# Trigonometry and Right Angles

##### - Trigonometry Angles

Using all the tricks in the proverbial math bag that have been gathered in the course of this series on trigonometry, it is possible to determine what the measures of the sides and angles of a right triangle trigonometrically; *without measuring them*. This article will demonstrate how to solve a triangle but first it will start with a question. Given a right triangle, where one angle is 90° and another angle $ \alpha $, which must be $ 0° \le \alpha \le 90° $: what is the value of $ \alpha $ if the $ \text{sin}\ \alpha = \frac{\sqrt{2}}{2} $?

## It’s Simple, Memorize It

The answer is simple, just look at this chart or memorize and it and bam: the answer is the angle $ \alpha $ must be 45°. Simple, but it can be confusing to the novice, or at least it was to this author. Most explainations tend to boil down to: since $ \text{sin ANSWER°} = \text{some value} $, then $ \alpha $ is ANSWER°, it’s like using a word to define itself. Now, that said, after time one does start to memorize these values which does make solving certain problems go much quicker; however, it is worth noting that sometimes it is easy to forget that novices do not have the experience, and therefore don’t recall these values as immediately as those with more time with them.

$\alpha$ | $ sin\alpha $ | $ cos\alpha $ | $ tan\alpha $ |
---|---|---|---|

30° | $ \frac{1}{2} $ | $ \frac{\sqrt{3}}{2} $ | $ \frac{\sqrt{3}}{3} $ |

45° | $ \frac{\sqrt{2}}{2} $ | $ \frac{\sqrt{2}}{2} $ | $ 1 $ |

60° | $ \frac{\sqrt{3}}{2} $ | $ \frac{1}{2} $ | $ \sqrt{3} $ |

Avoiding the easy road, how then, can the angle $ \alpha $ be found? The answer is unfortunately a mystery that will remain unsolved until Calculus. In the meantime, another way to remember these values is to recall the 45-45-90 and 30-60-90 triangles from the last article. Then, using the trigonometric ratios, the angles can be determined visually.

## Inverse Functions

For a function to be a function, any given input value can have only one output value. However, for angles, they can keep going round and round the unit circle adding $ 2\pi $ every time, while looking very much like the same angles. In order to avoid violating this aspect of functions, the angles must be restricted. In this case, only $ \text{sin} $, $ \text{cos} $, and $ \text{tan} $ apply to right angles, so the other trigonometric functions can be ignored.

- $ \text{sin}^{-1}(x) = \alpha $ provided $ \text{sin}\ \alpha = x $ and $ -90° \le \alpha \le 90° $
- $ \text{cos}^{-1}(x) = \alpha $ provided $ \text{cos}\ \alpha = x $ and $ ° \le \alpha \le 180° $
- $ \text{tan}^{-1}(x) = \alpha $ provided $ \text{tan}\ \alpha = x $ and $ -90° \lt \alpha \lt 90° $

## Right Triangles

In the last article in this series, the trigonometric functions were defined in terms of angles in standard position on a coordinate system. What about angles that aren’t in standard position? This section will look at how to solve the trigonometric function for a subset of angles, *the acute angles of a right triangle*.

Solving the trigonometric functions for a triangle previously depended on the $ \text{(x,y)} $ point on the terminal side of an angle, where the length from the origin to $ \text{(x,y)} $ was $ r $. Instead, we can think of these sides in terms of their relation to the angle in question, in this case $ \alpha $. The $ x $ and $ y $ values now represent the **adjacent** and **opposite** sides of the angle $ \alpha $. The length $ r $ is always opposite of the 90 angle and is known as the **hypotenuse**. The trigonometric ratios can now be reimagined in these relationships.

$ \text{sin}\ \alpha = \frac{\text{opp}}{\text{hyp}} $ | $ \text{cos}\ \alpha = \frac{adj}{hyp} $ | $ \text{tan}\ \alpha = \frac{opp}{adj} $ |

$ \text{csc}\ \alpha = \frac{hyp}{opp} $ | $ \text{sec}\ \alpha = \frac{hyp}{adj} $ | $ \text{cot}\ \alpha = \frac{adj}{opp} $ |

In right angles the two acute angles are generally labeled alpha ($ \alpha $) and beta ($ \beta $), with their opposite sides labeled $ a $ and $ b $, respectively. The side across from the 90 angle, gamma ($ \gamma $), is $ c $.

## Solving Right Triangles

Using all the tools in the triogonometric back learned so far, its time to solve a right triangle. In this case, the triangle will be the one in *Figure 2*. Given that $ c = 10 $ and one of the angles that isn’t 90° is 45°: what is the degree of the other angle and the length of $ a $ and $ b $?

First, finding the 3rd angle is simple. In this case, it might be easy to recall that when one angle of a triangle is 45°, so is the other. However, the sure way to find this for any triangle is to subtract 90° and 45° from 180°, the total of all three angles for all triangles, leaving the third angle.

\[\beta = 180° - 90° - 45° = 45°\]Now that all three angles are known, its time to find the length of $ a $ and $ b $, starting with side $ a $. To solve this problem, look at what information is available. In this case, the angles are known, as well as the hypotenuse. The trigonometric functions provide a ratio between the different sides of the triangle and one of the acute angles. Which function is the right one? The answer is simple: *its the one that includes the given and wanted information*. In this case, given the information that the hypotenuse is $ 10 $ and that the wanted information is $ a $, which trigonometric function satisifies these requirements?

In this case, the sine includes both the hypotenuse and opposite sides. Take a look at this in equation form, then solving for $ a $:

\[\text{sin}\ 45° = \frac{a}{10}\] \[10 * \text{sin}\ 45° = \frac{a}{\cancel{10}} * \cancel{10}\] \[a = 10 * \text{sin}\ 45°\]So the sine provides a solution for the side $ a $, but since this is a special type of triangle, the 45-45-90, it is possible to get a more simplified and exact answer. To do so, simply replace $ \text{sin}\ 45° $ with the value previously established for this trigonometric function at the beginning of the article, $ \text{sin}\ 45° = \frac{\sqrt{2}}{2} $:

\[a = 10 * \frac{2\sqrt{2}}{2} = 10\sqrt{2}\]So the exact value of $ a $ is $ 10\sqrt{2} $. In this case, finding $ b $ is simple, knowing that both sides that are not the hypotenuse in a 45-45-90 triangle are the same length means that $ b $ is also $ 10\sqrt{2} $. However, for other types of triangles $ b $ can be found in much the same way, either by finding the sine of angle $ \beta $ or with the cosine of $ \alpha $. The key is to remember that the right function to use is the one that includes the given and wanted information.

## Conclusion

This article tied information from the previous ones to show how the sides and angles of a triangle can be found without directly measuring them all. It turns out that this technique can be applied in many ways. One of the most well known applications is finding the height of exceptional tall objects that would otherwise be too tall to measure. How is this accomplished? By simply creating a triangle between the base of the object, its highest point, and the tip of the shadow it casts. Measuring the length of the shadow on the ground and the angle at which the light source must be focused to create it are much easier tasks than climbing the height of the measured object.

The next article will take a quick look at the Fundamental Identity and how Reference Angles can make measuring other angles a simpler task.