θ)/cos θ)

To help simplify this down, put tan θ into terms of sin θ and cos θ:

- (sin θ/cos θ)/((cos θ + sin θ)/cos θ)

We can now multiply by the reciprocal to get rid of the bottom denominator:

- ((sin θ/cos θ)*cos θ)/(((cos θ + sin θ)/cos θ)*cos θ)

Note that this is still being divided by sin θ + cos θ. When we multiply, the cos θ ’s will cancel each other out, so we are left with:

- sin θ/(sin θ + cos θ)

Which just happens to be the answer we were trying to prove!

While this may seem complicated, you only need to practice this more. Let's get an example that is a little more complicated:

Prove (make the left side look like the right without touching the right side)

- (2 sin θ cos θ)/(sin²θ - cos²θ + 1) = cot θ

When we look at an equation like this, it may seem impossible. Lets go through the basics. When we look at the bottom, we see that we have a -cos²θ and a +1. When we look at the Pythagorean relationships table, 1 - cos²θ is equal to sin²θ:

- (2 cos θ sin θ)/(sin²θ + sin²θ)

We can then add the two sin²θ together:

- (2 sin θ cos θ)/(2 sin²θ)

We can then simplify this further into:

- cos θ / sin θ

We can do this because sin θ will cross out to have a single sin θ (2/2²=1/2, right?) and then the twos will cross out. That leaves with cos θ / sin θ, or cot θ !

### Problems for Practice[edit | edit source]

- Prove (make the right side look like the left without touching the left side)
- (1 - sin θ)/(cos θ) = cos θ /(1 + sin θ)

- Prove (make the left side look like the right without touching the right side)
- (cot θ - tan θ)/(tan θ cos²θ) = csc²θ - sec²θ

- Prove (make the left side look like the right without touching the right side)
- tan θ/(1 + sec θ) + (1 + sec θ)/tan θ = 2 csc θ

- Prove (make the left side look like the right without touching the right side)
- tan²θ/(1 + tan²θ) = sin²θ

- Geometry Main Page
- Motivation
- Introduction
- Geometry/Chapter 1 - HS Definitions and Reasoning (Introduction)
- Geometry/Chapter 1/Lesson 1 Introduction
- Geometry/Chapter 1/Lesson 2 Reasoning
- Geometry/Chapter 1/Lesson 3 Undefined Terms
- Geometry/Chapter 1/Lesson 4 Axioms/Postulates
- Geometry/Chapter 1/Lesson 5 Theorems
- Geometry/Chapter 1/Vocabulary Vocabulary

- Geometry/Chapter 2 Proofs
- Geometry/Chapter 3 Logical Arguments
- Geometry/Chapter 4 Congruence and Similarity
- Geometry/Chapter 5 Triangle: Congruence and Similiarity
- Geometry/Chapter 6 Triangle: Inequality Theorem
- Geometry/Chapter 7 Parallel Lines, Quadrilaterals, and Circles
- Geometry/Chapter 8 Perimeters, Areas, Volumes
- Geometry/Chapter 9 Prisms, Pyramids, Spheres
- Geometry/Chapter 10 Polygons
- Geometry/Chapter 11
- Geometry/Chapter 12 Angles: Interior and Exterior
- Geometry/Chapter 13 Angles: Complementary, Supplementary, Vertical
- Geometry/Chapter 14 Pythagorean Theorem: Proof
- Geometry/Chapter 15 Pythagorean Theorem: Distance and Triangles
- Geometry/Chapter 16 Constructions
- Geometry/Chapter 17 Coordinate Geometry
- Geometry/Chapter 18 Trigonometry
- Geometry/Chapter 19 Trigonometry: Solving Triangles
- Geometry/Chapter 20 Special Right Triangles
- Geometry/Chapter 21 Chords, Secants, Tangents, Inscribed Angles, Circumscribed Angles
- Geometry/Chapter 22 Rigid Motion
- Geometry/Appendix A Formulae
- Geometry/Appendix B Answers to problems
- Appendix C. Geometry/Postulates & Definitions
- Appendix D. Geometry/The SMSG Postulates for Euclidean Geometry