#set document(title: "1.1 Angles and Triangles", author: "OpenStax / XYZ Homework") #set page(width: 8.5in, height: auto, margin: 1in) #import "@preview/cetz:0.5.2" #set text(font: ("STIX Two Text", "Libertinus Serif", "New Computer Modern"), size: 10.5pt, lang: "en") #show math.equation: set text(font: ("STIX Two Math", "New Computer Modern Math")) #set par(justify: true, leading: 0.62em, spacing: 0.9em) #set enum(spacing: 1.1em) // room between list items so tall inline fractions don't collide #set list(spacing: 1.1em) #set table(stroke: 0.5pt + rgb("#c7ccd3")) #let BLUE = rgb("#183B6F") // brand navy — section bars + example/solution labels (white on navy 11.09:1) #let ORANGE = rgb("#A94509") // brand primary-700 — AA-safe deep orange for TEXT (5.93:1 on white; raw brand #F37021 is 2.94:1 and must never carry text) #let RED = rgb("#DC2626") // brand error-600 #let GREEN = rgb("#059669") // brand success-600 (decoration only; small green text uses green-text #007942) #show heading.where(level: 1): it => block(width: 100%, above: 0pt, below: 16pt, fill: gradient.linear(BLUE, rgb("#2C5AA0")), inset: (x: 14pt, y: 12pt), radius: 3pt, text(fill: white, weight: "bold", size: 19pt, it.body)) #show heading.where(level: 2): it => block(width: 100%, above: 18pt, below: 10pt, fill: BLUE, inset: (x: 10pt, y: 6pt), radius: 2pt, text(fill: white, weight: "bold", size: 12pt, it.body)) #show heading.where(level: 3): it => text(fill: ORANGE, weight: "bold", size: 12.5pt, it.body) #show heading.where(level: 4): it => text(fill: BLUE, weight: "bold", size: 10.5pt, it.body) #let examplebox(label, title, body) = block(width: 100%, breakable: true, fill: rgb("#EFF1F5"), stroke: 0.5pt + rgb("#CFDDF0"), radius: 4pt, inset: 10pt, above: 12pt, below: 12pt)[ #block(below: 6pt)[#box(fill: BLUE, inset: (x: 6pt, y: 2pt), radius: 2pt, text(fill: white, weight: "bold", size: 8.5pt, label)) #h(0.4em) #strong[#title]] #body] // rail = decorative left rule (raw brand token); labelcolor = AA-safe label text shade #let notebox(label, rail, labelcolor, tint, body) = block(width: 100%, breakable: true, fill: tint, stroke: (left: 3pt + rail), inset: (left: 10pt, rest: 8pt), radius: (right: 4pt), above: 11pt, below: 11pt)[ #text(fill: labelcolor, weight: "bold", size: 7.5pt, tracking: 0.5pt)[#upper(label)] #linebreak() #body] #let solutionbox(body) = block(above: 4pt, below: 8pt)[ #text(fill: BLUE, weight: "bold", size: 8.5pt)[Solution] #linebreak() #body] #let figph(msg) = block(width: 100%, height: 60pt, fill: rgb("#f6f7f9"), stroke: (paint: rgb("#c7ccd3"), dash: "dashed"), radius: 4pt, inset: 10pt)[ #align(center + horizon, text(fill: rgb("#889"), style: "italic", size: 9pt, msg))] // Standardize inlined figure sizes: measure the natural CeTZ canvas, then scale to a // consistent envelope (aspect-aware; see build_typst.py FIG_* constants). Unlike the // print preamble, dimensions are FLOORED: in an editor a user can trim a figure to a // degenerate 1-D shape (a bare line), and w/h or tw/w would then divide by zero. #let _STD_W = 3.5 #let _WIDE_W = 5.6 #let _MAX_H = 3.4 #let _ASPECT_WIDE = 2.2 #let _UPSCALE_MAX = 1.15 #let stdfig(body) = context { let m = measure(body) let w = calc.max(m.width / 1in, 0.01) let h = calc.max(m.height / 1in, 0.01) let tw = if w / h > _ASPECT_WIDE { _WIDE_W } else { _STD_W } let s = calc.min(tw / w, _MAX_H / h, _UPSCALE_MAX) align(center, box(scale(x: s * 100%, y: s * 100%, reflow: true, body))) } #show figure: set block(breakable: false) #set figure(gap: 8pt) #show figure.caption: set text(size: 8.5pt, fill: rgb("#555")) == 1.1#h(0.6em)Angles and Triangles Historically, trigonometry began as the study of triangles and their properties. Let's review some definitions and facts from geometry. - We measure angles in degrees. - One full rotation is #math.equation(block: false, alt: "360 degrees")[$360^(∘)$], as shown below. - Half a full rotation is #math.equation(block: false, alt: "180 degrees")[$180^(∘)$] and is called a #strong[straight angle]. - One quarter of a full rotation is #math.equation(block: false, alt: "90 degrees")[$90^(∘)$] and is called a #strong[right angle]. #figure(figph[quadrantal angles], alt: "quadrantal angles", caption: none) === Triangles If you tear off the corners of any triangle and line them up, as shown below, they will always form a straight angle. #figure(figph[tear corners off triangle], alt: "tear corners off triangle", caption: none) #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #emph[Sum of angles in a triangle] 1. The sum of the angles in a triangle is #math.equation(block: false, alt: "180 degrees")[$180^(∘)$]. ] #examplebox("Example 1")[][ Two of the angles in the triangle at right are #math.equation(block: false, alt: "25 degrees")[$25^(∘)$] and #math.equation(block: false, alt: "115 degrees")[$115^(∘)$]. Find the third angle. #figure(figph[triangle with 25 and 115 angles], alt: "triangle with 25 and 115 angles", caption: none) #solutionbox[ To find the third angle, we write an equation. #math.equation(block: true, alt: "x plus 25 plus 115, equals 180, Simpify the left side.; x plus 140, equals 180, Subtract 140 from both sides.; x, equals 40")[$x + 25 + 115 & = 180 & & "Simpify the left side." \ x + 140 & = 180 & & "Subtract 140 from both sides." \ x & = 40 & &$] The third angle is #math.equation(block: false, alt: "40 degrees")[$40^(∘)$]. ] ] Find each of the angles in the triangle at right. #figure(figph[triangle with angles x, 2x, and 2x-15], alt: "triangle with angles x, 2x, and 2x-15", caption: none) Because the sum of the angles is #math.equation(block: false, alt: "180 degrees")[$180^(∘)$], we know that #math.equation(block: false, alt: "x plus 2 x plus 2 x minus 15 equals 180")[$" " x + 2 x + 2 x − 15 = 180$]. Solving this equation gives #math.equation(block: false, alt: "x equals 39 degrees , 2 x equals 78 degrees , 2 x minus 15 equals 63 degrees")[$" " x = 39^(∘) , " " " " 2 x = 78^(∘) , " " " " 2 x − 15 = 63^(∘)$] Some special categories of triangles are particularly useful. Most important of these are the #strong[right triangles]. #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #emph[Right triangle] 2. A right triangle has one angle of #math.equation(block: false, alt: "90 degrees")[$90^(∘)$]. ] #examplebox("Example 2")[][ One of the smaller angles of a right triangle is #math.equation(block: false, alt: "34 degrees")[$34^(∘)$]. What is the third angle? #figure(figph[right triangle with 34 angle], alt: "right triangle with 34 angle", caption: none) #solutionbox[ The sum of the two smaller angles in a right triangle is #math.equation(block: false, alt: "90 degrees")[$90^(∘)$]. So #math.equation(block: true, alt: "x plus 34, equals 90, Subtract 34 from both sides.; x, equals 56")[$x + 34 & = 90 & & "Subtract 34 from both sides." \ x & = 56 & &$] The unknown angle must be #math.equation(block: false, alt: "56 degrees")[$56^(∘)$]. ] ] Two angles of a triangle are #math.equation(block: false, alt: "35 degrees")[$35^(∘)$] and #math.equation(block: false, alt: "45 degrees")[$45^(∘)$]. Can it be a right triangle? No, the third angle must be #math.equation(block: false, alt: "100 degrees")[$100^(∘)$]. An #strong[equilateral] triangle has all three sides the same length. #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #emph[Angles of equilateral triangle] 3. All of the angles of an equilateral triangle are equal. ] #examplebox("Example 3")[][ All three sides of a triangle are 4 feet long. Find the angles. #figure(figph[equilateral triangle], alt: "equilateral triangle", caption: none) #solutionbox[ The triangle is equilateral, so all of its angles are equal. Thus #math.equation(block: true, alt: "3 x, equals 180, Divide both sides by 3.; x, equals 60")[$3 x & = 180 & & "Divide both sides by 3." \ x & = 60 & &$] Each of the angles is #math.equation(block: false, alt: "60 degrees")[$60^(∘)$]. ] ] Find #math.equation(block: false, alt: "x")[$x$], #math.equation(block: false, alt: "y")[$y$], and #math.equation(block: false, alt: "z")[$z$] in the triangle at right. #figure(figph[equilateral triangle with side 8], alt: "equilateral triangle with side 8", caption: none) The third angle is #math.equation(block: false, alt: "x equals 60 degrees")[$x = 60^(∘)$], so the triangle is equilateral and #math.equation(block: false, alt: "y equals 8 , z equals 8")[$" " y = 8 , " " z = 8$]. An #strong[isosceles] triangle has two sides of equal length. The angle between the equal sides is the #strong[vertex angle]. The other two angles are the #strong[base angles]. #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #emph[Base angles of an isosceles triangle] 4. The base angles of an isosceles triangle are equal. ] #examplebox("Example 4")[][ Find #math.equation(block: false, alt: "x")[$x$] and #math.equation(block: false, alt: "y")[$y$] in the triangle at right. #figure(figph[isosceles triangle with side 12, base angle 38], alt: "isosceles triangle with side 12, base angle 38", caption: none) #solutionbox[ The triangle is isosceles, so the base angles are equal. Therefore, #math.equation(block: false, alt: "y equals 38 degrees")[$y = 38^(∘)$]. To find the vertex angle, we solve #math.equation(block: true, alt: "x plus 38 plus 38, equals 180; x plus 76, equals 180, Subtract 76 from both sides.; x, equals 104")[$x + 38 + 38 & = 180 & & \ x + 76 & = 180 & & "Subtract 76 from both sides." \ x & = 104 & &$] The vertex angle is #math.equation(block: false, alt: "104 degrees")[$104^(∘)$]. ] ] Find #math.equation(block: false, alt: "x")[$x$] and #math.equation(block: false, alt: "y")[$y$] in the figure at right. #figure(figph[isosceles triangle with side 9, base angle 20], alt: "isosceles triangle with side 9, base angle 20", caption: none) The vertex angle is #math.equation(block: false, alt: "180 degrees minus open parenthesis 20 degrees plus 20 degrees close parenthesis equals 140 degrees")[$180^(∘) − ( 20^(∘) + 20^(∘) ) = 140^(∘)$], and the triangle is isosceles so #math.equation(block: false, alt: "y equals 9")[$" " y = 9$]. === Angles In addition to the facts about triangles reviewed above, there are several useful properties of angles. - Two angles that add to #math.equation(block: false, alt: "180 degrees")[$180^(∘)$] are called #strong[supplementary]. - Two angles that add to #math.equation(block: false, alt: "90 degrees")[$90^(∘)$] are called #strong[complementary]. - Angles between #math.equation(block: false, alt: "0 degrees")[$0^(∘)$] and #math.equation(block: false, alt: "90 degrees")[$90^(∘)$] are called #strong[acute]. - Angles between #math.equation(block: false, alt: "90 degrees")[$90^(∘)$] and #math.equation(block: false, alt: "180 degrees")[$180^(∘)$] are called #strong[obtuse]. #figure(figph[types of angles], alt: "types of angles", caption: none) #examplebox("Example 5")[][ In the figure at right, - #math.equation(block: false, alt: "∠")[$upright(∠)$]#math.equation(block: false, alt: "A O C")[$A O C$] and #math.equation(block: false, alt: "∠")[$upright(∠)$]#math.equation(block: false, alt: "B O C")[$B O C$] are supplementary. - #math.equation(block: false, alt: "∠")[$upright(∠)$]#math.equation(block: false, alt: "D O E")[$D O E$] and #math.equation(block: false, alt: "∠")[$upright(∠)$]#math.equation(block: false, alt: "B O E")[$B O E$] are complementary. - #math.equation(block: false, alt: "∠")[$upright(∠)$]#math.equation(block: false, alt: "A O C")[$A O C$] is obtuse, - and #math.equation(block: false, alt: "∠")[$upright(∠)$]#math.equation(block: false, alt: "B O C")[$B O C$] is acute. #figure(figph[types of angles], alt: "types of angles", caption: none) ] In trigonometry we often use lower-case Greek letters to represent unknown angles (or, more specifically, the measure of the angle in degrees). In the next exercise, we use the Greek letters #math.equation(block: false, alt: "α")[$α$] (alpha), #math.equation(block: false, alt: "β")[$β$] (beta), and #math.equation(block: false, alt: "γ")[$γ$] (gamma). In the figure, #math.equation(block: false, alt: "α")[$α$], #math.equation(block: false, alt: "β")[$β$], and #math.equation(block: false, alt: "γ")[$γ$] denote the measures of the angles in degrees. + Find the measure of angle #math.equation(block: false, alt: "α")[$α$]. + Find the measure of angle #math.equation(block: false, alt: "β")[$β$]. + Find the measure of angle #math.equation(block: false, alt: "γ")[$γ$]. + What do you notice about the measures of the angles? #figure(figph[straight angles with 50 degrees], alt: "straight angles with 50 degrees", caption: none) Angle #math.equation(block: false, alt: "α")[$α$] is supplementary to #math.equation(block: false, alt: "50 degrees")[$50^(∘)$] so #math.equation(block: false, alt: "α equals 130 degrees")[$α = 130^(∘)$]. Then #math.equation(block: false, alt: "β")[$β$] is supplementary to #math.equation(block: false, alt: "α")[$α$] and #math.equation(block: false, alt: "γ")[$γ$] is supplementary to #math.equation(block: false, alt: "β")[$β$], giving us #math.equation(block: false, alt: "β equals 50 degrees , γ equals 130 degrees .")[$β = 50^(∘) , " " γ = 130^(∘) .$] Non-adjacent angles formed by the intersection of two straight lines are called #strong[vertical angles]. In the previous exercise, the angles labeled #math.equation(block: false, alt: "α")[$α$] and #math.equation(block: false, alt: "γ")[$γ$] are vertical angles, as are the angles labeled #math.equation(block: false, alt: "β")[$β$] and #math.equation(block: false, alt: "50 degrees")[$50^(∘)$]. #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #emph[Vertical Angles] 5. Vertical angles are equal. ] #examplebox("Example 6")[][ Explain why #math.equation(block: false, alt: "α equals β")[$α = β$] in the triangle at right. #figure(figph[isosceles triangle with angles alpha and beta], alt: "isosceles triangle with angles alpha and beta", caption: none) #solutionbox[ Because they are the base angles of an isosceles triangle, #math.equation(block: false, alt: "θ")[$θ$] (theta) and #math.equation(block: false, alt: "ϕ")[$ϕ$] (phi) are equal. Also, #math.equation(block: false, alt: "α equals θ")[$α = θ$] because they are vertical angles, and similarly #math.equation(block: false, alt: "β equals ϕ")[$β = ϕ$]. Therefore, #math.equation(block: false, alt: "α equals β")[$α = β$] because they are equal to equal quantities. ] ] Find all the unknown angles in the figure at right. (You will find a list of all the Greek letters and their names at the end of this section.) #figure(figph[triangle with external angle 150], alt: "triangle with external angle 150", caption: none) #math.equation(block: false, alt: "α equals 40 degrees")[$α = 40^(∘)$] because of vertical angles, and #math.equation(block: false, alt: "β")[$β$] is its supplement at #math.equation(block: false, alt: "40 degrees")[$40^(∘)$]. #math.equation(block: false, alt: "δ equals 65 degrees")[$δ = 65^(∘)$], the supplement of #math.equation(block: false, alt: "115 degrees")[$115^(∘)$], and #math.equation(block: false, alt: "γ equals 180 minus open parenthesis 65 plus 40 close parenthesis equals 75 degrees")[$" " γ = 180 − ( 65 + 40 ) = 75^(∘)$]. A line that intersects two parallel lines forms eight angles, as shown in the figure below. There are four pairs of vertical angles, and four pairs of #strong[corresponding angles], or angles in the same position relative to the transversal on each of the parallel lines. For example, the angles labeled 1 and 5 are corresponding angles, as are the angles labeled 4 and 8. Finally, angles 3 and 6 are called #strong[alternate interior angles], and so are angles 4 and 5. #figure(figph[parallel lines with transversal], alt: "parallel lines with transversal", caption: none) #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #emph[Paralles lines cut by a transversal] 6. If parallel lines are intersected by a transversal, the alternate interior angles are equal. Corresponding angles are also equal. ] #examplebox("Example 7")[][ The parallelogram #math.equation(block: false, alt: "A B C D")[$A B C D$] shown at right is formed by the intersection of two sets of parallel lines. Show that the opposite angles of the parallelogram are equal. #figure(figph[parallelogram], alt: "parallelogram", caption: none) #solutionbox[ Angles 1 and 2 are equal because they are alternate interior angles, and angles 2 and 3 are equal because they are corresponding angles. Therefore angles 1 and 3, the opposite angles of the parallelogram, are equal. Similarly, you can show that angles 4, 5, and 6 are equal. ] ] Show that the adjacent angles of a parallelogram are supplementary. (You can use angles 1 and 4 in the parallelogram of the previous example.) Note that angles 2 and 6 are supplementary because they form a straight angle. Angle 1 equals angle 2 because they are alternate interior angles, and similarly angle 4 equals angle 5. Angle 5 equals angle 6 because they are corresponding angles. Thus, angle 4 equals angle 6, and angle 1 equals angle 2. So angles 4 and 1 are supplementary because 2 and 6 are. #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ In the Section 1.1 Summary, you will find a list of vocabulary words and a summary of the facts from geometry that we reviewed in this section. You will also find a set of study questions to test your understanding, and a list of skills to practice in the homework problems. ] Lower Case Letters in the Greek Alphabet #figure(table( columns: 3, align: left, inset: 6pt, [Greek Alphabet], [], [], [#math.equation(block: false, alt: "α alpha")[$α " " " " " " " " "alpha"$]], [#math.equation(block: false, alt: "β beta")[$β " " " " " " " " "beta"$]], [#math.equation(block: false, alt: "γ gamma")[$γ " " " " " " " " "gamma"$]], [#math.equation(block: false, alt: "δ delta")[$δ " " " " " " " " "delta"$]], [#math.equation(block: false, alt: "ϵ epsilon")[$ϵ " " " " " " " " "epsilon"$]], [#math.equation(block: false, alt: "ζ zeta")[$ζ " " " " " " " " "zeta"$]], [#math.equation(block: false, alt: "η eta")[$η " " " " " " " " "eta"$]], [#math.equation(block: false, alt: "θ theta")[$θ " " " " " " " " "theta"$]], [#math.equation(block: false, alt: "ι iota")[$ι " " " " " " " " "iota"$]], [#math.equation(block: false, alt: "κ kappa")[$κ " " " " " " " " "kappa"$]], [#math.equation(block: false, alt: "λ lambda")[$λ " " " " " " " " "lambda"$]], [#math.equation(block: false, alt: "μ mu")[$μ " " " " " " " " "mu"$]], [#math.equation(block: false, alt: "ν nu")[$ν " " " " " " " " "nu"$]], [#math.equation(block: false, alt: "ξ xi")[$ξ " " " " " " " " "xi"$]], [#math.equation(block: false, alt: "o omicron")[$o " " " " " " " " "omicron"$]], [#math.equation(block: false, alt: "π pi")[$π " " " " " " "pi"$]], [#math.equation(block: false, alt: "ρ rho")[$ρ " " " " " " " " "rho"$]], [#math.equation(block: false, alt: "σ sigma")[$σ " " " " " " " " "sigma"$]], [#math.equation(block: false, alt: "τ tau")[$τ " " " " " " " " "tau"$]], [#math.equation(block: false, alt: "υ upsilon")[$υ " " " " " " " " "upsilon"$]], [#math.equation(block: false, alt: "ϕ phi")[$ϕ " " " " " " " " "phi"$]], [#math.equation(block: false, alt: "χ chi")[$χ " " " " " " "chi"$]], [#math.equation(block: false, alt: "ψ psi")[$ψ " " " " " " "psi"$]], [#math.equation(block: false, alt: "ω omega")[$ω " " " " " " "omega"$]], )) Review the following skills you will need for this section. #notebox("Algebra Refresher", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Solve the equation. 1. #math.equation(block: false, alt: "x minus 8 equals 19 minus 2 x")[$x − 8 = 19 − 2 x #hide($display(frac(1, 2))$)$] 3. #math.equation(block: false, alt: "13 x plus 5 equals 2 x minus 28")[$13 x + 5 = 2 x − 28$] 2. #math.equation(block: false, alt: "2 x minus 9 equals 12 minus x")[$2 x − 9 = 12 − x #hide($display(frac(1, 2))$)$] 4. #math.equation(block: false, alt: "9 plus 9 x equals minus 7 plus x")[$9 + 9 x = − 7 + x$] Solve the system. 5. #math.equation(block: false, alt: "5 x minus 2 y, equals minus 13; 2 x plus 3 y, equals minus 9")[$5 x − 2 y & = − 13 & & \ 2 x + 3 y & = − 9 & &$] 6. #math.equation(block: false, alt: "4 x plus 3 y, equals 9; 3 x plus 2 y, equals 8")[$4 x + 3 y & = 9 & & \ 3 x + 2 y & = 8 & &$] #math.equation(block: true, alt: "bar")[$underline(#h(2em) #h(2em) #h(2em) #h(2em))$] Algebra Refresher Answers + #math.equation(block: false, alt: "9")[$9$] + #math.equation(block: false, alt: "7")[$7$] + #math.equation(block: false, alt: "minus 3")[$− 3$] + #math.equation(block: false, alt: "minus 2")[$− 2$] + #math.equation(block: false, alt: "x equals minus 3 , y equals minus 1")[$x = − 3 , y = − 1$] + #math.equation(block: false, alt: "x equals 6 , y equals minus 5")[$x = 6 , y = − 5$] ] === Section 1.1 Summary ==== Vocabulary - Right angle - Straight angle - Right triangle - Equilateral triangle - Isosceles triangle - Vertex angle - Base angle - Supplementary - Complementary - Acute - Obtuse - Vertical angles - Transversal - Corresponding angles - Alternate interior angles ==== Concepts #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #emph[Facts from Geometry] 1. The sum of the angles in a triangle is #math.equation(block: false, alt: "180 degrees")[$180^(∘)$]. 2. A right triangle has one angle of #math.equation(block: false, alt: "90 degrees")[$90^(∘)$]. 3. All of the angles of an equilateral triangle are equal. 4. The base angles of an isosceles triangle are equal. 5. Vertical angles are equal. 6. If parallel lines are intersected by a transversal, the alternate interior angles are equal. Corresponding angles are also equal. ] ==== Study Questions + Is it possible to have more than one obtuse angle in a triangle? Why or why not? + Draw any quadrilateral (a four-sided polygon) and divide it into two triangles by connecting two opposite vertices by a diagonal. What is the sum of the angles in your quadrilateral? + What is the difference between a vertex angle and vertical angles? + Can two acute angles be supplementary? + Choose any two of the eight angles formed by a pair of parallel lines cut by a transversal. Those two angles are either equal or \_\_\_\_\_\_\_ . ==== Skills Practice each skill in the Homework Problems listed. + Sketch a triangle with given properties \#1–6 + Find an unknown angle in a triangle \#7–12, 17–20 + Find angles formed by parallel lines and a transversal \#13–16, 35–44 + Find exterior angles of a triangle \#21–24 + Find angles in isosceles, equilateral, and right triangles \#25–34 + State reasons for conclusions \#45–48 === Homework 1.1 For Problems 1–6, sketch and label a triangle with the given properties. An isosceles triangle with vertex angle 30° #figure(figph[isosceles triangle with vertex angle 30°], alt: "isosceles triangle with vertex angle 30°", caption: none) A scalene triangle with one obtuse angle (#strong[Scalene] means three unequal sides.) A right triangle with legs 4 and 7 #figure(figph[right triangle with legs 4 and 7], alt: "right triangle with legs 4 and 7", caption: none) An isosceles right triangle An isosceles triangle with one obtuse angle #figure(figph[An isosceles triangle with one obtuse angle], alt: "An isosceles triangle with one obtuse angle", caption: none) A right triangle with one angle 20° For Problems 7–20, find each unknown angle. #figure(figph[triangle theta], alt: "triangle theta", caption: none) #math.equation(block: true, alt: "θ equals 108.8 degrees")[$θ = 108.8^(∘)$] #figure(figph[triangle phi], alt: "triangle phi", caption: none) #figure(figph[triangle alpha], alt: "triangle alpha", caption: none) #math.equation(block: true, alt: "α equals 29 degrees")[$α = 29^(∘)$] #figure(figph[triangle gamma], alt: "triangle gamma", caption: none) #figure(figph[triangle beta], alt: "triangle beta", caption: none) #math.equation(block: true, alt: "β equals 77 degrees")[$β = 77^(∘)$] #figure(figph[triangle omega], alt: "triangle omega", caption: none) #figure(figph[triangle alpha], alt: "triangle alpha", caption: none) #math.equation(block: true, alt: "α equals 12 degrees")[$α = 12^(∘)$] #figure(figph[triangle beta], alt: "triangle beta", caption: none) #figure(figph[triangle theta], alt: "triangle theta", caption: none) #math.equation(block: true, alt: "θ equals 65 degrees")[$θ = 65^(∘)$] #figure(figph[triangle phi], alt: "triangle phi", caption: none) #figure(figph[triangle theta], alt: "triangle theta", caption: none) #math.equation(block: true, alt: "θ equals 12 degrees")[$θ = 12^(∘)$] #figure(figph[triangle alpha], alt: "triangle alpha", caption: none) #figure(figph[triangle psi], alt: "triangle psi", caption: none) #math.equation(block: true, alt: "ψ equals 73 degrees")[$ψ = 73^(∘)$] #figure(figph[triangle beta], alt: "triangle beta", caption: none) In Problems 21 and 22, the angle labeled #math.equation(block: false, alt: "ϕ")[$ϕ$] is called an #strong[exterior angle] of the triangle, formed by one side and the extension of an adjacent side. Find #math.equation(block: false, alt: "ϕ")[$ϕ$]. #figure(figph[ext angle], alt: "ext angle", caption: none) #math.equation(block: true, alt: "ϕ equals 88 degrees")[$ϕ = 88^(∘)$] #figure(figph[ext angle], alt: "ext angle", caption: none) In parts (a) and (b), find the exterior angle #math.equation(block: false, alt: "ϕ")[$ϕ$]. + #figure(figph[ext angle], alt: "ext angle", caption: none) + #figure(figph[ext angle], alt: "ext angle", caption: none) + Find an algebraic expression for #math.equation(block: false, alt: "ϕ")[$ϕ$]. #linebreak() #figure(figph[ext angle], alt: "ext angle", caption: none) + Use your answer to part (c) to write a rule for finding an exterior angle of a triangle. + #math.equation(block: false, alt: "ϕ equals 120 degrees")[$ϕ = 120^(∘)$] + #math.equation(block: false, alt: "ϕ equals 160 degrees")[$ϕ = 160^(∘)$] + #math.equation(block: false, alt: "ϕ equals α plus β")[$ϕ = α + β$] + An exterior angle is equal to the sum of the opposite interior angles. + Find the three exterior angles of the triangle. What is the sum of the exterior angles? #figure(figph[ext angles], alt: "ext angles", caption: none) + Write an algebraic expression for each exterior angle in terms of one of the angles of the triangle. What is the sum of the exterior angles? #figure(figph[ext angles], alt: "ext angles", caption: none) In Problems 25 and 26, the figures inscribed are #strong[regular polygons], which means that all their sides are the same length, and all the angles have the same measure. Find the angles #math.equation(block: false, alt: "θ")[$θ$] and #math.equation(block: false, alt: "ϕ")[$ϕ$]. #figure(figph[pentagon], alt: "pentagon", caption: none) #math.equation(block: true, alt: "θ equals 72 degrees , ϕ equals 54 degrees")[$θ = 72^(∘) , ϕ = 54^(∘)$] #figure(figph[hexagon], alt: "hexagon", caption: none) In problems 27 and 28, #math.equation(block: false, alt: "△ A B C")[$upright(△) A B C$] is equilateral. Find the unknown angles. #figure(figph[triangles], alt: "triangles", caption: none) #math.equation(block: true, alt: "θ equals 100 degrees , ϕ equals 30 degrees")[$θ = 100^(∘) , ϕ = 30^(∘)$] #figure(figph[triangles], alt: "triangles", caption: none) #figure(figph[triangles], alt: "triangles", caption: none) + #math.equation(block: false, alt: "2 θ plus 2 ϕ equals")[$2 θ + 2 ϕ = " "$]\_\_\_\_\_\_\_\_ + #math.equation(block: false, alt: "θ plus ϕ equals")[$θ + ϕ = " "$]\_\_\_\_\_\_\_\_ + #math.equation(block: false, alt: "△ A B C")[$upright(△) A B C$] is$" " " " " "$\_\_\_\_\_\_\_\_ + #math.equation(block: false, alt: "180 degrees")[$180^(∘)$] + #math.equation(block: false, alt: "90 degrees")[$90^(∘)$] + a right triangle Find #math.equation(block: false, alt: "α")[$α$] and #math.equation(block: false, alt: "β")[$β$]. #figure(figph[triangles], alt: "triangles", caption: none) #figure(figph[circle], alt: "circle", caption: none) + Explain why #math.equation(block: false, alt: "∠ O A B")[$upright(∠) O A B$] and #math.equation(block: false, alt: "∠ A B O")[$upright(∠) A B O$] are equal in measure. + Explain why #math.equation(block: false, alt: "∠ O B C")[$upright(∠) O B C$] and #math.equation(block: false, alt: "∠ B C O")[$upright(∠) B C O$] are equal in measure. + Explain why #math.equation(block: false, alt: "∠ A B C")[$upright(∠) A B C$] is a right angle. (Hint: Use Problem 29.) + They are base angles of an isosceles triangle. + They are base angles of an isosceles triangle. + #math.equation(block: false, alt: "∠ O A B")[$upright(∠) O A B$] corresponds to #math.equation(block: false, alt: "θ")[$θ$] of Problem 29, and #math.equation(block: false, alt: "∠ O B C")[$upright(∠) O B C$] corresponds to #math.equation(block: false, alt: "ϕ")[$ϕ$] of Problem 29. #figure(figph[circle], alt: "circle", caption: none) + Compare #math.equation(block: false, alt: "θ")[$θ$] with #math.equation(block: false, alt: "α plus β")[$α + β$]. (Hint: What do you know about supplementary angles and the sum of angles in a triangle? + Compare #math.equation(block: false, alt: "α")[$α$] and #math.equation(block: false, alt: "β")[$β$]. + Explain why the #strong[inscribed angle] #math.equation(block: false, alt: "∠ B A O")[$upright(∠) B A O$] is half the size of the #strong[central angle] #math.equation(block: false, alt: "∠ B O D")[$upright(∠) B O D$]. Find #math.equation(block: false, alt: "α")[$α$] and #math.equation(block: false, alt: "β")[$β$]. #figure(figph[equil triangle], alt: "equil triangle", caption: none) #math.equation(block: true, alt: "α equals 30 degrees , β equals 60 degrees")[$α = 30^(∘) , β = 60^(∘)$] Find #math.equation(block: false, alt: "α")[$α$] and #math.equation(block: false, alt: "β")[$β$]. #figure(figph[square], alt: "square", caption: none) In Problems 35–44, arrows on a pair of lines indicate that they are parallel. Find #math.equation(block: false, alt: "x")[$x$] and #math.equation(block: false, alt: "y")[$y$] . #figure(figph[parallel lines], alt: "parallel lines", caption: none) #math.equation(block: true, alt: "x equals 47 degrees , y equals 133 degrees")[$x = 47^(∘) , y = 133^(∘)$] #figure(figph[parallel lines], alt: "parallel lines", caption: none) #figure(figph[parallel lines], alt: "parallel lines", caption: none) #math.equation(block: true, alt: "x equals 60 degrees , y equals 15 degrees")[$x = 60^(∘) , y = 15^(∘)$] #figure(figph[parallel lines], alt: "parallel lines", caption: none) #figure(figph[parallel lines], alt: "parallel lines", caption: none) #math.equation(block: true, alt: "x equals 100 degrees , y equals 16 degrees")[$x = 100^(∘) , y = 16^(∘)$] #figure(figph[parallel lines], alt: "parallel lines", caption: none) #figure(figph[parallel lines], alt: "parallel lines", caption: none) #math.equation(block: true, alt: "x equals 90 degrees , y equals 55 degrees")[$x = 90^(∘) , y = 55^(∘)$] #figure(figph[parallel lines], alt: "parallel lines", caption: none) #figure(figph[parallel lines], alt: "parallel lines", caption: none) #math.equation(block: true, alt: "x equals 50 degrees , y equals 80 degrees")[$x = 50^(∘) , y = 80^(∘)$] #figure(figph[parallel lines], alt: "parallel lines", caption: none) + Among the angles labeled 1 through 5 in the figure at right, find two pairs of equal angles. #figure(figph[parallel lines], alt: "parallel lines", caption: none) + #math.equation(block: false, alt: "∠ 4 plus ∠ 2 plus ∠ 5 equals")[$upright(∠) 4 + upright(∠) 2 + upright(∠) 5 = " "$]\_\_\_\_\_\_\_\_ + Use parts (a) and (b) to explain why the sum of the angles of a triangle is #math.equation(block: false, alt: "180 degrees")[$180^(∘)$] + #math.equation(block: false, alt: "∠ 1 equals ∠ 4 , ∠ 3 equals ∠ 5")[$upright(∠) 1 = upright(∠) 4 , upright(∠) 3 = upright(∠) 5$] + #math.equation(block: false, alt: "180 degrees")[$180^(∘)$] + In the equation #math.equation(block: false, alt: "∠ 4 plus ∠ 2 plus ∠ 5 equals 180 degrees ,")[$upright(∠) 4 + upright(∠) 2 + upright(∠) 5 = 180^(∘) ,$] substitute #math.equation(block: false, alt: "∠ 1")[$upright(∠) 1$] for #math.equation(block: false, alt: "∠ 4")[$upright(∠) 4$], and substitute #math.equation(block: false, alt: "∠ 3")[$upright(∠) 3$] for #math.equation(block: false, alt: "∠ 5")[$upright(∠) 5$] to conclude that the sum of the angles in the triangle is #math.equation(block: false, alt: "180 degrees")[$180^(∘)$]. + In the figure below, find #math.equation(block: false, alt: "θ")[$θ$], and justify your answer. #figure(figph[parallel lines], alt: "parallel lines", caption: none) + Write an algebraic expression for #math.equation(block: false, alt: "θ")[$θ$] in the figure below. #figure(figph[parallel lines], alt: "parallel lines", caption: none) #math.equation(block: false, alt: "A B C D")[$A B C D$] is a rectangle. The diagonals of a rectangle bisect each other. In the figure, #math.equation(block: false, alt: "∠ A Q D equals 130 degrees")[$upright(∠) A Q D = 130^(∘)$]. Find the angles labeled 1 through 5 in order, and give a reason for each answer. #figure(figph[rectangle], alt: "rectangle", caption: none) #math.equation(block: false, alt: "∠ 1 equals 130 degrees")[$upright(∠) 1 = 130^(∘)$] because vertical angles are equal. #math.equation(block: false, alt: "∠ 2 equals 50 degrees")[$upright(∠) 2 = 50^(∘)$] because it makes a straight angle with a #math.equation(block: false, alt: "130 degrees")[$130^(∘)$] angle. #math.equation(block: false, alt: "∠ 3 equals 65 degrees")[$upright(∠) 3 = 65^(∘)$] because it is a base angle of an isosceles triangle whose vertex angle is #math.equation(block: false, alt: "50 degrees")[$50^(∘)$]. #math.equation(block: false, alt: "∠ 4 equals 65 degrees")[$upright(∠) 4 = 65^(∘)$] for the same reason. #math.equation(block: false, alt: "∠ 5 equals 25 degrees")[$upright(∠) 5 = 25^(∘)$] because it is complementary to #math.equation(block: false, alt: "∠ 4")[$upright(∠) 4$]. A tangent meets the radius of a circle at a right angle. In the figure,#math.equation(block: false, alt: "∠ A O B equals 140 degrees")[$upright(∠) A O B = 140^(∘)$]. Find the angles labeled 1 through 5 in order, and give a reason for each answer. #figure(figph[circle with tangents], alt: "circle with tangents", caption: none)