Використання sin, cos, tan для знаходження сторін трикутника

SOH-CAH-TOA

We are going to use the trigonometric functions of sine, cosine, and tangent only with right triangles (triangles that contain a $90^{\circ}$ angle).

These functions will allow us to figure out the lengths of a triangle's sides.

Recall that the hypotenuse is the side of the triangle across from the right angle.

NOTE: we are going to be using calculators for these types of problems. Since we've got our angles written in degrees ($^{\circ}$), MAKE SURE YOUR CALCULATOR IS IN THE DEGREES MODE, NOT RADIANS.

In order to help remember which equation is which, many people use the abbreviation SOH-CAH-TOA.

SOH: Sine of the angle equals Opposite side length divided by Hypotenuse

CAH: Cosine of the angle equals Adjacent side length divided by Hypotenuse

TOA: Tangent of the angle equals Opposite side length divided by Adjacent side length

In the set of equations above, $\theta$ represents the measurement of the angle. Let's look at an example to see how to use them.

Let's say that I want to figure out the length of the side of the triangle opposite the $30^{\circ}$ angle. This is where we can use trigonometry.

$$\sin(30)=\frac{x}{8}$$

$$\sin(30)=\frac{x}{8}$$ $$.5=\frac{x}{8}$$ $$x=4$$

SOH-CAH-TOA is in the stars!

SOH-CAH-TOA isn't just some silly memory rule to help you pass your math class. It is used in the real world! Below is an example of how it can be used to find the distance between earth and a star (ROSS $154$) around $10$ lightyears away! That means it takes $10$ years for light to travel from earth to the star.

In order to solve for this distance we need to know two things:

1) The Parallax Angle ($\theta$)

2) The Distance from Earth to the Sun ($1$ AU)

Parallax Angle: half the angle the star has been observed to move in six months (from winter to summer).

1 Astronomical Unit (AU): the distance from the Earth to the Sun

The Parallax Angle for our star of interest, ROSS 154 is $9.3533x10^{-5}$ (ugly! right?)

Alright! Now we can use what we know about the SOH-CAH-TOA rule to find the distance to the star ROSS $154$ in Astronomical Units. Spoiler: it's super far away!

Recall:

$$\sin(\theta) = \frac{\text{Opposite}}{\text{Hypotenuse}}$$ $$\theta = 9.3533x10^{-5}$$

Plug in your theta (the parallax angle)

$$\sin (9.3533x10^{-5}) = \frac{1AU}{\text{Distance to star}}$$

Solve for the Distance to star

$$\text{Distance to star} = \frac{1AU}{\sin (9.3533x10^{-5}}$$ $$\text{Distance to star} = 612,572.8835 AU$$

So our star is about $612,572$ farther from us than the sun!

So, you’ve been cruising along through Algebra 1. You’ve learned everything from factoring polynomials to dealing with inequalities, with your trusty graphing calculator at your side the entire time. But now that you’re starting to delve into a bit of geometry and trigonometry, you’ll be seeing some fancy terms for the first time. What’s a sine? Does a cosine just coexist with a sine or something? What in the world is going on?

We’ll get to all of the juicy details soon, but let’s establish some basics first. The three trigonometric functions you’ll need to know for now are sine, cosine, and tangent. But before we proceed, we’ll have to give some names to the sides of this right triangle with angle $\theta$.

Now that we’ve got some names to go with the sides of that right triangle, let’s finally find out what our trigonometric functions are! Essentially, they’re ratios of the sides of a right triangle, as seen in the following image:

We’re just dividing different side lengths by each other. Not too hard at all!

The mnemonic SOH-CAH-TOA is used to remember these ratios -- sine is opposite over hypotenuse, cosine is adjacent over hypotenuse, and tangent is opposite over adjacent.

Here’s some food for thought: if we label the other acute angle of the right triangle $\beta$, we can also write out the trigonometric functions in terms of $\beta$.

Now that you know your trigonometric functions, let's try using them in the context of some real number crunching. You've got this down pat, so let's finish on a high note!

Now you know your trig functions! Those sines, cosines, and tangents are now in your arsenal of Algebra 1 skills. Trigonometry is incredibly important both in future math classes and in the real world, so good luck with your future mathematical endeavors!

(Video) Basic Trigonometry to Work out Unknown Sides

by tecmath

In this video by tecmath it is shown how to find the missing sides of a triangle using trigonometric functions.

$\large{\text{Summary}}$

Trigonometry is not as scary as it sounds! To solve these problems, you need to be able to label the sides of a right triangle based on the angle theta.

The steps for solving these problems are:

• Label sides
• Work out trig function
• Substitute values
• Solve

(Video) Special Triangles

by MySecretMathTutor

This video by MySecretMathTutor explores helpful strategies in applying SOH-CAH-TOA to special triangles.

$\large{\text{Summary}}$

Special triangles are triangles with specific sets of angles that form consistent relationships with its sides.

An implementation of the approach discussed in the video is finding these patterns within shapes such as squares and equilateral triangles. Though these don’t appear to have angles that work within the $30-60-90$ or $45-45-90$ agenda, these shapes can be divided to facilitate the rules of the special triangles.

For example, consider a square with sides “$x$” and an equilateral triangle with sides “$z$.”

A square contains all $90$ degree angles, but if we sketch in a line that bisects angles in opposite corners, we produce two $45-45-90$ triangles within the square. Since we know all of the square’s sides are the same, both sides opposite of the $45$ degree angles remain as “$x$.” To find the hypotenuse, we simply apply the formula of the special triangle that was explained in the video above, in which the hypotenuse is $x\sqrt{2}$.

By dropping a line that bisects the uppermost $60$-degree angle within the equilateral triangle and is perpendicular to the base, we can produce a similar problem as the one above. This line creates two $30-60-90$ triangles. The hypotenuse, in this case, remains as $z$, and the base, due to its bisection, becomes $\frac{z}{2}$. The bisecting line, based on the pattern of the $30-60-90$ triangle (which uses the Pythagorean theorem), comes out to length $\frac{x\sqrt{3}}{2}$.