Chapter+10

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 * Chapter 10-Applications of Trigonometric Functions **

Example one : Solving a right triangle  ​ What you already know: On a right triangle,  Sine = Opposite / Hypotenuse Cosine = Adjacent / Hypotenuse Tangent = Opposite / Adjacent 

In this chapter, you will use these facts to algebraically solve and find the missing length or angles on a triangle.

Example: If b = 2,​and Alpha (α) = 40 degrees, find β, a, and c.

We can find the missing sides using Sine, Cosine, or Tangent. First we have 2, and alpha is forty. We also know that Tan (θ) = Opposite / adjacent. If we plug in the numbers they gave use, we get:

// I did tangent of the angle 40, which is a/B. //

// Multiply both sides by 2 //

Now we need to find the length of c, which is the exact same process. Instead of tangent, I’m going to use Cos 40. If possible, use the given numbers in case you make a mistake I your calculations.

// Multiply by c //

// Divide by Cos 40 //

To find β, we know that α + β = 90, on a right triangle. 40 + β = 90, β = 50

Example two: Solving a right triangle.

Solve the following right triangle:



We know that a = 3, and b =2. Since it is a right triangle, we can use the Pythagorean Theorem.

Now we have all the sides, and all we need to do is find the angles. To find the Angles, we need to use the inverse functions like , Both functions have a restricted domain. Sin-1 is between 0 and 2 pi, and Cos-1 is between pi and negative pi, and tan-1 is between 0 and pi.

Use the inverse tan on both sides.

Solve. To find β we know α + β = 90, so 90-56.3 = β β = 33.7 Example Three: Angle of Elevation and Depression

Picture yourself walking down the side walk. You look up to see a bird at the top of a building. Whenever you look up or the hypotenuse is facing up, that is an angle of elevation. Now let’s say you’re the bird and you’re looking down at a person. That would be angle of depression, because the hypotenuse is facing downward.

You would solve these in pretty much the same way, but the triangle may be flipped around or in different units. **MAKE SURE TO CHECK YOUR UNITS, BEFORE AND AFTER YOU START.**


 * = 10.2 - Law of Sines =

Law of Sines: Sin (**α** )/A = Sin ( β ) / B = Sin ( γ ) / C The Law of Sines can be used to find missing sides and angles on any kind of triangle when given certain types of data. the types of data are as follow:

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 * 1) SAA - Side-Angle-Angle
 * 2) ASA - Angle-Side-Angle
 * 3) SSA - Side-Side-Angle (This is called the ambiguous case, and has multiple scenarios.)


 * SAA

For this triangle we are given A, Alpha, and Gamma. First, let's find out what Beta is.

ß = 180 - (15 + 32) ß = 180 - 47 ß = 133°

Now that we know Beta, we can find B using A and Alpha.

Sin 32/5 = Sin 133/B B(sin 32) = 5 sin 133 B = 5 sin 133/sin 32 //(When entering this in a calculator, do 5 sin 133 and sin 32 separately)// B = 6.9

Using Gamma, we can also find C.

Sin 32/5 = Sin 15/C C sin32 = 5 sin 15 C = 5 sin 15/sin32 C = 2.4

We now haw a complete triangle.

== Once again, we are going to find our missing angle, wh ich in this case is Alpha. α = 180 - (36 + 47) α = 180 - 83 α = 97 Now that we have alpha, we can start to solve the triangle using the same method we did before.
 * ASA

Let's start solving for b.

Sin 97/7 = Sin 47/b bSin 97 = 7Sin 47 b = 7Sin 47 / Sin 97 b = 5.2

Now we will start solving for c.

Sin 97/7 = Sin 36/c cSin 97 = 7Sin 36 c = 7Sin 36/ Sin 97 c = 4.1

We now have a solved triangle.


 * SSA

This situation is a little bit tricky. Because we only have one fixed angle (ß) and one free length (c), there is the possibility for one, two, or no triangles. The key to finding out what kind of situation we have is the height of the triangle. The situations are as follows:


 * 1) No triangle if b < h
 * 2) Right triangle if b = h
 * 3) One obtuse triangle if b > h and b > a
 * 4) Two triangles if b > h and b < a

For cases two, we can use our formula from 10.1 and for three, we can solve the triangle like normal. For case one, we do not need to solve at all. And for case four, we need to solve each triangle independently, except for one of the angles, which I'll describe later.

To find the height, we can split this triangle into two right triangles.

At this point, we can use basic trigonometry to find the height. Since we are given an angle and the Hypotenuse, we can use Sine of 50.

Sin 50 = h/8 8Sin 50= h h = 6.1

Since b > h and is > a, we have two triangles, one with a swept outside, and one with a swept inside. First, let's solve the swept out triangle. This triangle can be solved like a regular triangle. Since there are two triangles, I will be following all variables with the number 1, and the second triangle with the number 2

Sin 50/7 = Sin ß1/8 8Sin 50 /7 = Sin ß1 ß1 = 61

 γ1 = 180 - (50 + 61) γ1 = 69

Sin 50/7 = Sin 69 /C1 C1 = 7Sin 69 /Sin 50 C1 = 8.5

Now that we have that triangle solved, we can start solving the other one. Since the angles of ß between the two triangles are supplementary, we can find ß2 by using the following:

ß2 = 180 - ß1 ß2 = 180 - 61 ß2 = 119

Now that we know both α2 and ß2, we can easily find γ2

γ2 = 180 - (50 + 119) γ2 = 180 - 169 γ2 = 11

We can now solve like normal.

Sin 50/7 = Sin 11/C2 C2 = 7Sin 11 /Sin 50 C2 = 1.7

Now we have both triangles solved, and have successfully conquered the ambiguous case.

If you're not entirely sure of your answer, you can check your answers at this website(Scroll down until you see the text boxes).

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**10.3-Law of Cosines**
Section 10.3 is very similar to section 10.2, except for the fact that we will be dealing with SSS and SAS triangles. For these cases, we will use the law of cosines.

These are the equations to find sides a, b, and c respectively. To find the angles, a little rearranging of the functions is necessary.

Now for the moment you’ve all been waiting for… the examples!

Example A

SSS Triangle Your approach to a SSS triangle should be to find an angle using the law of cosines. Then, you can use the law of sines to find the remaining angles, or if you enjoy using the law of cosines, you can still use that. Whatever floats your boat.

Ok so in this example a=5, b=8, and c=6. Let’s find angle γ. cosγ = (25 + 64 – 36) ÷ 80 cosγ = .6625 To find gamma after this step we need to do the inverse cosine of both sides to cancel out the cosine on the left side. cos-1 .6625 ≈ 48.5 γ ≈ 48.5

Since you already know how to use the law of sines, I will use the law of cosines to finish the problem. You can use either, but I find the law of sines faster.

cosβ = 25 + 36 – 64 ÷ 60 cosβ = -.05 cos-1 -.05 ≈ 92.8 β ≈ 92.8° 180 – 92.8 – 48.5 = 38.7 = α Example B

SAS Triangle

Your approach to a SAS triangle should be to find the remaining side, which should be opposite the angle that you know. From there you can use the law of sines or the law of cosines to find the other two angles. <span style="font-family: 'Times New Roman','serif'; font-size: 12pt;"> In this problem, b = 4, c = 6 and α = 48. Side A is our first priority.

a^2 = 4^2 + 6^2 – 2 (4)(6)cos48 a^2 = 52 – (48)cos48 a^2 = 52 – 32.1 a^2 = 19.9 a = 4.46

Now, we find the two remaining angles. cosγ = 19.9 + 16 – 36 ÷ 35.68 cosγ = -.01÷ 35.68 cosγ = -.0028 cos-1 -.0028 = γ γ = 90.2 180 – 48 – 90.2 = 41.8 = β

=<span style="color: #0000ff; display: block; font-family: 'Arial Black',Gadget,sans-serif; font-size: 20pt; text-align: center;">10.4- Area of a Non-Right Triangle =

In this section, we will be finding the area of two kinds of triangles-SAS triangles, where two sides and an angle are given, and SSS triangles, where all three sides and no angles are given. The formula most of us are familiar with is






 * where b= (normally just a leg, but in this case...) the base of the triangle and h=the height.**

**.**, you can get this,
 * To find the area of a SAS triangle, you must first understand that**
 * [[image:equ1.JPG width="105" height="59"]],**
 * from which you can derive the equation**
 * If you plug the value of //h// into

**

**
 * which simplifies into


 * (equ. 1)

If you drop the height for the two other sides, then you get: (equ.2) and (equ. 3)

Let's do 2 examples.**


 * Ex. 1-Find the area of a triangle for which //a//= 5, //b//=3, and** //**Y(gamma)**//**=30 degrees.

To find the area, all you would do is plug in the numbers for //a//, //b//, and //Y//, to get:



which equals 7.5 sq. units.

Ex.2-



Find the area, and try to do it yourself. Highlight the text to find the procedure and solution.

To do this, all you have to do is plug the appropriate values into the appropriate equation, which in this case would be equ.1. After you do this, you get A=(1/2)(5)(3)sin (37 degrees), which is approximately 4.5. **

**Finding The Area Of SSS Triangles-**

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 * To find the area of a SSS triangle, you must use Heron's Formula, which is**
 * where S=

For how the Heron's formula is derived, check out this video. media type="youtube" key="AreOMvPCxv8" height="385" width="640" Here is part 2. media type="youtube" key="nZu7IZLhJRI" height="385" width="640" Ex.1- Find the area of a triangle where //a//=3, //b//=5, and //c//=7.

First, you want to determine what //S// would equal, which in this case would be which would equal half of 15 which would be 7.5.

The next step is to plug everything in, which would result in**

**
 * which equals
 * and that would be in sq. units, of course.**


 * If you get a picture, then just write down the values for a, b, and c (you can figure out which is which by the opposite angle) and solve from there.

Practice Problem 1

Find the area of this triangle.

Once again, highlight the black areas to reveal the answer.

To find the area, we use Heron's formula again.

Therefore, //S//=(1/2)(12+11+3) //S//=(1/2)(26) //S=//13

So, after you define the value of S, you can solve for the area. //sqrt=square root// ** **A=sqrt[13(13-3)(13-12)(13-11)] A=sqrt[13(10)(1)(2)] A=sqrt[260] A is approximately equal to 16.2 sq un.**

And that is the end.