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How to solve Right Triangle Trigonometry

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TRIGONOMETRY MICHAEL CORRAL1 Right Triangle Trigonometry Trigonometry is the study of the relations between the sides and angles of triangles. The word“trigonometry”isderivedfromtheGreekwordstrigono(τρ´ιγωνo),meaning“triangle”, ´ and metro (µǫτρ ω), meaning “measure”. Though the ancient Greeks, such as Hipparchus andPtolemy,usedtrigonometryintheirstudyofastronomybetweenroughly150 B.C. - A.D. 200, its history is much older. For example, the Egyptian scribe Ahmes recorded some rudi- mentary trigonometric calculations (concerning ratios of sides of pyramids) in the famous 1 Rhind Papyrus sometime around 1650 B.C. Trigonometry is distinguished from elementary geometry in part by its extensive use of certain functions of angles, known as the trigonometric functions. Before discussing those functions, we will review some basic terminology about angles. 1.1 Angles Recall the following definitions from elementary geometry: ◦ ◦ (a) An angle isacute if it is between 0 and 90 . ◦ (b) An angle is arightangle if it equals 90 . ◦ ◦ (c) An angle isobtuse if it is between 90 and 180 . ◦ (d) An angle is astraightangle if it equals 180 . (a) acute angle (b) right angle (c) obtuse angle (d) straight angle Figure1.1.1 Types of angles In elementary geometry, angles are always considered to be positive and not larger than ◦ 2 360 . For now we will only consider such angles. The following definitions will be used throughout the text: 1 Ahmes claimed that he copied the papyrus from a work that may date as far back as 3000 B.C. 2 ◦ Later in the text we will discuss negative angles and angles larger than 360 . 12 Chapter 1 • RightTriangleTrigonometry §1.1 ◦ ◦ (a) Two acute angles are complementary if their sum equals 90 . In other words, if 0 ≤ ◦ ◦ ∠A,∠B≤90 then∠A and∠B are complementary if∠A+∠B=90 . ◦ ◦ ◦ (b) Twoanglesbetween0 and180 aresupplementaryiftheirsumequals180 . Inother ◦ ◦ ◦ words, if 0 ≤∠A,∠B≤180 then∠A and∠B are supplementary if∠A+∠B=180 . ◦ ◦ (c) Twoanglesbetween0 and360 areconjugate(orexplementary)iftheirsumequals ◦ ◦ ◦ 360 . Inotherwords,if0 ≤∠A,∠B≤360 then∠Aand∠Bareconjugateif∠A+∠B= ◦ 360 . ∠B ∠B ∠A ∠A ∠A ∠B (a) complementary (b) supplementary (c) conjugate Figure1.1.2 Types of pairs of angles Instead of using the angle notation∠A to denote an angle, we will sometimes use just a capital letter by itself (e.g. A, B, C) or a lowercase variable name (e.g. x, y, t). It is also common to use letters (either uppercase or lowercase) from the Greek alphabet, shown in the table below, to represent angles: Table1.1 TheGreekalphabet Letters Name Letters Name Letters Name A α alpha I ι iota P ρ rho B β beta K κ kappa Σ σ sigma Γ γ gamma Λ λ lambda T τ tau ∆ δ delta M µ mu Υ υ upsilon E ǫ epsilon N ν nu Φ φ phi Z ζ zeta Ξ ξ xi X χ chi H η eta O o omicron Ψ ψ psi Θ θ theta Π π pi Ω ω omega ◦ In elementary geometry you learned that the sum of the angles in a triangle equals 180 , andthatanisoscelestriangleisatrianglewithtwosidesofequallength. Recallthatina righttriangle one of the angles is a right angle. Thus, in a right triangle one of the angles ◦ ◦ is 90 and the other two angles are acute angles whose sum is 90 (i.e. the other two angles are complementary angles).Angles • Section 1.1 3 Example1.1 For each triangle below, determine the unknown angle(s): E Y B ◦ 3α 53 ◦ ◦ α α 35 20 D F X Z A C Note: We will sometimes refer to the angles of a triangle by their vertex points. For example, in the first triangle above we will simply refer to the angle∠BAC as angle A. ◦ ◦ ◦ Solution: For triangle△ABC, A=35 andC=20 , and we know that A+B+C=180 , so ◦ ◦ ◦ ◦ ◦ ◦ ◦ 35 + B + 20 = 180 ⇒ B = 180 − 35 − 20 ⇒ B = 125 . ◦ ◦ For the right triangle△DEF, E=53 and F=90 , and we know that the two acute angles D and E are complementary, so ◦ ◦ ◦ ◦ D + E = 90 ⇒ D = 90 − 53 ⇒ D = 37 . Fortriangle△XYZ, theanglesareintermsofanunknownnumberα,butwedoknowthat X+Y+ ◦ Z=180 , which we can use to solve forα and then use that to solve for X,Y, and Z: ◦ ◦ ◦ ◦ ◦ ◦ ◦ α + 3α + α = 180 ⇒ 5α = 180 ⇒ α = 36 ⇒ X=36 , Y=3×36 =108 , Z=36 Example1.2 Thales’Theoremstatesthatif A,B,andC are(distinct)pointsonacirclesuchthatthelinesegment AB is a diameter of the circle, then the angle∠ACB is a right angle (see Figure 1.1.3(a)). In other words, the triangle△ABC is a right triangle. C C α β α β A B A B O O (a) (b) ◦ Figure1.1.3 Thales’ Theorem:∠ACB=90 Toprovethis,letO bethecenterofthecircleanddrawthelinesegmentOC,asinFigure1.1.3(b). Let α=∠BAC and β=∠ABC. Since AB is a diameter of the circle, OA and OC have the same length(namely,thecircle’sradius). Thismeansthat△OAC isanisoscelestriangle,andso∠OCA= ∠OAC = α. Likewise, △OBC is an isosceles triangle and ∠OCB=∠OBC = β. So we see that ◦ ◦ ∠ACB=α+β. Andsincetheanglesof△ABC mustaddupto180 ,weseethat180 =α+(α+β)+β= ◦ ◦ 2(α+β), soα+β=90 . Thus,∠ACB=90 . QED4 Chapter 1 • RightTriangleTrigonometry §1.1 In a right triangle, the side opposite the right angle is called thehy- B potenuse, and the other two sides are called itslegs. For example, in c Figure 1.1.4 the right angle is C, the hypotenuse is the line segment a AB, which has length c, and BC and AC are the legs, with lengths a andb,respectively. Thehypotenuseisalwaysthelongestsideofaright A b C triangle (see Exercise 11). Figure1.1.4 By knowing the lengths of two sides of a right triangle, the length of the third side can be determined by using thePythagoreanTheorem: Theorem 1.1. Pythagorean Theorem: The square of the length of the hypotenuse of a right triangle is equal to the sum of the squares of the lengths of its legs. Thus, if a right triangle has a hypotenuse of length c and legs of lengths a and b, as in Figure 1.1.4, then the Pythagorean Theorem says: 2 2 2 a + b = c (1.1) Let us prove this. In the right triangle△ABC in Figure 1.1.5(a) below, if we draw a line segmentfromthevertexCtothepointDonthehypotenusesuchthatCDisperpendicular to AB (that is, CD forms a right angle with AB), then this divides△ABC into two smaller triangles△CBD and△ACD, which are both similar to△ABC. B c C D B a b a d A b C C D A c−d D (a)△ABC (b)△CBD (c)△ACD Figure1.1.5 Similar triangles△ABC,△CBD,△ACD Recallthattrianglesaresimilariftheircorrespondinganglesareequal,andthatsimilarity implies that corresponding sides are proportional. Thus, since△ABC is similar to△CBD, by proportionality of corresponding sides we see that c a 2 AB is toCB (hypotenuses) as BC is toBD (vertical legs) ⇒ = ⇒ cd = a . a d Since△ABC is similar to△ACD, comparing horizontal legs and hypotenuses gives b c 2 2 2 2 2 2 2 = ⇒ b = c − cd = c − a ⇒ a + b = c . QED c−d b Note: The symbols⊥ and∼ denote perpendicularity and similarity, respectively. For exam- ple, in the above proof we had CD⊥AB and △ABC∼△CBD∼△ACD. c−d dAngles • Section 1.1 5 Example1.3 For each right triangle below, determine the length of the unknown side: B Y E 5 z a 2 1 1 e A 4 C D F X 1 Z Solution: For triangle△ABC, the Pythagorean Theorem says that 2 2 2 2 a + 4 = 5 ⇒ a = 25 − 16 = 9 ⇒ a = 3 . For triangle△DEF, the Pythagorean Theorem says that p 2 2 2 2 e + 1 = 2 ⇒ e = 4 − 1 = 3 ⇒ e = 3 . For triangle△XYZ, the Pythagorean Theorem says that p 2 2 2 2 1 + 1 = z ⇒ z = 2 ⇒ z = 2 . Example1.4 A 17 ft ladder leaning against a wall has its foot 8 ft from the base of the wall. At what height is the top of the ladder touching the wall? Solution: Let h be the height at which the ladder touches the wall. We can as- 17 h sume that the ground makes a right angle with the wall, as in the picture on the right. Then we see that the ladder, ground, and wall form a right triangle with a ◦ hypotenuse of length 17 ft (the length of the ladder) and legs with lengths 8 ft and 90 h ft. So by the Pythagorean Theorem, we have 8 2 2 2 2 h + 8 = 17 ⇒ h = 289 − 64 = 225 ⇒ h = 15 ft . Exercises For Exercises 1-4, find the numeric value of the indicated angle(s) for the triangle△ABC. ◦ ◦ ◦ ◦ 1. FindB if A=15 andC=50 . 2. FindC if A=110 andB=31 . ◦ 3. Find A andB ifC=24 , A=α, andB=2α. 4. Find A,B, andC if A=β andB=C=4β. ForExercises5-8,findthenumericvalueoftheindicatedangle(s)fortherighttriangle△ABC,with C being the right angle. ◦ 5. FindB if A=45 . 6. Find A andB if A=α andB=2α. 2 7. Find A andB if A=φ andB=φ . 8. Find A andB if A=θ andB=1/θ. 9. A car goes 24 miles due north then 7 miles due east. What is the straight distance between the car’s starting point and end point?6 Chapter 1 • RightTriangleTrigonometry §1.1 10. One end of a rope is attached to the top of a pole 100 ft high. If the rope is 150 ft long, what is the maximum distance along the ground from the base of the pole to where the other end can be attached? You may assume that the pole is perpendicular to the ground. 2 2 2 11. Prove that the hypotenuse is the longest side in every right triangle. (Hint: Is a +b a ?) 12. Can a right triangle have sides with lengths 2, 5, and 6? Explain your answer. 2 2 2 13. If the lengths a, b, and c of the sides of a right triangle are positive integers, with a +b =c , then they form what is called a Pythagorean triple. The triple is normally written as (a,b,c). For example, (3,4,5) and (5,12,13) are well-known Pythagorean triples. (a) Show that (6,8,10) is a Pythagorean triple. (b) Show that if (a,b,c) is a Pythagorean triple then so is (ka,kb,kc) for any integer k0. How would you interpret this geometrically? 2 2 2 2 (c) Show that (2mn,m −n ,m +n ) is a Pythagorean triple for all integers mn0. (d) Thetripleinpart(c)isknownasEuclid’sformulaforgeneratingPythagoreantriples. Write down the first ten Pythagorean triples generated by this formula, i.e. use: m=2 and n=1; m=3 and n=1, 2; m=4 and n=1, 2, 3; m=5 and n=1, 2, 3, 4. 14. This exercise will describe how to draw a line through any point outside a circle such that the lineintersectsthecircleatonlyonepoint. Thisiscalledatangentlinetothecircle(seethepicture on the left in Figure 1.1.6), a notion which we will use throughout the text. tangent line A • O P C Figure1.1.6 On a sheet of paper draw a circle of radius 1 inch, and call the center of that circle O. Pick a point P which is 2.5 inches away from O. Draw the circle which has OP as a diameter, as in the pictureontherightinFigure1.1.6. Let A beoneofthepointswherethiscircleintersectsthefirst circle. Draw the line through P and A. In general the tangent line through a point on a circle is perpendicular to the line joining that point to the center of the circle (why?). Use this fact to explain why the line you drew is the tangent line through A and to calculate the length of PA. Does it match the physical measurement of PA? C 15. Supposethat△ABC isatrianglewithside AB adiameterofacircle with center O, as in the picture on the right, and suppose that the ◦ vertexC liesonthecircle. Nowimaginethatyourotatethecircle180 around its center, so that△ABC is in a new position, as indicated by A B thedashedlinesinthepicture. ExplainhowthispictureprovesThales’ O Theorem. not tangentopposite TrigonometricFunctionsofanAcuteAngle • Section 1.2 7 1.2 Trigonometric Functions of an Acute Angle Consider a right triangle △ABC, with the right angle at C and B with lengths a, b, and c, as in the figure on the right. For the acute c angle A, call the leg BC its opposite side, and call the leg AC its a adjacentside. Recall that the hypotenuse of the triangle is the side AB. The ratios of sides of a right triangle occur often enough in prac- A b C tical applications to warrant their own names, so we define the six adjacent trigonometricfunctions of A as follows: Table1.2 Thesixtrigonometricfunctionsof A Nameoffunction Abbreviation Definition opposite side a sine A sin A = = hypotenuse c adjacent side b cosine A cos A = = hypotenuse c opposite side a tangent A tan A = = adjacent side b hypotenuse c cosecant A csc A = = opposite side a hypotenuse c secant A sec A = = adjacent side b adjacent side b cotangent A cot A = = opposite side a We will usually use the abbreviated names of the functions. Notice from Table 1.2 that the pairs sin A and csc A, cos A and sec A, and tan A and cot A are reciprocals: 1 1 1 csc A = sec A = cot A = sin A cos A tan A 1 1 1 sin A = cos A = tan A = csc A sec A cot A hypotenuse8 Chapter 1 • RightTriangleTrigonometry §1.2 Example1.5 Fortherighttriangle△ABCshownontheright,findthevaluesofallsixtrigono- B metric functions of the acute angles A andB. 5 3 Solution: Thehypotenuseof△ABC haslength5. Forangle A,theoppositeside BC has length 3 and the adjacent side AC has length 4. Thus: A 4 C opposite 3 adjacent 4 opposite 3 sin A = = cos A = = tan A = = hypotenuse 5 hypotenuse 5 adjacent 4 hypotenuse 5 hypotenuse 5 adjacent 4 csc A = = sec A = = cot A = = opposite 3 adjacent 4 opposite 3 For angleB, the opposite side AC has length 4 and the adjacent sideBC has length 3. Thus: opposite 4 adjacent 3 opposite 4 sin B = = cos B = = tan B = = hypotenuse 5 hypotenuse 5 adjacent 3 hypotenuse 5 hypotenuse 5 adjacent 3 csc B = = sec B = = cot B = = opposite 4 adjacent 3 opposite 4 Notice in Example 1.5 that we did not specify the units for the lengths. This raises the possibility that our answers depended on a triangle of a specific physical size. For example, suppose that two different students are reading this textbook: one in the United States and one in Germany. The American student thinks that the lengths 3, 4, and 5 in Example 1.5 are measured in inches, while the German student thinks that they are measured in centimeters. Since 1 in≈ 2.54 cm, the students are using triangles of different physical sizes (see Figure 1.2.1 below, not drawn to scale). B B 5 ′ ′ 3 B B 5 3 A ′ ′ ′ ′ 4 C A 4 C A C A C (a) Inches (b) Centimeters (c) Similar triangles ′ ′ ′ Figure1.2.1 △ABC ∼ △A B C ′ ′ ′ If the American triangle is △ABC and the German triangle is △A B C , then we see ′ ′ ′ from Figure 1.2.1 that △ABC is similar to △A B C , and hence the corresponding anglesTrigonometricFunctionsofanAcuteAngle • Section 1.2 9 are equal and the ratios of the corresponding sides are equal. In fact, we know that com- mon ratio: the sides of△ABC are approximately 2.54 times longer than the corresponding ′ ′ ′ sides of△A B C . So when the American student calculates sin A and the German student ′ 3 calculates sin A , they get the same answer: ′ ′ BC AB BC B C ′ ′ ′ ′ △ABC ∼ △A B C ⇒ = ⇒ = ⇒ sin A = sin A ′ ′ ′ ′ ′ ′ B C A B AB A B ′ Likewise, the other values of the trigonometric functions of A and A are the same. In fact, ourargumentwasgeneralenoughtoworkwithanysimilarrighttriangles. Thisleadsusto the following conclusion: When calculating the trigonometric functions of an acute angle A, you may useany right triangle which has A as one of the angles. Since we defined the trigonometric functions in terms of ratios of sides, you can think of the units of measurement for those sides as canceling out in those ratios. This means that the values of the trigonometric functions are unitless numbers. So when the American student calculated 3/5 as the value of sin A in Example 1.5, that is the same as the 3/5 that the German student calculated, despite the different units for the lengths of the sides. Example1.6 ◦ Find the values of all six trigonometric functions of 45 . 1 B ◦ Solution: Since we may use any right triangle which has 45 as one of the p angles,usethesimplestone: takeasquarewhosesidesareall1unitlongand 2 1 1 divide it in half diagonally, as in the figure on the right. Since the two legs of the triangle△ABC have the same length, △ABC is an isosceles triangle, ◦ ◦ 45 which means that the angles A and B are equal. So since A+B=90 , this ◦ 1 A C means that we must have A= B= 45 . By the Pythagorean Theorem, the length c of the hypotenuse is given by p 2 2 2 c = 1 + 1 = 2 ⇒ c = 2 . Thus, using the angle A we get: opposite 1 adjacent 1 opposite 1 ◦ ◦ ◦ sin 45 = = p cos 45 = = p tan 45 = = = 1 hypotenuse hypotenuse adjacent 1 2 2 p p hypotenuse hypotenuse adjacent 1 ◦ ◦ ◦ csc 45 = = 2 sec 45 = = 2 cot 45 = = = 1 opposite adjacent opposite 1 Note that we would have obtained the same answers if we had used any right triangle similar to p △ABC. For example, if we multiply each side of △ABC by 2, then we would have a similar p p ◦ 2 triangle with legs of length 2 and hypotenuse of length 2. This would give us sin 45 = , which 2 p 2 1 p p p equals = as before. The same goes for the other functions. 2· 2 2 3 We will use the notation AB to denote the length of a line segment AB.10 Chapter 1 • RightTriangleTrigonometry §1.2 Example1.7 ◦ Find the values of all six trigonometric functions of 60 . B ◦ Solution: Since we may use any right triangle which has 60 as one of the angles, we will use a simple one: take a triangle whose sides are all 2 ◦ units long and divide it in half by drawing the bisector from one vertex to 30 2 2 p the opposite side, as in the figure on the right. Since the original triangle 3 was an equilateral triangle (i.e. all three sides had the same length), its ◦ ◦ ◦ three angles were all the same, namely 60 . Recall from elementary ge- 60 60 ometry that the bisector from the vertex angle of an equilateral triangle A 1 C 1 to its opposite side bisects both the vertex angle and the opposite side. So ◦ 2 as in the figure on the right, the triangle △ABC has angle A=60 and ◦ ◦ angle B=30 , which forces the angle C to be 90 . Thus,△ABC is a right triangle. We see that the hypotenuse has length c= AB=2 and the leg AC has length b= AC=1. By the Pythagorean Theorem, the length a of the legBC is given by p 2 2 2 2 2 2 a + b = c ⇒ a = 2 − 1 = 3 ⇒ a = 3 . Thus, using the angle A we get: p p p opposite 3 adjacent 1 opposite 3 ◦ ◦ ◦ sin 60 = = cos 60 = = tan 60 = = = 3 hypotenuse 2 hypotenuse 2 adjacent 1 hypotenuse 2 hypotenuse adjacent 1 ◦ ◦ ◦ csc 60 = = p sec 60 = = 2 cot 60 = = p opposite adjacent opposite 3 3 ◦ Notice that, as a bonus, we get the values of all six trigonometric functions of 30 , by using angle ◦ B=30 in the same triangle△ABC above: p opposite 1 adjacent 3 opposite 1 ◦ ◦ ◦ sin 30 = = cos 30 = = tan 30 = = p hypotenuse 2 hypotenuse 2 adjacent 3 p p hypotenuse hypotenuse 2 adjacent 3 ◦ ◦ ◦ csc 30 = = 2 sec 30 = = p cot 30 = = = 3 opposite adjacent opposite 1 3 Example1.8 2 Aisanacuteanglesuchthatsin A= . Findthevaluesoftheothertrigonometric B 3 functions of A. 3 2 Solution: In general it helps to draw a right triangle to solve problems of this type. The reason is that the trigonometric functions were defined in terms of A b C ratios of sides of a right triangle, and you are given one such function (the sine, 2 in this case) already in terms of a ratio: sin A= . Since sin A is defined as 3 opposite , use 2 as the length of the side opposite A and use 3 as the length of the hypotenuse in a hypotenuse 2 right triangle△ABC (see the figure above), so that sin A= . The adjacent side to A has unknown 3 length b, but we can use the Pythagorean Theorem to find it: p 2 2 2 2 2 + b = 3 ⇒ b = 9 − 4 = 5 ⇒ b = 5TrigonometricFunctionsofanAcuteAngle • Section 1.2 11 We now know the lengths of all sides of the triangle△ABC, so we have: p adjacent 5 opposite 2 cos A = = tan A = = p hypotenuse 3 adjacent 5 p hypotenuse 3 hypotenuse 3 adjacent 5 csc A = = sec A = = cot A = = p opposite 2 adjacent opposite 2 5 You may have noticed the connections between the sine and cosine, secant and cosecant, and tangent and cotangent of the complementary angles in Examples 1.5 and 1.7. General- izing those examples gives us the following theorem: Theorem1.2. CofunctionTheorem: If A andB arethecomplementaryacuteanglesina right triangle△ABC, then the following relations hold: sin A = cos B sec A = csc B tan A = cot B sin B = cos A sec B = csc A tan B = cot A We say that the pairs of functions sin,cos, sec,csc, and tan,cot arecofunctions. So sine and cosine are cofunctions, secant and cosecant are cofunctions, and tangent and cotangentarecofunctions. Thatishowthefunctionscosine,cosecant,andcotangentgotthe “co” in their names. The Cofunction Theorem says that any trigonometric function of an acute angle is equal to its cofunction of the complementary angle. Example1.9 ◦ ◦ Writeeachofthefollowingnumbersastrigonometricfunctionsofananglelessthan45 : (a)sin 65 ; ◦ ◦ (b) cos 78 ;(c) tan 59 . ◦ ◦ ◦ ◦ Solution: (a) The complement of 65 is 90 −65 =25 and the cofunction of sin is cos, so by the ◦ ◦ Cofunction Theorem we know that sin 65 =cos 25 . ◦ ◦ ◦ ◦ ◦ ◦ (b) The complement of 78 is 90 −78 =12 and the cofunction of cos is sin, so cos 78 =sin 12 . ◦ ◦ ◦ ◦ ◦ ◦ (c) The complement of 59 is 90 −59 =31 and the cofunction of tan is cot, so tan 59 =cot 31 . p ◦ 2a 60 a 2 ◦ a 45 a ◦ 30 ◦ 45 p a 3 a (a)45−45−90 (b) 30−60−90 Figure1.2.2 Two general right triangles (any a0) ◦ ◦ ◦ The angles 30 , 45 , and 60 arise often in applications. We can use the Pythagorean Theorem to generalize the right triangles in Examples 1.6 and 1.7 and see what any 45− 45−90 and 30−60−90 right triangles look like, as in Figure 1.2.2 above.12 Chapter 1 • RightTriangleTrigonometry §1.2 Example1.10 ◦ Find the sine, cosine, and tangent of 75 . D ◦ ◦ ◦ Solution: Since75 =45 +30 ,placea30−60−90righttriangle p △ADB with legs of length 3 and 1 on top of the hypotenuse of a 45−45−90 right triangle △ABC whose hypotenuse has 1 p length 3,asinthefigureontheright. FromFigure1.2.2(a)we know that the length of each leg of△ABC is the length of the q p p B 3 3 p hypotenuse divided by 2. So AC=BC= = . Draw DE F 2 2 perpendicular to AC, so that△ADE is a right triangle. Since 2 ◦ ◦ ◦ ∠BAC=45 and∠DAB=30 , we see that∠DAE=75 since q itisthesumofthosetwoangles. Thus,weneedtofindthesine, 3 cosine, and tangent of∠DAE. 2 ◦ Noticethat∠ADE=15 ,sinceitisthecomplementof∠DAE. ◦ And∠ADB=60 , since it is the complement of∠DAB. Draw BF perpendicular to DE, so that △DFB is a right triangle. ◦ ◦ ◦ E 45 Then∠BDF=45 ,sinceitisthedifferenceof∠ADB=60 and A C q ◦ ◦ ∠ADE=15 . Also,∠DBF=45 since it is the complement of 3 2 ∠BDF. ThehypotenuseBD of△DFBhaslength1and△DFB 1 p is a 45−45−90 right triangle, so we know that DF=FB= . 2 Now,weknowthatDE⊥AC andBC⊥AC,soFE andBC areparallel. Likewise,FB andEC are both perpendicular to DE and hence FB is parallel to EC. Thus, FBCE is a rectangle, since∠BCE q 1 3 p is a right angle. So EC=FB= and FE=BC= . Hence, 2 2 q q p p 1 3 3 + 1 3 1 3 − 1 p p p p DE = DF + FE = + = and AE = AC − EC = − = . Thus, 2 2 2 2 2 2 p p p 3+1 3−1 3+1 p p p p p p p p p ◦ DE 2 6+ 2 ◦ AE 2 6− 2 ◦ DE 2 6+ 2 p p p sin 75 = = = , cos 75 = = = , and tan 75 = = = . AD 2 4 AD 2 4 AE 3−1 6− 2 p 2 p p ◦ 4 ◦ 4 ◦ 6− 2 p p p p p p Note: Taking reciprocals, we get csc 75 = , sec 75 = , and cot 75 = . 6+ 2 6− 2 6+ 2 Exercises For Exercises 1-10, find the values of all six trigonometric functions of B angles A andB in the right triangle△ABC in Figure 1.2.3. c a 1. a=5, b=12, c=13 2. a=8, b=15, c=17 A b C 3. a=7, b=24, c=25 4. a=20, b=21, c=29 p Figure1.2.3 5. a=9, b=40, c=41 6. a=1, b=2, c= 5 7. a=1, b=3 8. a=2, b=5 9. a=5, c=6 10. b=7, c=8 For Exercises 11-18, find the values of the other five trigonometric functions of the acute angle A given the indicated value of one of the functions. p ◦ 3 30TrigonometricFunctionsofanAcuteAngle • Section 1.2 13 3 2 2 2 p 11. sin A= 12. cos A= 13. cos A= 14. sin A= 4 3 4 10 5 7 15. tan A= 16. tan A=3 17. sec A= 18. csc A=3 9 3 For Exercises 19-23, write the given number as a trigonometric function of an acute angle less than ◦ 45 . ◦ ◦ ◦ ◦ ◦ 19. sin 87 20. sin 53 21. cos 46 22. tan 66 23. sec 77 For Exercises 24-28, write the given number as a trigonometric function of an acute angle greater ◦ than 45 . ◦ ◦ ◦ ◦ ◦ 24. sin 1 25. cos 13 26. tan 26 27. cot 10 28. csc 43 ◦ ◦ 29. In Example 1.7 we found the values of all six trigonometric functions of 60 and 30 . ◦ ◦ ◦ ◦ ◦ ◦ (a) Does sin 30 + sin 30 = sin 60 ? (b) Does cos 30 + cos 30 = cos 60 ? ◦ ◦ ◦ ◦ ◦ ◦ (c) Does tan 30 + tan 30 = tan 60 ? (d) Does 2sin 30 cos 30 = sin 60 ? 30. For an acute angle A, can sin A be larger than 1? Explain your answer. 31. For an acute angle A, can cos A be larger than 1? Explain your answer. 32. For an acute angle A, can sin A be larger than tan A? Explain your answer. 33. If A andB are acute angles and AB, explain why sin Asin B. 34. If A andB are acute angles and AB, explain why cos Acos B. 35. Prove the Cofunction Theorem (Theorem 1.2). (Hint: Draw a right triangle and label the angles andsides.) ◦ 36. Use Example 1.10 to find all six trigonometric functions of 15 . A 37. In Figure 1.2.4, CB is a diameter of a circle with a radius of D 2 cm and centerO,△ABC is a right triangle, andCD p has length 3 cm. p (a) Find sin A. (Hint: UseThales’Theorem.) 3 (b) Find the length of AC. C B 2 O (c) Find the length of AD. (d) Figure 1.2.4 is drawn to scale. Use a protractor to measure the angle A, then use your calculator to find the sine of that angle. Is the calculator result close to your answer from part(a)? Figure1.2.4 Note: Make sure that your calculator is in degree mode. 1 38. In Exercise 37, verify that the area of△ABC equals AB·CD. Why does this make sense? 2 1 39. In Exercise 37, verify that the area of△ABC equals AB·AC sin A. 2 1 2 40. In Exercise 37, verify that the area of△ABC equals (BC) cot A. 214 Chapter 1 • RightTriangleTrigonometry §1.3 1.3 Applications and Solving Right Triangles Throughoutitsearlydevelopment,trigonometrywasoftenusedasameansofindirectmea- surement,e.g. determininglargedistancesorlengthsbyusingmeasurementsofanglesand small,knowndistances. Today,trigonometryiswidelyusedinphysics,astronomy,engineer- ing, navigation, surveying, and various fields of mathematics and other disciplines. In this section we will see some of the ways in which trigonometry can be applied. Your calculator should be in degree mode for these examples. Example1.11 ◦ A person stands 150 ft away from a flagpole and measures an angle of elevation of 32 from his horizontallineofsighttothetopoftheflagpole. Assumethattheperson’seyesareaverticaldistance of 6 ft from the ground. What is the height of the flagpole? Solution: Thepictureontherightdescribesthesituation. Weseethat the height of the flagpole is h+6 ft, where h h ◦ ◦ = tan 32 ⇒ h = 150 tan 32 = 150(0.6249) = 94 . ◦ 32 150 150 ◦ How did we know that tan 32 =0.6249? By using a calculator. And 6 sincenoneofthenumbersweweregivenhaddecimalplaces,werounded offtheanswerfor htothenearestinteger. Thus,theheightoftheflagpoleis h+6=94+6= 100 ft . Example1.12 Apersonstanding400ftfromthebaseofamountainmeasurestheangleofelevationfromtheground ◦ to the top of the mountain to be 25 . The person then walks 500 ft straight back and measures the ◦ angle of elevation to now be 20 . How tall is the mountain? Solution: We will assume that the ground is flat and not in- clinedrelativetothebaseofthemountain. Lethbetheheight ofthemountain,andlet xbethedistancefromthebaseofthe mountain to the point directly beneath the top of the moun- h tain, as in the picture on the right. Then we see that ◦ ◦ 20 25 h ◦ ◦ = tan 25 ⇒ h = (x+400) tan 25 , and x 500 400 x+400 h ◦ ◦ = tan 20 ⇒ h = (x+900) tan 20 , so x+400+500 ◦ ◦ (x+400) tan 25 = (x+900) tan 20 , since they both equal h. Use that equation to solve for x: ◦ ◦ 900 tan 20 − 400 tan 25 ◦ ◦ ◦ ◦ x tan 25 − x tan 20 = 900 tan 20 − 400 tan 25 ⇒ x = = 1378ft ◦ ◦ tan 25 − tan 20 Finally, substitute x into the first formula for h to get the height of the mountain: ◦ h = (1378+400) tan 25 = 1778(0.4663) = 829 ftApplicationsandSolvingRightTriangles • Section 1.3 15 Example1.13 ◦ A blimp 4280 ft above the ground measures an angle of depression of 24 from its horizontal line of sight to the base of a house on the ground. Assuming the ground is flat, how far away along the ground is the house from the blimp? Solution: Let x be the distance along the ground from the blimp ◦ 24 to the house, as in the picture to the right. Since the ground and the blimp’shorizontallineofsightareparallel,weknowfromelementary 4280 geometry that the angle of elevation θ from the base of the house to θ the blimp is equal to the angle of depression from the blimp to the ◦ base of the house, i.e. θ=24 . Hence, x 4280 4280 ◦ = tan 24 ⇒ x = = 9613 ft . ◦ x tan 24 Example1.14 ◦ Anobserveratthetopofamountain3milesabovesealevelmeasuresanangleofdepressionof2.23 to the ocean horizon. Use this to estimate the radius of the earth. 4 Solution: We will assume that the earth is a sphere. Let r be A B ◦ the radius of the earth. Let the point A represent the top of the 2.23 3 mountain, and let H be the ocean horizon in the line of sight from H A, as in Figure 1.3.1. Let O be the center of the earth, and let B be a point on the horizontal line of sight from A (i.e. on the line r perpendicular toOA). Letθ be the angle∠AOH. θ Since A is 3 miles above sea level, we have OA= r+3. Also, ◦ OH= r. Now since AB⊥OA, we have ∠OAB= 90 , so we see O ◦ ◦ ◦ that∠OAH=90 −2.23 =87.77 . We see that the line through A and H isatangentlinetothesurfaceoftheearth(consideringthe surface as the circle of radius r through H as in the picture). So ◦ by Exercise 14 in Section 1.1, AH⊥OH and hence∠OHA=90 . ◦ Since the angles in the triangle △OAH add up to 180 , we have Figure1.3.1 ◦ ◦ ◦ ◦ θ=180 −90 −87.77 =2.23 . Thus, OH r r ◦ cosθ = = ⇒ = cos 2.23 , OA r+3 r+3 so solving for r we get ◦ ◦ ◦ r = (r + 3) cos 2.23 ⇒ r − r cos 2.23 = 3 cos 2.23 ◦ 3 cos 2.23 ⇒ r = ◦ 1 − cos 2.23 ⇒ r = 3958.3 miles . Note: This answer is very close to the earth’s actual (mean) radius of 3956.6 miles. 4 Of course it is not perfectly spherical. The earth is an ellipsoid, i.e. egg-shaped, with an observed ellipticity of 1/297 (a sphere has ellipticity 0). See pp. 26-27 in W.H. MUNK AND G.J.F MACDONALD,TheRotationofthe Earth: AGeophysicalDiscussion, London: Cambridge University Press, 1960. r16 Chapter 1 • RightTriangleTrigonometry §1.3 Example1.15 As another application of trigonometry to astronomy, we will find the distance B from the earth to the sun. Let O be the center of the earth, let A be a point on the equator, and let B represent an object (e.g. a star) in space, as in the α picture on the right. If the earth is positioned in such a way that the angle ◦ ∠OAB=90 , then we say that the angleα=∠OBA is theequatorialparallax of the object. The equatorial parallax of the sun has been observed to be ap- ◦ proximately α=0.00244 . Use this to estimate the distance from the center of the earth to the sun. A O Solution: Let B be the position of the sun. We want to find the length of OB. We will use the actual radius of the earth, mentioned at the end of Example ◦ 1.14, to getOA=3956.6 miles. Since∠OAB=90 , we have OA OA 3956.6 = sinα ⇒ OB = = = 92908394 , ◦ OB sinα sin 0.00244 so the distance from the center of the earth to the sun is approximately 93 million miles . Note: The earth’s orbit around the sun is an ellipse, so the actual distance to the sun varies. ◦ Intheaboveexampleweusedaverysmallangle(0.00244 ). Adegreecanbedividedinto smallerunits: aminuteisone-sixtiethofadegree,andasecondisone-sixtiethofaminute. ′ ′′ ◦ ◦ ′ Thesymbolforaminuteis andthesymbolforasecondis . Forexample,4.5 =4 30. And ◦ ◦ ′ ′′ 4.505 =4 30 18 : 30 18 ◦ ′ ′′ ◦ 4 30 18 = 4 + + degrees = 4.505 60 3600 ◦ ′′ In Example 1.15 we used α=0.00244 ≈8.8 , which we mention only because some angle measurement devices do use minutes and seconds. Example1.16 ′ ′′ An observer on earth measures an angle of 32 4 from one visible B edge of the sun to the other (opposite) edge, as in the picture on the right. Use this to estimate the radius of the sun. ′ ′′ E 32 4 S Solution: Let the point E be the earth and let S be the center of the sun. The observer’s lines of sight to the visible edges of the sun A are tangent lines to the sun’s surface at the points A and B. Thus, ◦ ∠EAS=∠EBS=90 . Theradiusofthesunequals AS. Clearly AS=BS. SosinceEB=EA (why?), 1 1 ′ ′′ ′ ′′ thetriangles△EAS and△EBS aresimilar. Thus,∠AES=∠BES= ∠AEB= (32 4 )=16 2 = 2 2 ◦ (16/60)+(2/3600)=0.26722 . Now, ES is the distance from the surface of the earth (where the observer stands) to the cen- ter of the sun. In Example 1.15 we found the distance from the center of the earth to the sun to be 92,908,394 miles. Since we treated the sun in that example as a point, then we are justi- fied in treating that distance as the distance between the centers of the earth and sun. So ES= 92908394− radius of earth=92908394−3956.6=92904437.4 miles. Hence, AS ◦ ◦ sin(∠AES) = ⇒ AS = ES sin 0.26722 = (92904437.4) sin 0.26722 = 433,293 miles . ES Note: This answer is close to the sun’s actual (mean) radius of 432,200 miles.ApplicationsandSolvingRightTriangles • Section 1.3 17 Youmayhavenoticedthatthesolutionstotheexampleswehaveshownrequiredatleast one right triangle. In applied problems it is not always obvious which right triangle to use, which is why these sorts of problems can be difficult. Often no right triangle will be immediately evident, so you will have to create one. There is no general strategy for this, but remember that a right triangle requires a right angle, so look for places where you can form perpendicular line segments. When the problem contains a circle, you can create right 5 anglesbyusingtheperpendicularityofthetangentlinetothecircleatapoint withtheline that joins that point to the center of the circle. We did exactly that in Examples 1.14, 1.15, and 1.16. Example1.17 ThemachinetooldiagramontherightshowsasymmetricV-block, in which one circular roller sits on top of a smaller circular roller. Each roller touches both slanted sides of the V-block. Find the di- O ameter d ofthelargeroller,giventheinformationinthediagram. d Solution: The diameter d of the large roller is twice the radius D ◦ OB, so we need to find OB. To do this, we will show that△OBC 37 is a right triangle, then find the angle∠BOC, and then find BC. B C The lengthOB will then be simple to determine. P Since the slanted sides are tangent to each roller, ∠ODA = ◦ E ∠PEC= 90 . By symmetry, since the vertical line through the ◦ centers of the rollers makes a 37 angle with each slanted side, ◦ A we have ∠OAD=37 . Hence, since △ODA is a right triangle, ◦ ∠DOA is the complement of∠OAD. So∠DOA=53 . ◦ Since the horizontal line segment BC is tangent to each roller, ∠OBC =∠PBC = 90 . Thus, ◦ △OBC is a right triangle. And since∠ODA=90 , we know that△ODC is a right triangle. Now, OB=OD (since they each equal the radius of the large roller), so by the Pythagorean Theorem we haveBC=DC: 2 2 2 2 2 2 BC = OC − OB = OC − OD = DC ⇒ BC = DC ∼ Thus, △OBC and △ODC are congruent triangles (which we denote by △OBC △ODC), since = their corresponding sides are equal. Thus, their corresponding angles are equal. So in particular, ◦ ∠BOC=∠DOC. We know that∠DOB=∠DOA=53 . Thus, ◦ ◦ 53 = ∠DOB = ∠BOC + ∠DOC=∠BOC + ∠BOC = 2∠BOC ⇒ ∠BOC = 26.5 . ◦ Likewise, since BP=EP and∠PBC=∠PEC=90 ,△BPC and△EPC are congruent right trian- gles. Thus,BC=EC. ButweknowthatBC=DC,andweseefromthediagramthatEC+DC=1.38. Thus,BC+BC=1.38 and soBC=0.69. We now have all we need to findOB: BC BC 0.69 = tan∠BOC ⇒ OB = = = 1.384 ◦ OB tan∠BOC tan 26.5 Hence, the diameter of the large roller is d=2×OB=2(1.384)= 2.768 . 5 This will often be worded asthelinethatistangenttothecircle. 1.38connecting rod 18 Chapter 1 • RightTriangleTrigonometry §1.3 Example1.18 A slider-crank mechanism is shown in Figure 1.3.2 below. As the piston moves downward the con- necting rod rotates the crank in the clockwise direction, as indicated. piston A a C b c B θ r O Figure1.3.2 Slider-crank mechanism The point A is the center of the connecting rod’s wrist pin and only moves vertically. The point B is the center of the crank pin and moves around a circle of radius r centered at the point O, which is directly below A and does not move. As the crank rotates it makes an angle θ with the line OA. The instantaneous center of rotation of the connecting rod at a given time is the point C where the horizontal line through A intersects the extended line through O and B. From Figure 1.3.2 we see ◦ that∠OAC=90 , and we let a= AC, b= AB, and c=BC. In the exercises you will show that for ◦ ◦ 0 θ90 , p 2 2 2 p b − r (sinθ) 2 2 2 c = and a = r sinθ + b − r (sinθ) tanθ . cosθ crankApplicationsandSolvingRightTriangles • Section 1.3 19 For some problems it may help to remember that when a right tri- r angle has a hypotenuse of length r and an acute angle θ, as in the r sinθ pictureontheright,theadjacentsidewillhavelength rcosθ andthe θ opposite side will have length rsin θ. You can think of those lengths r cosθ as the horizontal and vertical “components” of the hypotenuse. Notice in the above right triangle that we were given two pieces of information: one of the acute angles and the length of the hypotenuse. From this we determined the lengths of the other two sides, and the other acute angle is just the complement of the known acute angle. Ingeneral,atrianglehassixparts: threesidesandthreeangles. Solvingatriangle means finding the unknown parts based on the known parts. In the case of a right triangle, ◦ one part is always known: one of the angles is 90 . Example1.19 Solve the right triangle in Figure 1.3.3 using the given B information: c a ◦ (a) c=10, A=22 Solution: The unknown parts are a, b, andB. Solving yields: A b C ◦ a = c sin A = 10 sin 22 = 3.75 Figure1.3.3 ◦ b = c cos A = 10 cos 22 = 9.27 ◦ ◦ ◦ ◦ B = 90 − A = 90 − 22 = 68 ◦ (b) b=8, A=40 Solution: The unknown parts are a, c, andB. Solving yields: a ◦ = tan A ⇒ a = b tan A = 8 tan 40 = 6.71 b b b 8 = cos A ⇒ c = = = 10.44 ◦ c cos A cos 40 ◦ ◦ ◦ ◦ B = 90 − A = 90 − 40 = 50 (c) a=3, b=4 Solution: The unknown parts are c, A, andB. By the Pythagorean Theorem, p p p 2 2 2 2 c = a + b = 3 + 4 = 25 = 5 . ✄ a 3 −1 Now, tan A= = =0.75. So how do we find A? There should be a key labeled tan on your ✂ ✁ b 4 calculator, which works like this: give it a number x and it will tell you the angle θ such that tanθ=x. In our case, we want the angle A such that tan A=0.75: ✄ −1 Enter: 0.75 Press: tan Answer: 36.86989765 ✂ ✁ ◦ ◦ ◦ ◦ ◦ This tells us that A=36.87 , approximately. ThusB=90 −A=90 −36.87 =53.13 . ✄ ✄ −1 −1 Note: The sin and cos keys work similarly for sine and cosine, respectively. These keys use ✂ ✁ ✂ ✁ theinversetrigonometricfunctions, which we will discuss in Chapter 5.