Geometry/Chapter 19
Contents
Solving Right Triangles[edit]
In order to solve a right triangle using Trig, a simple acronym is used: SOHCAHTOA; 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 SOHCAHTOA, except you use (opp/hyp), (adj/hyp) and (opp/adj)
Example Problems[edit]
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]
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 similair 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 = 1sin²θ. Easy, right?
Problems for Practice[edit]
Solve these equations using the Pythagorean relationships:
 tan²θ + 1 =
 sec²θ  1 =
 csc²θ  cot²θ =
 sin²θ + cos²θ =
 sec²θ  tan²θ =
 sin²θ  1 =
Verifying More Complex Identities[edit]
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]
 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 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