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== 10.8#h(0.6em)Right Triangle Trigonometry


  #figure(figph[Pythagoras is shown writing in a book as a young man presents him with a tablet showing a diagrammatic representation of a lyre above a drawing of the sacred tetractys.], alt: "Pythagoras is shown writing in a book as a young man presents him with a tablet showing a diagrammatic representation of a lyre above a drawing of the sacred tetractys.", caption: [In the lower left corner of the fresco #emph[The School of Athens] by Raphael, the figure in white writing in the book represents Pythagoras. Alongside him, to the right, the figure with the long, light-brown hair is said to depict Archimedes.])


  
    
=== Learning Objectives


    After completing this section, you should be able to:


    + Apply the Pythagorean Theorem to find the missing sides of a right triangle.
+ Apply the #math.equation(block: false, alt: "30 degrees - 60 degrees - 90 degrees")[$30^(∘) "-" 60^(∘) "-" 90^(∘)$] and #math.equation(block: false, alt: "45 degrees - 45 degrees - 90 degrees")[$45^(∘) "-" 45^(∘) "-" 90^(∘)$] right triangle relationships to find the missing sides of a triangle.
+ Apply trigonometric ratios to find missing parts of a right triangle.
+ Solve application problems involving trigonometric ratios.


  
  This is another excerpt from Raphael’s #emph[The School of Athens.] The man writing in the book represents Pythagoras, the namesake of one of the most widely used formulas in geometry, engineering, architecture, and many other fields, the Pythagorean Theorem. However, there is evidence that the theorem was known as early as 1900–1100 BC by the Babylonians. The Pythagorean Theorem is a formula used for finding the lengths of the sides of right triangles.


  Born in Greece, Pythagoras lived from 569–500 BC. He initiated a cult-like group called the Pythagoreans, which was a secret society composed of mathematicians, philosophers, and musicians. Pythagoras believed that everything in the world could be explained through numbers. Besides the Pythagorean Theorem, Pythagoras and his followers are credited with the discovery of irrational numbers, the musical scale, the relationship between music and mathematics, and many other concepts that left an immeasurable influence on future mathematicians and scientists.


  The focus of this section is on right triangles. We will look at how the #strong[Pythagorean Theorem] is used to find the unknown sides of a right triangle, and we will also study the #emph[special] triangles, those with set ratios between the lengths of sides. By ratios we mean the relationship of one side to another side. When you think about ratios, you should think about fractions. A fraction is a ratio, the ratio of the numerator to the denominator. Finally, we will preview trigonometry. We will learn about the basic trigonometric functions, sine, cosine and tangent, and how they are used to find not only unknown sides but unknown angles, as well, with little information.


  
    
=== Pythagorean Theorem


    The Pythagorean Theorem is used to find unknown sides of right triangles. The theorem states that the sum of the squares of the two legs of a right triangle equals the square of the hypotenuse (the longest side of the right triangle).


    #notebox("Formula", rgb("#059669"), rgb("#007942"), rgb("#EAF3EC"))[
The Pythagorean Theorem states


      #math.equation(block: true, alt: "a squared plus b squared equals c squared")[$a^(2) + b^(2) = c^(2)$]
      where #math.equation(block: false, alt: "a")[$a$] and #math.equation(block: false, alt: "b")[$b$] are two sides (legs) of a right triangle and #math.equation(block: false, alt: "c")[$c$] is the hypotenuse, as shown.


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    For example, given that side #math.equation(block: false, alt: "a equals 6 ,")[$a = 6 ,$] and side #math.equation(block: false, alt: "b equals 8 ,")[$b = 8 ,$] we can find the measure of side #math.equation(block: false, alt: "c")[$c$] using the Pythagorean Theorem. Thus,


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    #examplebox("Example 1")[Using the Pythagorean Theorem][
Find the length of the missing side of the triangle .


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      #solutionbox[
Using the Pythagorean Theorem, we have


        #math.equation(block: true, alt: "open parenthesis 6 close parenthesis squared plus b squared, equals, open parenthesis 14 close parenthesis squared; 36 plus b squared, equals, 196; b squared, equals, 196 minus 36; b squared, equals, 160; b, equals, plus or minus the square root of 160; equals, 4 the square root of 10 equals 12.65")[$attach(( 6 ), t: 2) + b^(2) & = & attach(( 14 ), t: 2) \ 36 + b^(2) & = & 196 \ b^(2) & = & 196 − 36 \ b^(2) & = & 160 \ b & = & ± sqrt(160) \  & = & 4 sqrt(10) = 12.65$]
        When we take the square root of a number, the answer is usually both the positive and negative root. However, lengths cannot be negative, which is why we only consider the positive root.
]
]


    
  
  
    
=== Distance


    The applications of the Pythagorean Theorem are countless, but one especially useful application is that of distance. In fact, the distance formula stems directly from the theorem. It works like this:


    In , the problem is to find the distance between the points #math.equation(block: false, alt: "open parenthesis minus 3 , minus 1 close parenthesis")[$( − 3 , − 1 )$] and #math.equation(block: false, alt: "open parenthesis 3 , 2 close parenthesis .")[$( 3 , 2 ) .$] We call the length from point #math.equation(block: false, alt: "open parenthesis minus 3 , minus 1 close parenthesis")[$( − 3 , − 1 )$] to point #math.equation(block: false, alt: "open parenthesis 3 , minus 1 close parenthesis")[$( 3 , − 1 )$] side #math.equation(block: false, alt: "a")[$a$], and the length from point #math.equation(block: false, alt: "open parenthesis 3 , minus 1 close parenthesis")[$( 3 , − 1 )$] to point #math.equation(block: false, alt: "open parenthesis 3 , 2 close parenthesis")[$( 3 , 2 )$] side #math.equation(block: false, alt: "b")[$b$]. To find side #math.equation(block: false, alt: "c")[$c$], we use the distance formula and we will explain it relative to the Pythagorean Theorem. The distance formula is #math.equation(block: false, alt: "d equals the square root of open parenthesis x sub 2 minus x sub 1 close parenthesis squared plus open parenthesis y sub 2 minus y sub 1 close parenthesis squared ,")[$d = sqrt(attach(( x_(2) − x_(1) ), t: 2) + attach(( y_(2) − y_(1) ), t: 2)) ,$] such that #math.equation(block: false, alt: "open parenthesis x sub 2 minus x sub 1 close parenthesis")[$( x_(2) − x_(1) )$] is a substitute for #math.equation(block: false, alt: "a")[$a$] in the Pythagorean Theorem and is equal to #math.equation(block: false, alt: "3 minus open parenthesis minus 3 close parenthesis equals 6 ;")[$3 − ( − 3 ) = 6 ;$] and #math.equation(block: false, alt: "open parenthesis y sub 2 minus y sub 1 close parenthesis")[$( y_(2) − y_(1) )$] is a substitute for #math.equation(block: false, alt: "b")[$b$] in the Pythagorean Theorem and is equal to #math.equation(block: false, alt: "2 minus open parenthesis minus 1 close parenthesis equals 3.")[$2 − ( − 1 ) = 3.$] When we plug in these numbers to the distance formula, we have


    #math.equation(block: true, alt: "d, equals, the square root of open parenthesis 3 minus open parenthesis minus 3 close parenthesis close parenthesis squared plus open parenthesis 2 minus open parenthesis minus 1 close parenthesis close parenthesis squared; equals, the square root of open parenthesis 6 close parenthesis squared plus open parenthesis 3 close parenthesis squared equals the square root of 36 plus 9; equals, the square root of 45 equals 3 the square root of 5 equals 6.7")[$d & = & sqrt(attach(( 3 − ( − 3 ) ), t: 2) + attach(( 2 − ( − 1 ) ), t: 2)) \  & = & sqrt(attach(( 6 ), t: 2) + attach(( 3 ), t: 2)) = sqrt(36 + 9) \  & = & sqrt(45) = 3 sqrt(5) = 6.7$]
    Thus, #math.equation(block: false, alt: "d equals c")[$d = c$], the hypotenuse, in the Pythagorean Theorem.


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    #examplebox("Example 2")[Calculating Distance Using the Distance Formula][
You live on the corner of First Street and Maple Avenue, and work at Star Enterprises on Tenth Street and Elm Drive . You want to calculate how far you walk to work every day and how it compares to the actual distance (as the crow flies). Each block measures 200 ft by 200 ft.


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      #solutionbox[
You travel 7 blocks south and 9 blocks west. If each block measures 200 ft by 200 ft, then #math.equation(block: false, alt: "9 open parenthesis 200 close parenthesis plus 7 open parenthesis 200 close parenthesis equals 1,800 ft plus 1,400 ft equals 3,200 ft")[$9 ( 200 ) + 7 ( 200 ) = 1,800 #h(0.28em) "ft" + 1,400 #h(0.28em) "ft" = 3,200 #h(0.28em) "ft"$].


        As the crow flies, use the distance formula. We have


        #math.equation(block: true, alt: "d, equals, the square root of open parenthesis 1,800 minus 0 close parenthesis squared plus open parenthesis 1,400 minus 0 close parenthesis squared; equals, the square root of 3,240,000 plus 1,960,000; equals, the square root of 5,200,000; equals, 2280.4 ft")[$d & = & sqrt(attach(( 1,800 − 0 ), t: 2) + attach(( 1,400 − 0 ), t: 2)) \  & = & sqrt(3","240","000 + 1","960","000) \  & = & sqrt(5","200","000) \  & = & 2280.4 #h(0.28em) "ft"$]
]
]


    
    #examplebox("Example 3")[Calculating Distance with the Pythagorean Theorem][
The city has specific building codes for wheelchair ramps. Every vertical rise of 1 in requires that the horizontal length be 12 inches. You are constructing a ramp at your business. The plan is to make the ramp 130 inches in horizontal length and the slanted distance will measure approximately 132.4 inches . What should the vertical height be?


        #figure(
  stdfig([#cetz.canvas({
  import cetz.draw: *
  line((28.57pt, 28.57pt), (171.43pt, 28.57pt), (28.57pt, 85.71pt), close: true, fill: none, stroke: 0.7pt + black)
  content((107.03pt, 22.95pt), anchor: "west", text(size: 9pt)[130 inches])
  content((108.36pt, 60.49pt), anchor: "west", text(size: 9pt)[132.4 inches])
  content((19.75pt, 58.91pt), anchor: "east", text(size: 9pt)[b])
  line((28.57pt, 28.57pt), (36.57pt, 28.57pt), (36.57pt, 36.57pt), (28.57pt, 36.57pt), close: true, stroke: 0.6pt + black, fill: none)
})]),
  alt: "A right triangle with its legs marked b and 130 inches. The hypotenuse is marked 132.4 inches.", caption: none,
)


      
      #solutionbox[
The Pythagorean Theorem states that the horizontal length of the base of the ramp, side #emph[a], is 130 in. The length of #emph[c], or the length of the hypotenuse, is 132.4 in. The length of the height of the triangle is side #emph[b].


        Then, by the Pythagorean Theorem, we have:


        #math.equation(block: true, alt: "a squared plus b squared, equals, c squared; open parenthesis 130 close parenthesis squared plus b squared, equals, open parenthesis 132.4 close parenthesis squared; 16,900 plus b squared, equals, 17,529.76; b squared, equals, 17,529.76 minus 16,900; b squared, equals, 629.8; b, equals, the square root of 629.8 equals 25")[$a^(2) + b^(2) & = & c^(2) \ attach(( 130 ), t: 2) + b^(2) & = & attach(( 132.4 ), t: 2) \ 16,900 + b^(2) & = & 17,529.76 \ b^(2) & = & 17,529.76 − 16,900 \ b^(2) & = & 629.8 \ b & = & sqrt(629.8) = 25$]
        If you construct the ramp with a 25 in vertical rise, will it fulfill the building code? If not, what will have to change?


        The building code states 12 in of horizontal length for each 1 in of vertical rise. The vertical rise is 25 in, which means that the horizontal length has to be #math.equation(block: false, alt: "12 open parenthesis 25 close parenthesis equals 300 in .")[$12 ( 25 ) = 300 #h(0.28em) "in" .$] So, no, this will not pass the code. If you must keep the vertical rise at 25 in, what will the other dimensions have to be? Since we need a minimum of 300 in for the horizontal length:


        #math.equation(block: true, alt: "open parenthesis 300 close parenthesis squared plus open parenthesis 25 close parenthesis squared, equals, c squared; 90,625, equals, c squared; the square root of 90,625, equals, c equals 301 in")[$attach(( 300 ), t: 2) + attach(( 25 ), t: 2) & = & c^(2) \ 90,625 & = & c^(2) \ sqrt(90","625) & = & c = 301 #h(0.28em) "in"$]
        The new ramp will look like .


        #figure(
  stdfig([#cetz.canvas({
  import cetz.draw: *
  line((25.00pt, 25.00pt), (25.00pt, 125.00pt), (175.00pt, 25.00pt), close: true, fill: none, stroke: 0.7pt + black)
  content((16.46pt, 77.85pt), anchor: "east", text(size: 9pt)[25 inches])
  content((107.49pt, 79.99pt), anchor: "west", text(size: 9pt)[301 inches])
  content((105.40pt, 17.80pt), anchor: "north", text(size: 9pt)[300 inches])
  line((25.00pt, 25.00pt), (25.00pt, 33.00pt), (33.00pt, 33.00pt), (33.00pt, 25.00pt), close: true, stroke: 0.6pt + black, fill: none)
})]),
  alt: "A right triangle with its legs marked 25 inches and 300 inches. The hypotenuse is marked 301 inches.", caption: none,
)
]
]


    
  
  
    
=== #math.equation(block: false, alt: "30 degrees - 60 degrees - 90 degrees")[$30^(∘) "-" 60^(∘) "-" 90^(∘)$] Triangles


    In geometry, as in all fields of mathematics, there are always special rules for special circumstances. An example is the #emph[perfect square] rule in algebra. When expanding an expression like #math.equation(block: false, alt: "open parenthesis 2 x plus 5 y close parenthesis squared ,")[$attach(( 2 x + 5 y ), t: 2) ,$] we do not have to expand it the long way:


    #math.equation(block: true, alt: "open parenthesis 2 x plus 5 y close parenthesis squared, equals, open parenthesis 2 x plus 5 y close parenthesis open parenthesis 2 x plus 5 y close parenthesis; equals, open parenthesis 2 x close parenthesis squared plus 10 x y plus 10 x y plus open parenthesis 5 y close parenthesis squared; equals, 4 x squared plus 20 x y plus 25 y squared")[$attach(( 2 x + 5 y ), t: 2) & = & ( 2 x + 5 y ) ( 2 x + 5 y ) \  & = & attach(( 2 x ), t: 2) + 10 x y + 10 x y + attach(( 5 y ), t: 2) \  & = & 4 x^(2) + 20 x y + 25 y^(2)$]
    If we know the perfect square formula, given as


    #math.equation(block: true, alt: "open parenthesis a plus b close parenthesis squared equals a squared plus 2 a b plus b squared ,")[$attach(( a + b ), t: 2) = a^(2) + 2 a b + b^(2) ,$]
    we can skip the middle step and just start writing down the answer. This may seem trivial with problems like #math.equation(block: false, alt: "open parenthesis a plus b close parenthesis squared .")[$attach(( a + b ), t: 2) .$] However, what if you have a problem like #math.equation(block: false, alt: "open parenthesis 2 the square root of 3 plus 3 the cube root of 31.8 c close parenthesis squared ?")[$attach(( 2 sqrt(3) + 3 root(3, 31.8 c) ), t: 2) ?$] That is a different story. Nevertheless, we use the same perfect square formula. The same idea applies in geometry. There are special formulas and procedures to apply in certain types of problems. What is needed is to remember the formula and remember the kind of problems that fit. Sometimes we believe that because a formula is labeled #emph[special,] we will rarely have use for it. That assumption is incorrect. So, let us identify the #math.equation(block: false, alt: "30 degrees - 60 degrees - 90 degrees")[$30^(∘) "-" 60^(∘) "-" 90^(∘)$] triangle and find out why it is special.


    #figure(
  stdfig([#cetz.canvas({
  import cetz.draw: *
  line((28.57pt, 28.57pt), (171.43pt, 28.57pt), (28.57pt, 114.29pt), close: true, fill: none, stroke: 0.7pt + black)
  content((105.76pt, 21.66pt), anchor: "north", text(size: 9pt)[x√3])
  content((107.72pt, 76.06pt), anchor: "west", text(size: 9pt)[2x])
  content((19.95pt, 74.01pt), anchor: "east", text(size: 9pt)[x])
  line((28.57pt, 28.57pt), (36.57pt, 28.57pt), (36.57pt, 36.57pt), (28.57pt, 36.57pt), close: true, stroke: 0.6pt + black, fill: none)
  arc((171.43pt, 28.57pt), start: 149.04deg, stop: 180.00deg, radius: 14.00pt, anchor: "origin", stroke: 0.6pt + black)
  content((149.26pt, 34.71pt), text(size: 9pt)[30°])
  arc((28.57pt, 114.29pt), start: 270.00deg, stop: 329.04deg, radius: 14.00pt, anchor: "origin", stroke: 0.6pt + black)
  content((39.90pt, 94.27pt), text(size: 9pt)[60°])
})]),
  alt: "A right triangle with its legs marked x and x times square root of 3. The hypotenuse is marked 2 x. The angles at the top, bottom-left, and bottom-right are labeled 60 degrees, 90 degrees, and 30 degrees.", caption: [The #math.equation(block: false, alt: "3 0 degrees - 6 0 degrees - 9 0 degrees")[$3 0^(∘) "-" 6 0^(∘) "-" 9 0^(∘)$]],
)


    We see that the shortest side is opposite the smallest angle, and the longest side, the hypotenuse, will always be opposite the right angle. There is a set ratio of one side to another side for the #math.equation(block: false, alt: "30 degrees - 60 degrees - 90 degrees")[$30^(∘) "-" 60^(∘) "-" 90^(∘)$] triangle given as #math.equation(block: false, alt: "1 : the square root of 3 : 2 ,")[$1 : sqrt(3) : 2 ,$] or #math.equation(block: false, alt: "x : x the square root of 3 : 2 x .")[$x : x sqrt(3) : 2 x .$] Thus, you only need to know the length of one side to find the other two sides in a #math.equation(block: false, alt: "30 degrees - 60 degrees - 90 degrees")[$30^(∘) "-" 60^(∘) "-" 90^(∘)$] triangle.


    #examplebox("Example 4")[Finding Missing Lengths in a #math.equation(block: false, alt: "3 0 degrees - 6 0 degrees - 9 0 degrees")[$3 0^(∘) "-" 6 0^(∘) "-" 9 0^(∘)$] Triangle][
Find the measures of the missing lengths of the triangle .


        #figure(
  stdfig([#cetz.canvas({
  import cetz.draw: *
  line((0.00pt, 0.00pt), (0.00pt, 100.00pt), (173.20pt, 0.00pt), close: true, fill: none, stroke: 0.7pt + black)
  content((-8.65pt, 52.50pt), anchor: "east", text(size: 9pt)[a])
  content((94.39pt, 54.50pt), anchor: "west", text(size: 9pt)[10])
  content((92.49pt, -6.80pt), anchor: "north", text(size: 9pt)[b])
  line((0.00pt, 0.00pt), (0.00pt, 8.00pt), (8.00pt, 8.00pt), (8.00pt, 0.00pt), close: true, stroke: 0.6pt + black, fill: none)
  arc((173.20pt, 0.00pt), start: 150.00deg, stop: 180.00deg, radius: 14.00pt, anchor: "origin", stroke: 0.6pt + black)
  content((150.98pt, 5.95pt), text(size: 9pt)[30°])
})]),
  alt: "A right triangle with its legs marked a and b. The hypotenuse is marked 10. The angles at the bottom-left and bottom-right are labeled 90 degrees and 30 degrees.", caption: none,
)


      
      #solutionbox[
We can see that this is a #math.equation(block: false, alt: "30 degrees - 60 degrees - 90 degrees")[$30^(∘) "-" 60^(∘) "-" 90^(∘)$] triangle because we have a right angle and a #math.equation(block: false, alt: "30 degrees")[$30^(∘)$] angle. The remaining angle, therefore, must equal #math.equation(block: false, alt: "60 degrees .")[$60^(∘) .$] Because this is a special triangle, we have the ratios of the sides to help us identify the missing lengths. Side #math.equation(block: false, alt: "a")[$a$] is the shortest side, as it is opposite the smallest angle #math.equation(block: false, alt: "30 degrees ,")[$30^(∘) ,$] and we can substitute #math.equation(block: false, alt: "a equals x .")[$a = x .$] The ratios are #math.equation(block: false, alt: "x : x the square root of 3 : 2 x .")[$x : x sqrt(3) : 2 x .$] We have the hypotenuse equaling 10, which corresponds to side #math.equation(block: false, alt: "c")[$c$], and side #math.equation(block: false, alt: "c")[$c$] is equal to 2#math.equation(block: false, alt: "x")[$x$]. Now, we must solve for #math.equation(block: false, alt: "x")[$x$]:


        #math.equation(block: true, alt: "2 x, equals, 10; x, equals, 5")[$2 x & = & 10 \ x & = & 5$]
        Side #math.equation(block: false, alt: "b")[$b$] is equal to #math.equation(block: false, alt: "x the square root of 3")[$x sqrt(3)$] or #math.equation(block: false, alt: "5 the square root of 3 .")[$5 sqrt(3) .$] The lengths are #math.equation(block: false, alt: "5 , 5 the square root of 3 , 10.")[$5 , 5 sqrt(3) , 10.$]
]
]


    
    #examplebox("Example 5")[Applying #math.equation(block: false, alt: "3 0 degrees - 6 0 degrees - 9 0 degrees")[$3 0^(∘) "-" 6 0^(∘) "-" 9 0^(∘)$] Triangle to the Real World][
A city worker leans a 40-foot ladder up against a building at a #math.equation(block: false, alt: "30 degrees")[$30^(∘)$] angle to the ground . How far up the building does the ladder reach?


        #figure(
  stdfig([#cetz.canvas({
  import cetz.draw: *
  content((100.00pt, 147.33pt), text(font: sys.inputs.at("title-font", default: "STIX Two Text"), size: 11pt)[Ladder Against Building])
  line((0.00pt, 0.00pt), (177.78pt, 0.00pt), (0.00pt, 100.00pt), close: true, fill: none, stroke: 0.7pt + black)
  content((94.87pt, -6.73pt), anchor: "north", text(size: 9pt)[x])
  content((96.73pt, 54.41pt), anchor: "west", text(size: 9pt)[40])
  line((0.00pt, 0.00pt), (8.00pt, 0.00pt), (8.00pt, 8.00pt), (0.00pt, 8.00pt), close: true, stroke: 0.6pt + black, fill: none)
  arc((177.78pt, 0.00pt), start: 150.64deg, stop: 180.00deg, radius: 14.00pt, anchor: "origin", stroke: 0.6pt + black)
  content((155.53pt, 5.83pt), text(size: 9pt)[30°])
  content((111.11pt, -11.11pt), anchor: "center", text(size: 9pt)[Building])
  content((88.89pt, 55.56pt), anchor: "center", text(size: 9pt)[Ladder])
})]),
  alt: "An illustration shows a 40-foot ladder placed against a building at an angle of 30 degrees. The vertical distance from the ladder to the ground is marked x. The horizontal distance from the base of the building to the base of the ladder is unknown.", caption: none,
)


      
      #solutionbox[
We have a #math.equation(block: false, alt: "30 degrees - 60 degrees - 90 degrees")[$30^(∘) "-" 60^(∘) "-" 90^(∘)$] triangle, and the hypotenuse is 40 ft. This length is equal to 2#math.equation(block: false, alt: "x")[$x$], where #math.equation(block: false, alt: "x")[$x$] is the shortest side. If #math.equation(block: false, alt: "2 x equals 40")[$2 x = 40$], then #math.equation(block: false, alt: "x equals 20")[$x = 20$]. The ladder is leaning on the wall 20 ft up from the ground.
]
]


    
  
  
    
=== #math.equation(block: false, alt: "4 5 degrees - 4 5 degrees - 9 0 degrees")[$4 5^(∘) "-" 4 5^(∘) "-" 9 0^(∘)$] Triangles


    The #math.equation(block: false, alt: "45 degrees - 45 degrees - 90 degrees")[$45^(∘) "-" 45^(∘) "-" 90^(∘)$] triangle is another #emph[special] triangle such that with the measure of one side we can find the measures of all the sides. The two angles adjacent to the #math.equation(block: false, alt: "90 degrees")[$90^(∘)$] angle are equal, and each measures #math.equation(block: false, alt: "45 degrees .")[$45^(∘) .$] If two angles are equal, so are their opposite sides. The ratio among sides is #math.equation(block: false, alt: "1 : 1 : the square root of 2 ,")[$1 : 1 : sqrt(2) ,$] or #math.equation(block: false, alt: "x : x : x the square root of 2 ,")[$x : x : x sqrt(2) ,$] as shown.


    #figure(
  stdfig([#grid(columns: 2, gutter: 1.5em,
  [#cetz.canvas({
  import cetz.draw: *
  content((100.00pt, 214.00pt), text(font: sys.inputs.at("title-font", default: "STIX Two Text"), size: 11pt)[First triangle (legs 1 and 1)])
  line((40.00pt, 40.00pt), (120.00pt, 40.00pt), (40.00pt, 120.00pt), close: true, fill: none, stroke: 0.7pt + black)
  content((84.02pt, 31.95pt), anchor: "north", text(size: 9pt)[1])
  content((86.36pt, 86.36pt), anchor: "west", text(size: 9pt)[√2])
  content((31.95pt, 84.02pt), anchor: "east", text(size: 9pt)[1])
  line((40.00pt, 40.00pt), (48.00pt, 40.00pt), (48.00pt, 48.00pt), (40.00pt, 48.00pt), close: true, stroke: 0.6pt + black, fill: none)
  arc((120.00pt, 40.00pt), start: 135.00deg, stop: 180.00deg, radius: 14.00pt, anchor: "origin", stroke: 0.6pt + black)
  content((98.75pt, 48.80pt), text(size: 9pt)[45°])
  arc((40.00pt, 120.00pt), start: 270.00deg, stop: 315.00deg, radius: 14.00pt, anchor: "origin", stroke: 0.6pt + black)
  content((48.80pt, 98.75pt), text(size: 9pt)[45°])
})],
  [#cetz.canvas({
  import cetz.draw: *
  content((100.00pt, 214.00pt), text(font: sys.inputs.at("title-font", default: "STIX Two Text"), size: 11pt)[Second triangle (legs x and x)])
  line((28.57pt, 28.57pt), (142.86pt, 28.57pt), (28.57pt, 142.86pt), close: true, fill: none, stroke: 0.7pt + black)
  content((89.74pt, 20.52pt), anchor: "north", text(size: 9pt)[x])
  content((92.08pt, 92.08pt), anchor: "west", text(size: 9pt)[x√2])
  content((20.52pt, 89.74pt), anchor: "east", text(size: 9pt)[x])
  line((28.57pt, 28.57pt), (36.57pt, 28.57pt), (36.57pt, 36.57pt), (28.57pt, 36.57pt), close: true, stroke: 0.6pt + black, fill: none)
  arc((142.86pt, 28.57pt), start: 135.00deg, stop: 180.00deg, radius: 14.00pt, anchor: "origin", stroke: 0.6pt + black)
  content((121.61pt, 37.37pt), text(size: 9pt)[45°])
  arc((28.57pt, 142.86pt), start: 270.00deg, stop: 315.00deg, radius: 14.00pt, anchor: "origin", stroke: 0.6pt + black)
  content((37.37pt, 121.61pt), text(size: 9pt)[45°])
})]
)]),
  alt: "Two right triangles. In the first triangle, the legs measure 1 and 1. The hypotenuse measures the square root of 2. The angles measure 90 degrees, 45 degrees, and 45 degrees. In the second triangle, the legs measure x and x. The hypotenuse measures x times the square root of 2. The angles measure 90 degrees, 45 degrees, and 45 degrees.", caption: [#math.equation(block: false, alt: "4 5 degrees - 4 5 degrees - 9 0 degrees")[$4 5^(∘) "-" 4 5^(∘) "-" 9 0^(∘)$] Triangles],
)


    #examplebox("Example 6")[Finding Missing Lengths of a #math.equation(block: false, alt: "4 5 degrees - 4 5 degrees - 9 0 degrees")[$4 5^(∘) "-" 4 5^(∘) "-" 9 0^(∘)$] Triangle][
Find the measures of the unknown sides in the triangle .


        #figure(
  stdfig([#cetz.canvas({
  import cetz.draw: *
  content((100.00pt, 214.00pt), text(font: sys.inputs.at("title-font", default: "STIX Two Text"), size: 11pt)[Right Triangle])
  line((40.00pt, 40.00pt), (160.00pt, 40.00pt), (40.00pt, 160.00pt), close: true, fill: none, stroke: 0.7pt + black)
  content((104.02pt, 31.95pt), anchor: "north", text(size: 9pt)[3])
  content((106.36pt, 106.36pt), anchor: "west", text(size: 9pt)[c])
  content((31.95pt, 104.02pt), anchor: "east", text(size: 9pt)[a])
  line((40.00pt, 40.00pt), (48.00pt, 40.00pt), (48.00pt, 48.00pt), (40.00pt, 48.00pt), close: true, stroke: 0.6pt + black, fill: none)
  content((49.90pt, 49.90pt), text(size: 9pt)[90°])
  arc((160.00pt, 40.00pt), start: 135.00deg, stop: 180.00deg, radius: 14.00pt, anchor: "origin", stroke: 0.6pt + black)
  content((138.75pt, 48.80pt), text(size: 9pt)[45°])
  arc((40.00pt, 160.00pt), start: 270.00deg, stop: 315.00deg, radius: 14.00pt, anchor: "origin", stroke: 0.6pt + black)
  content((48.80pt, 138.75pt), text(size: 9pt)[45°])
  content((44.00pt, 44.00pt), anchor: "south-west", text(size: 9pt)[A])
  content((156.00pt, 44.00pt), anchor: "south-east", text(size: 9pt)[B])
  content((44.00pt, 156.00pt), anchor: "north-west", text(size: 9pt)[C])
})]),
  alt: "A right triangle. The legs measure a and 3. The hypotenuse measures c. The angles are marked 90 degrees, 45 degrees, and 45 degrees.", caption: none,
)


      
      #solutionbox[
Because we have a #math.equation(block: false, alt: "45 degrees - 45 degrees - 90 degrees")[$45^(∘) "-" 45^(∘) "-" 90^(∘)$] triangle, we know that the two legs are equal in length and the hypotenuse is a product of one of the legs and #math.equation(block: false, alt: "the square root of 2 .")[$sqrt(2) .$] One leg measures 3, so the other leg, #math.equation(block: false, alt: "a")[$a$], measures 3. Remember the ratio of #math.equation(block: false, alt: "x : x : x the square root of 2 .")[$x : x : x sqrt(2) .$] Then, the hypotenuse, #math.equation(block: false, alt: "c")[$c$], equals #math.equation(block: false, alt: "3 the square root of 2 .")[$3 sqrt(2) .$]
]
]


    
  
  
    
=== Trigonometry Functions


    Trigonometry developed around 200 BC from a need to determine distances and to calculate the measures of angles in the fields of astronomy and surveying. Trigonometry is about the relationships (or ratios) of angle measurements to side lengths in primarily right triangles. However, trigonometry is useful in calculating missing side lengths and angles in other triangles and many applications.


    #notebox("Checkpoint", rgb("#059669"), rgb("#007942"), rgb("#EAF3EC"))[
#emph[NOTE: You will need either a scientific calculator or a graphing calculator for this section. It must have the capability to calculate trigonometric functions and express angles in degrees.]
]


    Trigonometry is based on three functions. We title these functions using the following abbreviations:


    - #math.equation(block: false, alt: "sin equals sine")[$sin = "sine"$]
- #math.equation(block: false, alt: "cos equals cosine")[$cos = "cosine"$]
- #math.equation(block: false, alt: "tan equals tangent")[$tan = "tangent"$]


    Letting #math.equation(block: false, alt: "r equals the square root of x squared plus y squared ,")[$r = sqrt(x^(2) + y^(2)) ,$] which is the hypotenuse of a right triangle, we have . The functions are given in terms of #math.equation(block: false, alt: "x")[$x$], #math.equation(block: false, alt: "y")[$y$], and #math.equation(block: false, alt: "r")[$r$], and in terms of sides relative to the angle, like opposite, adjacent, and the hypotenuse.


    #figure(table(
  columns: 3, align: left, inset: 6pt,
  [#math.equation(block: false, alt: "sin θ equals the fraction y over r equals the fraction o p p over h y p")[$sin θ = frac(y, r) = frac(o p p, h y p)$]], [#math.equation(block: false, alt: "cos θ equals the fraction x over r equals the fraction a d j over h y p")[$cos θ = frac(x, r) = frac(a d j, h y p)$]], [#math.equation(block: false, alt: "tan θ equals the fraction y over x equals the fraction o p p over a d j")[$tan θ = frac(y, x) = frac(o p p, a d j)$]],
))


    We will be applying the sine function, cosine function, and tangent function to find side lengths and angle measurements for triangles we cannot solve using any of the techniques we have studied to this point. In , we have an illustration mainly to identify #math.equation(block: false, alt: "r")[$r$] and the sides labeled #math.equation(block: false, alt: "x")[$x$] and #math.equation(block: false, alt: "y")[$y$].


    #figure(
  stdfig([#cetz.canvas({
  import cetz.draw: *
  line((0.00pt, 0.00pt), (230.00pt, 0.00pt), stroke: 0.7pt + black)
  line((0.00pt, 0.00pt), (115.00pt, 85.00pt), stroke: 0.7pt + black)
  circle((0.00pt, 0.00pt), radius: 3.0pt, fill: black, stroke: none)
  circle((115.00pt, 85.00pt), radius: 3.0pt, fill: rgb("#D62128"), stroke: none)
  content((109.06pt, 93.04pt), anchor: "south-east", text(size: 8pt)[r])
  line((0.00pt, 0.00pt), (234.00pt, 0.00pt), stroke: 0.7pt + black, mark: (end: ">", fill: black, size: 0.18))
  line((0.00pt, 0.00pt), (0.00pt, 174.00pt), stroke: 0.7pt + black, mark: (end: ">", fill: black, size: 0.18))
  line((23.00pt, 0.00pt), (23.00pt, -3.00pt), stroke: 0.5pt + black)
  content((23.00pt, -5.00pt), anchor: "north", text(size: 8pt)[1])
  line((46.00pt, 0.00pt), (46.00pt, -3.00pt), stroke: 0.5pt + black)
  content((46.00pt, -5.00pt), anchor: "north", text(size: 8pt)[2])
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})]),
  alt: "Two rays are plotted on an x y coordinate plane. Both rays begin at the origin. The first ray lies on the positive x-axis. The second ray lies in the first quadrant and a point, (x, y) is marked on the ray. The angle made by the two rays is marked theta. The distance from the origin to the point along the ray is labeled r.", caption: [Angle #math.equation(block: false, alt: "θ")[$θ$]],
)


    An angle #math.equation(block: false, alt: "θ")[$θ$] sweeps out in a counterclockwise direction from the positive #math.equation(block: false, alt: "x")[$x$]-axis and stops when the angle reaches the desired measurement. That ray extending from the origin that marks #math.equation(block: false, alt: "θ degrees")[$θ^(∘)$] is called the terminal side because that is where the angle terminates. Regardless of the information given in the triangle, we can find all missing sides and angles using the trigonometric functions. For example, in , we will solve for the missing sides.


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    Let’s use the trigonometric functions to find the sides #math.equation(block: false, alt: "x")[$x$] and #math.equation(block: false, alt: "y")[$y$]. As long as your calculator mode is set to degrees, you do not have to enter the degree symbol. First, let’s solve for #math.equation(block: false, alt: "y")[$y$].


    We have #math.equation(block: false, alt: "sin θ equals the fraction y over r ,")[$sin θ = frac(y, r) ,$] and #math.equation(block: false, alt: "θ equals 60 degrees .")[$θ = 60^(∘) .$] Then,


    #math.equation(block: true, alt: "sin 60 degrees, equals, the fraction y over 2; 2 sin 60 degrees, equals, y; 1.732, equals, y; the square root of 3, equals, y")[$sin 60^(∘) & = & frac(y, 2) \ 2 sin 60^(∘) & = & y \ 1.732 & = & y \ sqrt(3) & = & y$]
    Next, let’s find #math.equation(block: false, alt: "x")[$x$]. This is the cosine function. We have #math.equation(block: false, alt: "cos θ equals the fraction x over r .")[$cos θ = frac(x, r) .$] Then,


    #math.equation(block: true, alt: "cos 60 degrees, equals, the fraction x over 2; 2 cos 60 degrees, equals, x; equals, 1")[$cos 60^(∘) & = & frac(x, 2) \ 2 cos 60^(∘) & = & x \  & = & 1$]
    Now we have all sides, #math.equation(block: false, alt: "1 , the square root of 3 , 2.")[$1 , sqrt(3) , 2.$] You can also check the sides using the #math.equation(block: false, alt: "30 degrees - 60 degrees - 90 degrees")[$30^(∘) "-" 60^(∘) "-" 90^(∘)$] ratio of #math.equation(block: false, alt: "1 : the square root of 3 : 2.")[$1 : sqrt(3) : 2.$]  is a list of common angles, which you should find helpful.


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  [#math.equation(block: false, alt: "sin 30 degrees equals the fraction 1 over 2")[$sin 30^(∘) = frac(1, 2)$]], [#math.equation(block: false, alt: "cos 30 degrees equals the fraction the square root of 3 over 2")[$cos 30^(∘) = frac(sqrt(3), 2)$]],
  [#math.equation(block: false, alt: "sin 45 degrees equals the fraction the square root of 2 over 2")[$sin 45^(∘) = frac(sqrt(2), 2)$]], [#math.equation(block: false, alt: "cos 45 degrees equals the fraction the square root of 2 over 2")[$cos 45^(∘) = frac(sqrt(2), 2)$]],
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  [#math.equation(block: false, alt: "sin 90 degrees equals 1")[$sin 90^(∘) = 1$]], [#math.equation(block: false, alt: "cos 90 degrees equals 0")[$cos 90^(∘) = 0$]],
))


    #examplebox("Example 7")[Using Trigonometric Functions][
Find the lengths of the missing sides for the triangle .


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)


      
      #solutionbox[
We have a #math.equation(block: false, alt: "55 degrees")[$55^(∘)$] angle, and the length of the triangle on the #math.equation(block: false, alt: "x")[$x$]-axis is 6 units.


        #strong[Step 1:] To find the length of #math.equation(block: false, alt: "r")[$r$], we can use the cosine function, as #math.equation(block: false, alt: "cos θ equals the fraction x over r .")[$cos θ = frac(x, r) .$] We manipulate this equation a bit to solve for #math.equation(block: false, alt: "r")[$r$]:


        #math.equation(block: true, alt: "cos open parenthesis 55 degrees close parenthesis, equals, the fraction 6 over r; r cos open parenthesis 55 degrees close parenthesis, equals, 6; r, equals, the fraction 6 over cos open parenthesis 55 degrees close parenthesis; r, equals, the fraction 6 over 0.5736 equals 10.46")[$cos ( 55^(∘) ) & = & frac(6, r) \ r cos ( 55^(∘) ) & = & 6 \ r & = & frac(6, cos ( 55^(∘) )) \ r & = & frac(6, 0.5736) = 10.46$]
        
        #strong[Step 2:] We can use the Pythagorean Theorem to find the length of #math.equation(block: false, alt: "y")[$y$]. Prove that your answers are correct by using other trigonometric ratios:


        #math.equation(block: true, alt: "6 squared plus y squared, equals, 10.46 squared; y squared, equals, 109.4 minus 36; y, equals, 8.57")[$6^(2) + y^(2) & = & 10.46^(2) \ y^(2) & = & 109.4 − 36 \ y & = & 8.57$]
        
        #strong[Step 3:] Now that we have #math.equation(block: false, alt: "y")[$y$], we can use the sine function to prove that #math.equation(block: false, alt: "r")[$r$] is correct. We have #math.equation(block: false, alt: "sin θ equals the fraction y over r .")[$sin θ = frac(y, r) .$]


        #math.equation(block: true, alt: "sin open parenthesis 55 degrees close parenthesis, equals, the fraction 8.57 over r; r sin open parenthesis 55 degrees close parenthesis, equals, 8.57; r, equals, the fraction 8.57 over sin open parenthesis 55 degrees close parenthesis; equals, the fraction 8.57 over 0.819 equals 10.46")[$sin ( 55^(∘) ) & = & frac(8.57, r) \ r sin ( 55^(∘) ) & = & 8.57 \ r & = & frac(8.57, sin ( 55^(∘) )) \  & = & frac(8.57, 0.819) = 10.46$]
]
]


    
    To find angle measurements when we have two side measurements, we use the inverse trigonometric functions symbolized as #math.equation(block: false, alt: "sin to the power minus 1 ,")[$sin^(− 1) ,$] #math.equation(block: false, alt: "cos to the power minus 1 ,")[$cos^(− 1) ,$] or #math.equation(block: false, alt: "tan to the power minus 1 .")[$tan^(− 1) .$] The –1 looks like an exponent, but it means inverse. For example, in the previous example, we had #math.equation(block: false, alt: "x equals 6")[$x = 6$] and #math.equation(block: false, alt: "r equals 10.46.")[$r = 10.46.$] To find what angle has these values, enter the values for the inverse cosine function #math.equation(block: false, alt: "cos to the power minus 1 open parenthesis the fraction x over r close parenthesis")[$cos^(− 1) ( frac(x, r) )$] in your calculator:


    #math.equation(block: true, alt: "cos to the power minus 1 open parenthesis the fraction 6 over 10.46 close parenthesis equals 55 degrees .")[$cos^(− 1) ( frac(6, 10.46) ) = 55^(∘) .$]
    You can also use the inverse sine function and enter the values of #math.equation(block: false, alt: "sin to the power minus 1 open parenthesis the fraction y over r close parenthesis")[$sin^(− 1) ( frac(y, r) )$] in your calculator given #math.equation(block: false, alt: "y equals 8.57")[$y = 8.57$] and #math.equation(block: false, alt: "r equals 10.46.")[$r = 10.46.$] We have


    #math.equation(block: true, alt: "sin to the power minus 1 open parenthesis the fraction 8.57 over 10.46 close parenthesis equals 55 degrees .")[$sin^(− 1) ( frac(8.57, 10.46) ) = 55^(∘) .$]
    Finally, we can also use the inverse tangent function. Recall #math.equation(block: false, alt: "tan θ equals the fraction y over x .")[$tan θ = frac(y, x) .$] We have


    #math.equation(block: true, alt: "tan to the power minus 1 open parenthesis the fraction 8.57 over 6 close parenthesis equals 55 degrees .")[$tan^(− 1) ( frac(8.57, 6) ) = 55^(∘) .$]
    
    #examplebox("Example 8")[Solving for Lengths in a Right Triangle][
Solve for the lengths of a right triangle in which #math.equation(block: false, alt: "θ equals 30 degrees")[$θ = 30^(∘)$] and #math.equation(block: false, alt: "r equals 6")[$r = 6$] .


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  line((-3.00pt, 153.00pt), (0.00pt, 153.00pt), stroke: 0.5pt + black)
  content((-5.00pt, 153.00pt), anchor: "east", text(size: 8pt)[9])
  line((-3.00pt, 170.00pt), (0.00pt, 170.00pt), stroke: 0.5pt + black)
  content((-5.00pt, 170.00pt), anchor: "east", text(size: 8pt)[10])
})]),
  alt: "A ray is plotted on an x y coordinate plane. The ray lies in the first quadrant and a vertical dashed line is extended from a point on the ray to meet the x-axis and it makes a right angle. This dashed line measures a. The horizontal distance from the origin to the line is marked b. The angle made by the ray with the x-axis is marked 30 degrees. The distance from the origin to the point along the ray is labeled 6.", caption: none,
)


      
      #solutionbox[
#strong[Step 1:] To find side #math.equation(block: false, alt: "a")[$a$], we use the sine function:


        #math.equation(block: true, alt: "sin 30 degrees, equals, the fraction a over 6; 6 sin 30 degrees, equals, a equals 3")[$sin 30^(∘) & = & frac(a, 6) \ 6 sin 30^(∘) & = & a = 3$]
        
        #strong[Step 2:] To find #math.equation(block: false, alt: "b")[$b$], we use the cosine function:


        #math.equation(block: true, alt: "cos 30 degrees, equals, the fraction b over 6; 6 cos 30 degrees, equals, b equals 5.196")[$cos 30^(∘) & = & frac(b, 6) \ 6 cos 30^(∘) & = & b = 5.196$]
        
        #strong[Step 3:] Since this is a #math.equation(block: false, alt: "30 degrees - 60 degrees - 90 degrees")[$30^(∘) "-" 60^(∘) "-" 90^(∘)$] triangle and side #math.equation(block: false, alt: "b")[$b$] should equal #math.equation(block: false, alt: "x the square root of 3 ,")[$x sqrt(3) ,$] if we input 3 for #math.equation(block: false, alt: "x")[$x$], we have #math.equation(block: false, alt: "b equals 3 the square root of 3 .")[$b = 3 sqrt(3) .$] Put this in your calculator and you will get #math.equation(block: false, alt: "3 the square root of 3 equals 5.196.")[$3 sqrt(3) = 5.196.$]
]
]


    
    #examplebox("Example 9")[Finding Altitude][
A small plane takes off from an airport at an angle of #math.equation(block: false, alt: "31.3 degrees")[$31.3^(∘)$] to the ground. About two-thirds of a mile (3,520 ft) from the airport is an 1,100-ft peak in the flight path of the plane . If the plane continues that angle of ascent, find its altitude when it is above the peak, and how far it will be above the peak.


        #figure(
  stdfig([#cetz.canvas({
  import cetz.draw: *
  line((16.67pt, 16.67pt), (183.33pt, 16.67pt), stroke: 0.7pt + black)
  line((183.33pt, 16.67pt), (183.33pt, 75.00pt), stroke: 0.7pt + black)
  line((16.67pt, 16.67pt), (166.67pt, 108.33pt), stroke: 0.7pt + black)
  line((166.67pt, 108.33pt), (160.24pt, 107.34pt), (162.85pt, 103.07pt), close: true, fill: black, stroke: 0.7pt + black)
  arc((16.67pt, 16.67pt), start: 0.00deg, stop: 31.43deg, radius: 14.00pt, anchor: "origin", stroke: 0.6pt + black)
  content((38.81pt, 22.90pt), text(size: 9pt)[31.3°])
  content((100.00pt, 8.33pt), anchor: "north", text(size: 9pt)[3520 feet])
  content((191.67pt, 45.83pt), anchor: "west", text(size: 9pt)[1100 feet])
})]),
  alt: "A horizontal line measures 3520 feet. A ray originates from the left end of the line and it makes an angle of 31.3 degrees. A vertical line measuring 1100 feet is on the right end of the line. A plane is flying above the line.", caption: none,
)


      
      #solutionbox[
To solve this problem, we use the tangent function:


        #math.equation(block: true, alt: "tan 31.3 degrees, equals, the fraction x over 3,520; 3,520 tan 31.3 degrees, equals, 2,140")[$tan 31.3^(∘) & = & frac(x, 3","520) \ 3,520 tan 31.3^(∘) & = & 2,140$]
        The plane’s altitude when passing over the peak is 2,140 ft, and it is 1,040 ft above the peak.
]
]


    
    #examplebox("Example 10")[Finding Unknown Sides and Angles][
Suppose you have two known sides, but do not know the measure of any angles except for the right angle . Find the measure of the unknown angles and the third side.


        #figure(
  stdfig([#cetz.canvas({
  import cetz.draw: *
  line((25.00pt, 25.00pt), (175.00pt, 25.00pt), (25.00pt, 125.00pt), close: true, fill: none, stroke: 0.7pt + black)
  content((105.40pt, 17.80pt), anchor: "north", text(size: 9pt)[6])
  content((107.49pt, 79.99pt), anchor: "west", text(size: 9pt)[c])
  content((16.46pt, 77.85pt), anchor: "east", text(size: 9pt)[4])
  arc((175.00pt, 25.00pt), start: 146.31deg, stop: 180.00deg, radius: 14.00pt, anchor: "origin", stroke: 0.6pt + black)
  content((152.99pt, 31.67pt), text(size: 9pt)[θ])
  line((25.00pt, 25.00pt), (33.00pt, 25.00pt), (33.00pt, 33.00pt), (25.00pt, 33.00pt), close: true, stroke: 0.6pt + black, fill: none)
})]),
  alt: "A right triangle with its legs marked 4 and 6. The hypotenuse is marked c. The angle made by the hypotenuse and the horizontal leg is marked theta.", caption: none,
)


      
      #solutionbox[
#strong[Step 1:] We can find the third side using the Pythagorean Theorem:


        #math.equation(block: true, alt: "6 squared plus 4 squared, equals, c squared; 52, equals, c squared; 2 the square root of 13, equals, c")[$6^(2) + 4^(2) & = & c^(2) \ 52 & = & c^(2) \ 2 sqrt(13) & = & c$]
        Now, we have all three sides.


        #strong[Step 2:] To find #math.equation(block: false, alt: "θ ,")[$θ ,$] we will first find #math.equation(block: false, alt: "sin θ .")[$sin θ .$]


        #math.equation(block: true, alt: "sin θ, equals, the fraction o p p over h y p; equals, the fraction 4 over 2 the square root of 13; equals, the fraction 2 over the square root of 13 .")[$sin θ & = & frac(o p p, h y p) \  & = & frac(4, 2 sqrt(13)) \  & = & frac(2, sqrt(13)) .$]
        The angle #math.equation(block: false, alt: "θ")[$θ$] is the angle whose sine is #math.equation(block: false, alt: "the fraction 2 over the square root of 13 .")[$frac(2, sqrt(13)) .$]


        #strong[Step 3:] To find #math.equation(block: false, alt: "θ")[$θ$], we use the inverse sine function:


        #math.equation(block: true, alt: "θ, equals, sin to the power minus 1 open parenthesis the fraction 2 over the square root of 13 close parenthesis; equals, 33.7 degrees")[$θ & = & sin^(− 1) ( frac(2, sqrt(13)) ) \  & = & 33.7^(∘)$]
        
        #strong[Step 4:] To find the last angle, we just subtract: #math.equation(block: false, alt: "180 degrees minus 90 degrees minus 33.7 degrees equals 56.3 degrees")[$180^(∘) − 90^(∘) − 33.7^(∘) = 56.3^(∘)$].
]
]


    
  
  
    
=== Angle of Elevation and Angle of Depression


    Other problems that involve trigonometric functions include calculating the #strong[angle of elevation] and the #strong[angle of depression]. These are very common applications in everyday life. The angle of elevation is the angle formed by a horizontal line and the line of sight from an observer to some object at a higher level. The angle of depression is the angle formed by a horizontal line and the line of sight from an observer to an object at a lower level.


    #examplebox("Example 11")[Finding the Angle of Elevation][
A guy wire of length 110 meters runs from the top of an antenna to the ground . If the angle of elevation of an observer to the top of the antenna is #math.equation(block: false, alt: "43 degrees ,")[$43^(∘) ,$] how high is the antenna?


        #figure(
  stdfig([#cetz.canvas({
  import cetz.draw: *
  content((100.00pt, 164.00pt), text(font: sys.inputs.at("title-font", default: "STIX Two Text"), size: 11pt)[Right Triangle - Antenna and Guy Wire])
  line((25.00pt, 25.00pt), (175.00pt, 25.00pt), (175.00pt, 125.00pt), close: true, fill: none, stroke: 0.7pt + black)
  content((183.54pt, 77.85pt), anchor: "west", text(size: 9pt)[height])
  content((92.51pt, 79.99pt), anchor: "east", text(size: 9pt)[110 m])
  arc((25.00pt, 25.00pt), start: 0.00deg, stop: 33.69deg, radius: 14.00pt, anchor: "origin", stroke: 0.6pt + black)
  content((47.01pt, 31.67pt), text(size: 9pt)[43°])
  line((175.00pt, 25.00pt), (167.00pt, 25.00pt), (167.00pt, 33.00pt), (175.00pt, 33.00pt), close: true, stroke: 0.6pt + black, fill: none)
  content((29.00pt, 29.00pt), anchor: "south-west", text(size: 9pt)[Observer])
  content((175.00pt, 75.00pt), anchor: "west", text(size: 9pt)[antenna])
})]),
  alt: "An illustration shows a right triangle. The vertical leg of the triangle represents the height of the antenna. The hypotenuse represents a guy wire of 110 meters. An observer is at the bottom-left vertex of the triangle. The bottom-left and bottom-right triangles are marked 43 degrees and 90 degrees.", caption: none,
)


      
      #solutionbox[
We are looking for the height of the tower. This corresponds to the #math.equation(block: false, alt: "y")[$y$]-value, so we will use the sine function:


        #math.equation(block: true, alt: "sin 43 degrees, equals, the fraction y over 110; 110 sin 43 degrees, equals, y; 75, equals, y")[$sin 43^(∘) & = & frac(y, 110) \ 110 sin 43^(∘) & = & y \ 75 & = & y$]
        The tower is 75 m high.
]
]


    
    #examplebox("Example 12")[Finding Angle of Elevation][
You are sitting on the grass flying a kite on a 50-foot string . The angle of elevation is #math.equation(block: false, alt: "60 degrees .")[$60^(∘) .$] How high above the ground is the kite?


        #figure(
  stdfig([#cetz.canvas({
  import cetz.draw: *
  content((63.64pt, 214.00pt), text(font: sys.inputs.at("title-font", default: "STIX Two Text"), size: 11pt)[Right Triangle with Kite])
  line((18.18pt, 18.18pt), (109.09pt, 18.18pt), (18.18pt, 175.64pt), close: true, fill: none, stroke: 0.7pt + black)
  content((68.14pt, 104.70pt), anchor: "south", text(size: 9pt)[50 feet])
  line((18.18pt, 18.18pt), (26.18pt, 18.18pt), (26.18pt, 26.18pt), (18.18pt, 26.18pt), close: true, stroke: 0.6pt + black, fill: none)
  arc((109.09pt, 18.18pt), start: 120.00deg, stop: 180.00deg, radius: 14.00pt, anchor: "origin", stroke: 0.6pt + black)
  content((89.17pt, 29.68pt), text(size: 9pt)[60°])
  circle((18.18pt, 175.64pt), radius: 1.8pt, fill: black, stroke: none)
  content((18.18pt, 171.64pt), anchor: "north", text(size: 9pt)[Kite])
})]),
  alt: "A right triangle. A kite is placed at the top of the triangle. The hypotenuse measures 50 feet. The angle made by the hypotenuse and the horizontal leg measures 60 degrees.", caption: none,
)


      
      #solutionbox[
We can solve this using the sine function, #math.equation(block: false, alt: "sin θ equals the fraction o p p over h y p .")[$sin θ = frac(o p p, h y p) .$]


        #math.equation(block: true, alt: "sin 60 degrees, equals, the fraction x over 50; 50 sin 60 degrees, equals, x; equals, 43.3 ft")[$sin 60^(∘) & = & frac(x, 50) \ 50 sin 60^(∘) & = & x \  & = & 43.3 #h(0.28em) "ft"$]
]
]


    
    #notebox("People in Mathematics", rgb("#183B6F"), rgb("#183B6F"), rgb("#EFF1F5"))[
#emph[Pythagoras and the Pythagoreans]


      The Pythagorean Theorem is so widely used that most people assume that Pythagoras (570–490 BC) discovered it. The philosopher and mathematician uncovered evidence of the right triangle concepts in the teachings of the Babylonians dating around 1900 BC. However, it was Pythagoras who found countless applications of the theorem leading to advances in geometry, architecture, astronomy, and engineering.


      Among his accolades, Pythagoras founded a school for the study of mathematics and music. Students were called the Pythagoreans, and the school’s teachings could be classified as a religious indoctrination just as much as an academic experience. Pythagoras believed that spirituality and science coexist, that the intellectual mind is superior to the senses, and that intuition should be honored over observation.


      Pythagoras was convinced that the universe could be defined by numbers, and that the natural world was based on mathematics. His primary belief was #emph[All is Number.] He even attributed certain qualities to certain numbers, such as the number 8 represented justice and the number 7 represented wisdom. There was a quasi-mythology that surrounded Pythagoras. His followers thought that he was more of a spiritual being, a sort of mystic that was all-knowing and could travel through time and space. Some believed that Pythagoras had mystical powers, although these beliefs were never substantiated.


      Pythagoras and his followers contributed more ideas to the field of mathematics, music, and astronomy besides the Pythagorean Theorem. The Pythagoreans are credited with the discovery of irrational numbers and of proving that the morning star was the planet Venus and not a star at all. They are also credited with the discovery of the musical scale and that different strings made different sounds based on their length. Some other concepts attributed to the Pythagoreans include the properties relating to triangles other than the right triangle, one of which is that the sum of the interior angles of a triangle equals #math.equation(block: false, alt: "180 degrees .")[$180^(∘) .$] These geometric principles, proposed by the Pythagoreans, were proven 200 years later by Euclid.
]


    #notebox("Who Knew?", rgb("#183B6F"), rgb("#183B6F"), rgb("#EFF1F5"))[
#emph[A Visualization of the Pythagorean Theorem]


      In , which is one of the more popular visualizations of the Pythagorean Theorem, we see that square #math.equation(block: false, alt: "a")[$a$] is attached to side #math.equation(block: false, alt: "a")[$a$]; square #math.equation(block: false, alt: "b")[$b$] is attached to side #math.equation(block: false, alt: "b")[$b$]; and the largest square, square #math.equation(block: false, alt: "c")[$c$], is attached to side #math.equation(block: false, alt: "c")[$c$]. Side #math.equation(block: false, alt: "a")[$a$] measures 3 cm in length, side #math.equation(block: false, alt: "b")[$b$] measures 4 cm in length, and side #math.equation(block: false, alt: "c")[$c$] measures 5 cm in length. By definition, the area of square #math.equation(block: false, alt: "a")[$a$] measures 9 square units, the area of square #math.equation(block: false, alt: "b")[$b$] measures 16 square units, and the area of square #math.equation(block: false, alt: "c")[$c$] measures 25 square units. Substitute the values given for the areas of the three squares into the Pythagorean Theorem and we have


      #math.equation(block: true, alt: "a squared plus b squared, equals, c squared; 3 squared plus 4 squared, equals, 5 squared; 9 plus 16, equals, 25")[$a^(2) + b^(2) & = & c^(2) \ 3^(2) + 4^(2) & = & 5^(2) \ 9 + 16 & = & 25$]
      Thus, the sum of the squares of the two legs of a right triangle is equal to the square of the hypotenuse, as stated in the Pythagorean Theorem.


      #figure(
  stdfig([#cetz.canvas({
  import cetz.draw: *
  content((92.31pt, 214.00pt), text(font: sys.inputs.at("title-font", default: "STIX Two Text"), size: 11pt)[Right Triangle with Squares on Each Side])
  line((61.54pt, 76.92pt), (123.08pt, 76.92pt), (61.54pt, 123.08pt), close: true, fill: none, stroke: 0.7pt + black)
  content((97.30pt, 69.43pt), anchor: "north", text(size: 9pt)[b = 4])
  content((99.51pt, 105.40pt), anchor: "west", text(size: 9pt)[c = 5])
  content((53.11pt, 103.16pt), anchor: "east", text(size: 9pt)[a = 3])
  line((61.54pt, 76.92pt), (123.08pt, 76.92pt), (123.08pt, 15.38pt), (61.54pt, 15.38pt), close: true, fill: none, stroke: 0.7pt + black)
  line((61.54pt, 76.92pt), (61.54pt, 123.08pt), (15.38pt, 123.08pt), (15.38pt, 76.92pt), close: true, fill: none, stroke: 0.7pt + black)
  line((123.08pt, 76.92pt), (169.23pt, 138.46pt), (107.69pt, 184.62pt), (61.54pt, 123.08pt), close: true, fill: none, stroke: 0.7pt + black)
  line((61.54pt, 76.92pt), (69.54pt, 76.92pt), (69.54pt, 84.92pt), (61.54pt, 84.92pt), close: true, stroke: 0.6pt + black, fill: none)
})]),
  alt: "A right triangle with its legs marked a equals 3 and b equals 4. The hypotenuse is marked c equals 5. A square is placed along each side.", caption: none,
)
]


  
  
  
  
    
=== Key Terms


    - right triangle
- sine
- cosine
- tangent


  
  
    
=== Key Concepts


    - The Pythagorean Theorem is applied to right triangles and is used to find the measure of the legs and the hypotenuse according the formula #math.equation(block: false, alt: "a squared plus b squared equals c squared ,")[$a^(2) + b^(2) = c^(2) ,$] where #emph[c] is the hypotenuse.
- To find the measure of the sides of a #emph[special] angle, such as a #math.equation(block: false, alt: "30 degrees - 60 degrees - 90 degrees")[$30^(∘) "-" 60^(∘) "-" 90^(∘)$] triangle, use the ratio #math.equation(block: false, alt: "x : x the square root of 3 : 2 x ,")[$x : x sqrt(3) : 2 x ,$] where each of the three sides is associated with the opposite angle and 2#math.equation(block: false, alt: "x")[$x$] is associated with the hypotenuse, opposite the #math.equation(block: false, alt: "90 degrees")[$90^(∘)$] angle.
- To find the measure of the sides of the second #emph[special] triangle, the #math.equation(block: false, alt: "45 degrees - 45 degrees - 90 degrees")[$45^(∘) "-" 45^(∘) "-" 90^(∘)$] triangle, use the ratio #math.equation(block: false, alt: "x : x : x the square root of 2 ,")[$x : x : x sqrt(2) ,$] where each of the three sides is associated with the opposite angle and #math.equation(block: false, alt: "x the square root of 2")[$x sqrt(2)$] is associated with the hypotenuse, opposite the #math.equation(block: false, alt: "90 degrees")[$90^(∘)$] angle.
- The primary trigonometric functions are #math.equation(block: false, alt: "sin θ equals the fraction o p p over h y p ,")[$sin θ = frac(o p p, h y p) ,$] #math.equation(block: false, alt: "cos θ equals the fraction a d j over h y p ,")[$cos θ = frac(a d j, h y p) ,$] and #math.equation(block: false, alt: "tan θ equals the fraction o p p over a d j .")[$tan θ = frac(o p p, a d j) .$]
- Trigonometric functions can be used to find either the length of a side or the measure of an angle in a right triangle, and in applications such as the angle of elevation or the angle of depression formed using right triangles.


  
  
    
=== Formula


    The Pythagorean Theorem states


    #math.equation(block: true, alt: "a squared plus b squared equals c squared")[$a^(2) + b^(2) = c^(2)$]
    where #math.equation(block: false, alt: "a")[$a$] and #math.equation(block: false, alt: "b")[$b$] are two sides (legs) of a right triangle and #math.equation(block: false, alt: "c")[$c$] is the hypotenuse.


  
  
    
=== Projects


    + One of the reasons so many formulas in geometry were discovered was because of the importance in finding measurements of lengths, areas, perimeter, and angles. Find at least five examples of how geometry can be used in practical applications today.
+ Who were the Pythagoreans? Why did this society exist? Explore what they did and discuss some of their beliefs.