Sect 10.2 Trigonometric Ratios
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1 86 Sect 0. Trigonometric Rtios Objective : Understnding djcent, Hypotenuse, nd Opposite sides of n cute ngle in right tringle. In right tringle, the otenuse is lwys the longest side; it is the side opposite of (or "cross from") the right ngle. The otenuse of the following tringles is highlighted in gry. The cent side of n cute ngle is the side of the tringle tht forms the cute ngle with the otenuse. The cent side is the side of the ngle tht is "cent" to or next to the otenuse. The cent side depends on which of the cute ngles one is using. The cent side of the following ngles (lbeled ) is highlighted in gry. The opposite side of n cute ngle is the side of the tringle tht is not side of the cute ngle. It is "cross from" or opposite of the cute ngle. The opposite side depends on which of the cute ngles one is using. The opposite side of the following ngles (lbeled ) is highlighted in gry.
2 87 For ech cute ngle in the following right tringles, identify the cent, otenuse, nd opposite side: R Ex. B Ex. Q C X For, For R, the cent side is C, the cent side is QR, the opposite side is BC, the opposite side is QX, nd the otenuse is B. nd the otenuse is RX. For B, For X the cent side is BC, the cent side is QX, the opposite side is C, the opposite side is QR, nd the otenuse is B. nd the otenuse is RX. Objective : Understnding Trigonometric Rtios or Functions. Recll tht two tringles re similr if they hve the sme shpe, but re not necessrily the sme size. In similr tringles, the corresponding ngles re equl. Thus, if BC HET, then m = m H, m B = m E, nd m C = m T. The corresponding sides re not equl. However, the rtios of corresponding sides re equl since one tringle is in proportion to the other tringle. Thus, if BC HET, then B E B BC C = = HE ET HT Now, let's consider the proportion B HE = BC ET. If we cross multiply, C we get (B) (ET) = (HE) (BC). Now, H T divide both sides by (BC) (ET) nd simplifying, we get: (B) (ET) (HE) (BC) B HE = = (BC) (ET) (BC) (ET) BC ET
3 88 Now consider ngles C nd T. B E If nd H re right ngles, then the proportion B BC = HE ET implies tht the rtio of the opposite side to the otenuse is the sme for ny right tringle if the corresponding cute C ngles hve the sme mesure. H T We will define this rtio s the Sine Rtio or Sine Function. The bbrevition for the sine function is "sin", but it is pronounced s "sign". Sine Rtio or Function Given n cute ngle in right tringle, then the sine rtio of, is the rtio of the opposite side of to the otenuse. sin = oppositeside otenuse = opp For ech cute ngle in the following right tringles, find the sine of the ngle: 6. in Ex. Ex cm 4.0 cm 9. in 5.0 in cm sin 5 = opp = 4. sin 40 = opp 4.0 = = = sin 55 = opp = 4.4 sin 50 = opp 4.0 = = = Since the opposite is never greter thn the otenuse, then the vlue of the sine function cnnot exceed. If the opposite side is equl to the otenuse, then the mesure of the ngle is 90 nd the tringle degenertes into verticl line. This mens tht sin 90 =. Similrly, if the opposite side hs zero length mening the sine rtio is 0, then the mesure of the ngle is 0 nd the tringle degenertes into horizontl line. Hence, sin 0 = 0. opp
4 We hve seen tht if the corresponding cute ngles of rights tringles hve the sme mesure, the rtio of the opposite side to the otenuse is fixed. We cn mke the sme rgument tht the rtio of the cent side to the otenuse is lso fixed. We will define this rtio s the Cosine Rtio or Cosine Function. The bbrevition for the cosine function is "cos", but it is pronounced s "co-sign". 89 Cosine Rtio or Function Given n cute ngle in right tringle, then the cosine rtio of, is the rtio of the cent side of to the otenuse. cos = centside otenuse = opp For ech cute ngle in the following right tringles, find the cosine of the ngle: 6. in Ex. 5 Ex cm 4.0 cm 9. in 5.0 in cm cos 5 = = 4.4 cos 40 = 4.0 = = = cos 55 = = 4. cos 50 = 4.0 = = = Since the cent is never greter thn the otenuse, then the vlue of the cosine function cnnot exceed. If the cent side is equl to the otenuse, then the mesure of the ngle is 0 nd the tringle degenertes into horizontl line. This mens tht cos 0 =. Similrly, if the cent side hs zero length mening the cosine rtio is 0, then the mesure of the ngle is 90 nd the tringle degenertes into verticl line. Hence, cos 90 = 0. We hve seen tht if the corresponding cute ngles of rights tringles hve the sme mesure, the rtio of the opposite side to the otenuse is
5 90 fixed nd the rtio of the cent side to the otenuse is fixed. We cn mke the sme rgument tht the rtio of the opposite side to the cent side is lso fixed. We will define this rtio s the Tngent Rtio or Tngent Function. The bbrevition for the tngent function is "tn", but it is pronounced s "tngent". Tngent Rtio or Function Given n cute ngle in right tringle, then the tngent rtio of, is the rtio of the opposite side of to the cent side of. tn = oppositeside centside = opp opp For ech cute ngle in the following right tringles, find the tngent of the ngle: 6. in Ex. 7 Ex cm 4.0 cm 9. in 5.0 in cm tn 5 = opp = 4. tn 40 = opp 4.4 = = 0.89 tn 55 = opp = 4.4 tn 50 = opp 4. =.47.4 =.9.9 = = When the opposite side hs zero length, then the mesure of the ngle is zero. The tngent rtio is the rtio of 0 to the cent side, which is equl to zero. Hence, tn 0 = 0. Similrly, if the cent side hs zero length, then the mesure of the ngle is 90. The tngent rtio is the rtio of the opposite side to 0, which is undefined. Thus, tn 90 is undefined. Unlike the sine nd cosine function, the vlues of the tngent function cn exceed one. There re some memories ids tht you cn use to remember the trigonometric rtios. If you remember the order sine, cosine, nd tngent, then you might find one of these phrses helpful:
6 9 Trigonometric Rtio sin = cos = tn = Rtio of sides opp opp Rted G Phrse oscr hd hep of pples Rted PG- Phrse oh h ll nother hour of lgebr If you wnt to include the trigonometric functions themselves, you cn use the following phrses: s-o-h-c--h-t-o- s o h c h t o i p y o d y p d n p p s j p n p j e o o i o g o s t n c t e s c i e e e e n i e t n n n t t n e u t u e t s s e e sinful oscr hd consumed heping tste of pples sinful oscr hd consumed heping tste of pples. i p y o d y p d n p p s j p n p j e o o i o g o s t n c t e s c i e e e e n i e t n n n t t n e u t u e t s s e e
7 Complete the following tble: Ex sin cos tn Recll tht sin 0 = 0, cos 0 =, nd tn 0 = 0. lso, sin 90 =, cos 90 = 0, nd tn 90 is undefined: sin 0 cos 0 tn 0 undef. One of our specil tringles from before ws tringle: cos 60 = = = sin 0 = opp = = cos 0 = = tn 0 = opp = sin 60 = opp = tn 60 = opp = = = = = sin 0 cos tn 0 undef. 0 nother of our specil tringles ws tringle:
8 sin 45 = opp = = cos 45 = = tn 45 = opp = = = sin 0 cos tn 0 Our lst specil tringle ws -4-5 tringle: undef cos 5. = = 5 = 5 sin 6.9 = opp = 5 = 5 cos 6.9 = = 4 5 = 4 5 tn 6.9 = opp sin 5. = opp tn 5. = opp = 4 = 4 = 4 5 = 4 5 = 4 = sin 0 cos tn undef.
9 Objective : Evluting trigonometric rtios on scientific clcultor. On scientific clcultor, SIN key is for the sine rtio, COS key is for the cosine rtio, nd TN key is for the tngent rtio. To evlute trigonometric rtio for specific ngle, you first wnt to be in the correct mode. If the ngle is in degree, then the clcultor needs to be in degree mode while if the ngle is in rdins, the clcultor needs to be in rdin mode. Next, type in the ngle nd then hit the pproprite trigonometric rtio. On some clcultors, you might need to enter the trigonometric rtio first nd then the ngle. Evlute the following (round to three significnt digits): 94 Ex. 0 cos 6 Ex. 0b tn 78. Ex. 0c sin 6 Ex. 0d sin 56 ' Ex. 0e tn 89 9' Ex. 0f cos 47 7' Ex. 0g sin. Ex. 0h tn π ) We put our clcultor in degree mode, type 6 nd hit the COS key: cos 6 = b) We put our clcultor in degree mode, type 78. nd hit the TN key: tn 78. = c) We put our clcultor in degree mode, type 6 nd hit the SIN key: sin 6 = d) We put our clcultor in degree mode, but we will need to convert the degrees nd minutes. We type 56. nd hit DMS DD key to get 56.8 Now, hit SIN key: sin 56 ' = e) We put our clcultor in degree mode, but we will need to convert the degrees nd minutes. We type nd hit DMS DD key to get Now, hit TN key: tn 89 9' = f) We put our clcultor in degree mode, but we will need to convert the degrees nd minutes. We type 47.7 nd hit DMS DD key to get 47.8 Now, hit COS key: cos 47 7' =
10 95 g) Since there is no degree mrk on the ngle, then the ngle is in rdins. We put our clcultor in rdin mode, type. nd hit the SIN key: sin. = h) Since there is no degree mrk on the ngle, then the ngle is in rdins. We put our clcultor in rdin mode, type π nd hit the TN key: tn π = Error. This mens tht tn π is undefined. Objective 4: Finding ngles using Trigonometric Rtio. If we know the rtio of two sides of right tringle, we cn use inverse trigonometric rtios or functions to find the ngle. The inverse trigonometric function produces n ngle for the nswer. Inverse Trigonometric Functions Given right tringle, we cn find the cute ngle using ny of the following: opp ) Inverse sine (rcsine) = sin ( opp ) ) Inverse cosine (rccosine) = cos ( ) ) Inverse tngent (rctngent) = tn ( opp ) On scientific clcultor, these functions re lbeled SIN, COS nd TN. They re usully directly bove the SIN, COS, nd TN keys. To find the ngle on the clcultor, first determine if you wnt the nswer in degrees or in rdins nd set your clcultor in the pproprite mode. Second, type the rtio of the two sides you re given nd hit the nd (or INV) nd then the pproprite trigonometric rtio key. On some scientific clcultors, you might hve to enter the rtio fter hitting the nd (or INV) nd the pproprite trigonometric rtio key. It is interesting to note tht the inverse sine nd the inverse tngent on scientific clcultor will produce n nswer between 90 nd 90 inclusively, wheres the inverse cosine will produce nd nswer between 0 nd 80 inclusively. For cute ngles
11 96 of right tringles, the clcultor will yield the desire result, but outside of tht context, one needs to be creful. Verify tht the cute ngles of -4-5 tringle re 6.9 nd 5. : Ex. 4 5 Since we hve ll three sides, we cn use ny of the inverse trigonometric functions. Let's strt with nd we will use the inverse sine function. m = sin ( opp ) = sin ( 5 ) = sin ( 5 ) Enter /5 nd hit nd SIN: sin ( ) = For B, let's use the inverse tngent function: m B = tn ( opp ) = tn ( 4 ) = tn ( 4 ) Enter 4/ nd hit nd TN: tn ( 4 ) = If we hd use the inverse cosine for B, we would hve derived the sme result: m B = cos ( ) = cos ( 5 ) = cos ( 5 ) Enter /5 nd hit nd COS: cos ( ) = Solve the following. Round to the nerest minute: Ex. sin x = 0.8 Ex. b cos x = 0.58 Ex. c tn x =.8 B
12 97 ) Since sin x = 0.8, then x = sin (0.8). Type 0.8 nd hit nd SIN: sin (0.8) = To convert to DMS, hit DD DMS (nd =) = 56 4'9", but 9" is < 0" so we round down. 56 4' b) Since cos x = 0.58, then x = cos (0.58). Type 0.58 nd hit nd COS: cos (0.58) = 58. To convert to DMS, hit DD DMS (nd =) 58. = 58 07'46"4, but 46" 0", so we round up 58 8' c) Since tn x =.8, then x = tn (.8). Type.8 nd hit nd TN: tn (.8) = 6. To convert to DMS, hit DD DMS (nd =) 6. = 6 '48", but 48" is 0" so we round up. 6 '
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