Trigonometry Short Course Tutorial PDF

Trigonometry An Overview of Important Topics 1 Contents Trigonometry – An Overview of Important Topics....................................................................................... 4 UNDERSTAND HOW ANGLES ARE MEASURED.....................................................

Trigonometry An Overview of Important Topics 1 Contents Trigonometry – An Overview of Important Topics....................................................................................... 4 UNDERSTAND HOW ANGLES ARE MEASURED............................................................................................. 6 Degrees..................................................................................................................................................... 7 Radians...................................................................................................................................................... 7 Unit Circle.................................................................................................................................................. 9 Practice Problems............................................................................................................................... 10 Solutions.............................................................................................................................................. 11 TRIGONOMETRIC FUNCTIONS.................................................................................................................... 12 Definitions of trig ratios and functions................................................................................................... 12 Khan Academy video 2........................................................................................................................ 14 Find the value of trig functions given an angle measure........................................................................ 15 Find a missing side length given an angle measure................................................................................ 19 Khan Academy video 3........................................................................................................................ 19 Find an angle measure using trig functions............................................................................................ 20 Practice Problems............................................................................................................................... 21 Solutions.............................................................................................................................................. 24 USING DEFINITIONS AND FUNDAMENTAL IDENTITIES OF TRIG FUNCTIONS............................................. 26 Fundamental Identities........................................................................................................................... 26 Khan Academy video 4........................................................................................................................ 28 Sum and Difference Formulas................................................................................................................. 29 Khan Academy video 5........................................................................................................................ 31 Double and Half Angle Formulas............................................................................................................ 32 Khan Academy video 6........................................................................................................................ 34 Product to Sum Formulas....................................................................................................................... 35 Sum to Product Formulas....................................................................................................................... 36 Law of Sines and Cosines........................................................................................................................ 37 Practice Problems............................................................................................................................... 39 Solutions.............................................................................................................................................. 42 UNDERSTAND KEY FEATURES OF GRAPHS OF TRIG FUNCTIONS................................................................ 43 Graph of the sine function (𝒚 = 𝒔𝒊𝒏 𝒙)................................................................................................ 44 Graph of the cosine function (𝒚 = 𝒄𝒐𝒔 𝒙)............................................................................................ 45 2  Key features of the sine and cosine function.......................................................................................... 46 Khan Academy video 7........................................................................................................................ 51 Graph of the tangent function (𝒚 = 𝒕𝒂𝒏 𝒙)......................................................................................... 52 Key features of the tangent function...................................................................................................... 53 Khan Academy video 8........................................................................................................................ 56 Graphing Trigonometric Functions using Technology............................................................................ 57 Practice Problems............................................................................................................................... 60 Solutions.............................................................................................................................................. 62 Rev. 05.06.2016-4 3 Trigonometry – An Overview of Important Topics So I hear you’re going to take a Calculus course? Good idea to brush up on your Trigonometry!! Trigonometry is a branch of mathematics that focuses on relationships between the sides and angles of triangles. The word trigonometry comes from the Latin derivative of Greek words for triangle (trigonon) and measure (metron). Trigonometry (Trig) is an intricate piece of other branches of mathematics such as, Geometry, Algebra, and Calculus. In this tutorial we will go over the following topics.  Understand how angles are measured o Degrees o Radians o Unit circle o Practice  Solutions  Use trig functions to find information about right triangles o Definition of trig ratios and functions o Find the value of trig functions given an angle measure o Find a missing side length given an angle measure o Find an angle measure using trig functions o Practice  Solutions  Use definitions and fundamental Identities of trig functions o Fundamental Identities o Sum and Difference Formulas o Double and Half Angle Formulas o Product to Sum Formulas o Sum to Product Formulas o Law of Sines and Cosines o Practice  Solutions 4  Understand key features of graphs of trig functions o Graph of the sine function o Graph of the cosine function o Key features of the sine and cosine function o Graph of the tangent function o Key features of the tangent function o Practice  Solutions Back to Table of Contents. 5 UNDERSTAND HOW ANGLES ARE MEASURED Since Trigonometry focuses on relationships of sides and angles of a triangle, let’s go over how angles are measured… Angles are formed by an initial side and a terminal side. An initial side is said to be in standard position when it’s vertex is located at the origin and the ray goes along the positive x axis. An angle is measured by the amount of rotation from the initial side to the terminal side. A positive angle is made by a rotation in the counterclockwise direction and a negative angle is made by a rotation in the clockwise direction. Angles can be measured two ways: 1. Degrees 2. Radians 6 Degrees A circle is comprised of 360°, which is called one revolution Degrees are used primarily to describe the size of an angle. The real mathematician is the radian, since most computations are done in radians. Radians 1 revolution measured in radians is 2π, where π is the constant approximately 3.14. How can we convert between the two you ask? Easy, since 360° = 2π radians (1 revolution) Then, 180° = π radians 𝜋 So that means that 1° = radians 180 7  180 And degrees = 1 radian 𝜋 Example 1 Convert 60° into radians 𝜋 𝜋 60𝜋 𝜋 60 ⋅ (1 degree) = 60 ⋅ = = radian 180 180 180 3 Example 2 Convert (-45°) into radians 𝜋 −45𝜋 𝜋 -45 ⋅ = =− radian 180 180 4 Example 3 3𝜋 Convert radian into degrees 2 3𝜋 180 3𝜋 180 540𝜋 ⋅ (1 radian) = ⋅ = = 270° 2 𝜋 2 𝜋 2𝜋 Example 4 7𝜋 Convert − radian into degrees 3 7𝜋 180 1260 − ⋅ = = 420° 3 𝜋 3 Before we move on to the next section, let’s take a look at the Unit Circle. 8 Unit Circle The Unit Circle is a circle that is centered at the origin and always has a radius of 1. The unit circle will be helpful to us later when we define the trigonometric ratios. You may remember from Algebra 2 that the equation of the Unit Circle is 𝑥² + 𝑦² = 1. Need more help? Click below for a Khan Academy video Khan Academy video 1 9 Practice Problems 10 Solutions Back to Table of Contents. 11 TRIGONOMETRIC FUNCTIONS Definitions of trig ratios and functions In Trigonometry there are six trigonometric ratios that relate the angle measures of a right triangle to the length of its sides. (Remember a right triangle contains a 90° angle) A right triangle can be formed from an initial side x and a terminal side r, where r is the radius and hypotenuse of the right triangle. (see figure below) The Pythagorean Theorem tells us that x² + y² = r², therefore r = √𝑥² + 𝑦². 𝜃 (theta) is used to label a non-right angle. The six trigonometric functions can be used to find the ratio of the side lengths. The six functions are sine (sin), cosine (cos), tangent (tan), cosecant (csc), secant (sec), and cotangent (cot). Below you will see the ratios formed by these functions. 𝑦 𝑜𝑝𝑝𝑜𝑠𝑖𝑡𝑒 𝑠𝑖𝑑𝑒 sin 𝜃 = , also referred to as 𝑟 ℎ𝑦𝑝𝑜𝑡𝑒𝑛𝑢𝑠𝑒 𝑥 𝑎𝑑𝑗𝑎𝑐𝑒𝑛𝑡 𝑠𝑖𝑑𝑒 cos 𝜃 = , also referred to as 𝑟 ℎ𝑦𝑝𝑜𝑡𝑒𝑛𝑢𝑠𝑒 𝑦 𝑜𝑝𝑝𝑜𝑠𝑖𝑡𝑒 𝑠𝑖𝑑𝑒 tan 𝜃 = , also referred to as 𝑥 𝑎𝑑𝑗𝑎𝑐𝑒𝑛𝑡 𝑠𝑖𝑑𝑒 These three functions have 3 reciprocal functions 𝑟 csc 𝜃 = , which is the reciprocal of sin 𝜃 𝑦 12  𝑟 sec 𝜃 = ,which is the reciprocal of cos 𝜃 𝑥 𝑥 cot 𝜃 = , which is the reciprocal of tan 𝜃 𝑦 You may recall a little something called SOH-CAH-TOA to help your remember the functions! SOH… Sine = opposite/hypotenuse …CAH… Cosine = adjacent/hypotenuse …TOA Tangent = opposite/adjacent Example: Find the values of the trigonometric ratios of angle 𝜃 Before we can find the values of the six trig ratios, we need to find the length of the missing side. Any ideas? Good call, we can use r = √𝑥² + 𝑦² (from the Pythagorean Theorem) r = √5² + 12² = √25 + 144 = √169 = 13 Now we can find the values of the six trig functions 𝑜𝑝𝑝𝑜𝑠𝑖𝑡𝑒 12 ℎ𝑦𝑝𝑡𝑜𝑒𝑛𝑢𝑠𝑒 13 sin θ = = csc θ = = ℎ𝑦𝑝𝑡𝑜𝑒𝑛𝑢𝑠𝑒 13 𝑜𝑝𝑝𝑜𝑠𝑖𝑡𝑒 12 𝑎𝑑𝑗𝑎𝑐𝑒𝑛𝑡 5 ℎ𝑦𝑝𝑜𝑡𝑒𝑛𝑢𝑠𝑒 13 cos θ = = sec θ = = ℎ𝑦𝑝𝑜𝑡𝑒𝑛𝑢𝑠𝑒 13 𝑎𝑑𝑗𝑎𝑐𝑒𝑛𝑡 5 𝑜𝑝𝑝𝑜𝑠𝑖𝑡𝑒 12 𝑎𝑑𝑗𝑎𝑐𝑒𝑛𝑡 5 tan θ = = cot θ = = 𝑎𝑑𝑗𝑎𝑐𝑒𝑛𝑡 5 𝑜𝑝𝑝𝑜𝑠𝑖𝑡𝑒 12 13 Example 5 a) Use the triangle below to find the six trig ratios First use Pythagorean Theorem to find the hypotenuse a² + b² = c², where a and b are legs of the right triangle and c is the hypotenuse 𝑜 8 4 1 5 sin 𝜃 = ℎ = 10 = 5 csc 𝜃 = sin 𝜃 = 4 6² + 8² = 𝑐² 𝑎 6 3 1 5 cos 𝜃 = = = sec 𝜃 = = 36 + 64 = 𝑐² ℎ 10 5 cos 𝜃 3 𝑜 8 4 1 3 100 = 𝑐² tan 𝜃 = = = cot 𝜃 = = 𝑎 6 3 tan 𝜃 4 √100 = √𝑐² 10 = 𝑐 Example 6 Use the triangle below to find the six trig ratios 1² + 𝑏² = (√5 )² 2 2√5 √5 sin 𝜃 = = csc 𝜃 = √5 5 2 1 + 𝑏² = 5 1 √5 √5 cos 𝜃 = = sec 𝜃 = = √5 √5 5 1 𝑏² = 4 2 1 𝑏 = 2 tan 𝜃 = 1 = 2 cot 𝜃 = 2 Need more help? Click below for a Khan Academy Video Khan Academy video 2 14 Find the value of trig functions given an angle measure Suppose you know the value of 𝜃 is 45°, how can this help you find the values of the six trigonometric functions? First way: You can familiarize yourself with the unit circle we talked about. An ordered pair along the unit circle (x, y) can also be known as (cos 𝜃, sin 𝜃), since the r value on the unit circle is always 1. So to find the trig function values √2 √2 for 45° you can look on the unit circle and easily see that sin 45° = , cos 45° = 2 2 With that information we can easily find the values of the reciprocal functions 2 2√2 csc 45° = = = √2 , sec 45° = √2 √2 2 We can also find the tangent and cotangent function values using the quotient identities 15  √2 sin 45° 2 tan 45° = = √2 =1 cos 45° 2 cot 45° = 1 Example 7 𝜋 1 1 Find sec ( ) = 𝜋 = √2 = √2 4 cos( 4 ) 2 Example 8 1 𝜋 √3 Find tan ( ) = 2 √3 = 6 3 2 Example 9 1 − √3 Find cot 240° = 2 √3 = − 3 2 Using this method limits us to finding trig function values for angles that are accessible on the unit circle, plus who wants to memorize it!!! Second Way: If you are given a problem that has an angle measure of 45°, 30°, or 60°, you are in luck! These angle measures belong to special triangles. If you remember these special triangles you can easily find the ratios for all the trig functions. Below are the two special right triangles and their side length ratios 16 How do we use these special right triangles to find the trig ratios? If the θ you are given has one of these angle measures it’s easy! Example 10 Example 11 Example 12 Find sin 30° Find cos 45° Find tan 60° 1 √2 √3 sin 30° = cos 45° = tan 60° = = √3 2 2 1 Third way: This is not only the easiest way, but also this way you can find trig values for angle measures that are less common. You can use your TI Graphing calculator. First make sure your TI Graphing calculator is set to degrees by pressing mode 17 Next choose which trig function you need After you choose which function you need type in your angle measure Example 13 Example 14 Example 15 cos 55° ≈ 0.5736 tan 0° = 0 sin 30° = 0.5 18 Find a missing side length given an angle measure Suppose you are given an angle measure and a side length, can you find the remaining side lengths? Yes. You can use the trig functions to formulate an equation to find missing side lengths of a right triangle. Example 16 𝑜 𝑥 First we know that sin 𝜃 = ℎ, therefore sin 30 = 5 Next we solve for x, 5 ⋅ sin 30 = 𝑥 Use your TI calculator to compute 5 ⋅ sin 30, And you find out 𝑥 = 2.5 Let’s see another example, Example 17 We are given information about the opposite and adjacent sides of the triangle, so we will use tan 16 tan 52 = 𝑥 16 𝑥= tan 52 𝑥 ≈ 12.5 Need more help? Click below for a Khan Academy video Khan Academy video 3 19 Find an angle measure using trig functions Wait a minute, what happens if you have the trig ratio, but you are asked to find the angles measure? Grab your TI Graphing calculator and notice that above the sin, cos, and tan buttons, there is 𝑠𝑖𝑛−1 , 𝑐𝑜𝑠 −1 , 𝑡𝑎𝑛−1. These are your inverse trigonometric functions, also known as arcsine, arccosine, and arctangent. If you use these buttons in conjunction with your trig ratio, you will get the angle measure for 𝜃! Let’s see some examples of this. Example 18 We know that tan 𝜃 = 6 8 So to find the value of θ, press 2nd tan on your calculator and then type in (8/6) 8 𝑡𝑎𝑛−1 ( ) ≈ 53.13 6 𝜃 ≈ 53.13° How about another Example 19 We are given information about the adjacent side and the hypotenuse, so we will use the cosine function 1 cos 𝜃 = 2 1 𝑐𝑜𝑠 −1 ( ) = 60 2 𝜃 = 60° 20 Practice Problems 21 22 23 Solutions 24 Back to Table of Contents. 25 USING DEFINITIONS AND FUNDAMENTAL IDENTITIES OF TRIG FUNCTIONS Fundamental Identities Reciprocal Identities sin 𝜃 = 1/(csc 𝜃) csc 𝜃 = 1/(sin 𝜃) cos 𝜃 = 1/(sec 𝜃) sec 𝜃 = 1/(cos 𝜃) tan 𝜃 = 1/(cot 𝜃) cot 𝜃 = 1/(tan 𝜃) Quotient Identities tan 𝜃 = (sin 𝜃)/(cos 𝜃) cot 𝜃 = (cos 𝜃)/(sin 𝜃) Pythagorean Identities sin²𝜃 + cos²𝜃 = 1 1+ tan²𝜃 = sec²𝜃 1+ cot²𝜃 = csc²𝜃 Negative Angle Identities sin(−𝜃) = − sin 𝜃 cos(−𝜃) = cos 𝜃 tan(−𝜃) = − tan 𝜃 csc(−𝜃) = − csc 𝜃 sec(−𝜃) = sec 𝜃 cot(−𝜃) = − cot 𝜃 Complementary Angle Theorem If two acute angles add up to be 90°, they are considered complimentary. The following are considered cofunctions: sine and cosine tangent and cotangent secant and cosecant The complementary angle theorem says that cofunctions of complimentary angles are equal. 26 Example 20) sin 54° = cos 36° How can we use these identities to find exact values of trigonometric functions? Follow these examples to find out! Examples 21-26 21) Find the exact value of the expression sin² 30° + cos² 30° Solution: Since sin² 𝜃 + cos² 𝜃 = 1, therefore sin² 30° + cos² 30° = 1 22) Find the exact value of the expression sin 45° tan 45° − cos 45° sin 45° Solution: Since ( ) = tan 45°, therefore tan 45° − tan 45° = 0 cos 45° 23) tan 35° ⋅ cos 35° ⋅ csc 35° sin 35° cos 35° 1 Solution: ⋅ ⋅ =1 cos 35° 1 sin 35° 24) tan 22° − cot 68° Solution: tan 22° = cot 68°, therefore cot 68° − cot 68° = 0 2 25) cot 𝜃 = − , find csc 𝜃, where 𝜃 is in quadrant II 3 Solution: Pick an identity that relates cotangent to cosecant, like the Pythagorean identity 1 + cot² 𝜃 = csc² 𝜃. 2 2 1 + (− ) = csc² θ 3 4 1 + = csc² 𝜃 9 13 = csc² 𝜃 9 13 √ = csc 𝜃 9 √13 = csc 𝜃 3 The positive square root is chosen because csc is positive in quadrant II 27  26) Prove the following identity is true cot 𝜃 ⋅ sin 𝜃 ∙ cos 𝜃 = cos² 𝜃 cos 𝜃 sin 𝜃 cos 𝜃 Solution: ∙ ∙ = cos² 𝜃 sin 𝜃 1 1 Need more help? Click below for a Khan Academy video Khan Academy video 4 28 Sum and Difference Formulas In this section we will use formulas that involve the sum or difference of two angles, call the sum and difference formulas. Sum and difference formulas for sines and cosines sin(𝛼 + 𝛽) = sin 𝛼 cos 𝛽 + cos 𝛼 sin 𝛽 sin(𝛼 − 𝛽) = sin 𝛼 cos 𝛽 − cos 𝛼 sin 𝛽 cos(𝛼 + 𝛽) = cos 𝛼 cos 𝛽 − sin 𝛼 sin 𝛽 cos(𝛼 − 𝛽) = cos 𝛼 cos 𝛽 + sin 𝛼 sin 𝛽 How do we use these formulas? Example 27 Find the exact value of cos 105° Well we can break 105° into 60° and 45° since those values are relatively easy to find the cosine of. Therefore cos 105° = cos(60° + 45°)= cos 60° cos 45° − sin 60° sin 45° Using the unit circle we obtain, 1 √2 √3 √2 =2∙ − ∙ 2 2 2 √2 √6 1 = − = 4 (√2 − √6) 4 4 Example 28 Find the exact value of sin 15° = sin(45° − 30°) = sin 45° cos 30° − cos 45° sin 30° √2 √3 √2 1 = ∙ 2 − ∙ 2 2 2 √6 √2 1 = 4 − 4 = 4 (√6 − √2) 29 Sum and difference formulas for tangent tan 𝛼+tan 𝛽 tan(𝛼 + 𝛽) = 1−tan 𝛼 tan 𝛽 tan 𝛼−tan 𝛽 tan(𝛼 − 𝛽) = 1+tan 𝛼 tan 𝛽 Example 29 Find the exact value of tan 75° tan 75° = tan(45° + 30°) √3 tan 45°+tan 30° 1+ 3+√3 = = 3 √3 = (rationalize the denominator) 1−tan 45° tan 30° 1− 3−√3 3 3 + √3 3 − √3 12 + 6√3 = ∙ = = 2 + √3 3 − √3 3 − √ 3 6 7𝜋 Example 30 Find tan ( ) 12 7𝜋 3𝜋 4𝜋 𝜋 𝜋 tan = tan ( + ) = tan ( + ) 12 12 12 4 3 𝜋 𝜋 tan +tan 1+√3 1+√3 1+√3 4+2√3 = 4 𝜋 3 𝜋 = = ∙ = = −2 − √3 1−tan ⋅tan 1−√3 1−√3 1+√3 −2 4 3 Cofunction Identities cos(90° − 𝜃) = sin 𝜃 sec(90° − 𝜃) = csc 𝜃 sin(90° − 𝜃) = cos 𝜃 csc(90° − 𝜃) = sec 𝜃 tan(90° − 𝜃) = cot 𝜃 cot(90° − 𝜃) = tan 𝜃 Example 31 Find cos 30° √3 cos 30° = sin(90° − 30°) = sin 60° = 2 30 Need more help? Click below for a Khan Academy video Khan Academy video 5 31 Double and Half Angle Formulas Below you will learn formulas that allow you to use the relationship between the six trig functions for a particular angle and find the trig values of an angle that is either half or double the original angle. Double Angle Formulas cos 2𝜃 = cos²𝜃 − sin²𝜃 =2cos²𝜃 − 1 = 1 − 2sin²𝜃 sin 2𝜃 = 2 sin 𝜃 cos 𝜃 2 tan 𝜃 tan 2𝜃 = 1 − tan² 𝜃 Half Angle Formulas 𝜃 1+cos 𝜃 cos = ±√ 2 2 𝜃 1−cos 𝜃 sin = ±√ 2 2 𝜃 1−cos 𝜃 𝜃 sin 𝜃 𝜃 1−cos 𝜃 tan = ±√ tan = tan = 2 1+cos 𝜃 2 1+cos 𝜃 2 sin 𝜃 Lets see these formulas in action! Example 32 Use the double angle formula to find the exact value of each expression sin 120° √3 sin 120° = sin 2(60°) = 2 sin 60° ⋅ cos 60° = 2 32 Example 33 5 3𝜋 tan 𝜃 = 𝑎𝑛𝑑 𝜋 < 𝜃 < , 𝐹𝑖𝑛𝑑 cos 2𝜃 12 2 First we need to find what the cos 𝜃 is. We know that tan 𝜃 is opposite leg over adjacent leg, so we need to find the hypotenuse since cos is adjacent over hypotenuse. We can use 𝑟 = √12² + 5² = 13 to find the length of the 12 hypotenuse. Now we know the cos 𝜃 =. Now use the double angle formula to 13 find cos 2𝜃. 5 2 25 50 119 cos 2𝜃 = 1 − 2sin² 𝜃 = 1 − 2 ( ) = 1 − 2 ( )=1− = 13 169 169 169 We take the positive answer since 𝜃 is in the third quadrant making the ratio a negative over a negative. Now lets try using the half angle formula Example 34 cos 15° √3 30 1 + cos 30 √1 + 2 √2 + √ 3 cos 15° = cos = ±√ =± = 2 2 2 2 Choose the positive root Example 35 4 𝜃 cos 𝜃 = − and 90° < 𝜃 < 180°. 𝐹𝑖𝑛𝑑 sin 5 2 First we use the Pythagorean Theorem to find the third side 42 + 𝑥 2 = 52 33 𝑥2 = 9 𝑥=3 3 sin 𝜃 = 5 4 9 𝜃 1 − (− ) sin = ± √ 5 = ±√ 5 = ±√ 9 = 3√10 2 2 2 10 10 Since sin is positive in the third quadrant we take the positive answer Need more help? Click below for a Khan Academy video Khan Academy video 6 34 Product to Sum Formulas 1 cos 𝐴 cos 𝐵 = [cos(𝐴 + 𝐵) + cos(𝐴 − 𝐵)] 2 1 sin 𝐴 sin 𝐵 = [cos(𝐴 − 𝐵) − cos(𝐴 + 𝐵)] 2 1 sin 𝐴 cos 𝐵 = [sin(𝐴 + 𝐵) + sin(𝐴 − 𝐵)] 2 1 cos 𝐴 sin 𝐵 = [sin(𝐴 + 𝐵) − sin(𝐴 − 𝐵)] 2 Example 36 Use the product-to-sum formula to change sin 75° sin 15° to a sum 1 1 sin 75° sin 15° = [cos(75° − 15°) − cos(75° + 15°)] = [cos 60° − cos 90°] 2 2 1 1 1 = [ − 0] = 2 2 4 35 Sum to Product Formulas 𝐴+𝐵 𝐴−𝐵 sin 𝐴 + sin 𝐵 = 2 sin ( ) cos ( ) 2 2 𝐴+𝐵 𝐴−𝐵 sin 𝐴 − sin 𝐵 = 2 cos ( ) sin ( ) 2 2 𝐴+𝐵 𝐴−𝐵 cos 𝐴 + cos 𝐵 = 2 cos ( ) cos ( ) 2 2 𝐴+𝐵 𝐴−𝐵 cos 𝐴 − cos 𝐵 = −2 sin ( ) sin ( ) 2 2 Example 37 Use the sum-to-product formula to change sin 70° − sin 30° into a product 70° + 30° 70° − 30° sin 70° − sin 30° = 2 cos ( ) sin ( ) = 2 cos 50° ⋅ sin 20° 2 2 36 Law of Sines and Cosines These laws help us to find missing information when dealing with oblique triangles (triangles that are not right triangles) Law of Sines sin 𝐴 sin 𝐵 sin 𝐶 = = 𝑎 𝑏 𝑐 You can use the Law of Sines when the problem is referring to two sets of angles and their opposite sides. Example 38 Find the length of AB. Round your answer to the nearest tenth. Since we are given information about an angle, the side opposite of that angle, another angle, and missing the side opposite of that angle, we can apply the Law of Sines. sin 92° sin 28° = 15 𝐴𝐵 Multiply both sides by the common denominator in order to eliminate the fractions. We do this so that we can solve for the unknown. This gives us, sin 92 ⋅ 𝐴𝐵 = sin 28 ⋅ 15 Then we can divide by sin 92. When we do this we find 𝐴𝐵 = 7 Law of Cosines 𝑐² = 𝑎² + 𝑏² − 2𝑎𝑏 cos 𝐶 𝑏² = 𝑎² + 𝑐² − 2𝑎𝑐 cos 𝐵 𝑎² = 𝑏² + 𝑐² − 2𝑏𝑐 cos 𝐴 You can use the Law of Cosines when the problem is referring to all three sides and only one angle. 37 Example 39 Find the length of AB. Round to the nearest tenth. Since all three sides of the triangle are referred to and information about one angle is given, we can use the Law of Cosines. Since AB is opposite of