.1 ANGLES AND THEIR MEASURE Given two interseting lines or line segments, the mount of rottion out the point of intersetion (the vertex) required to ring one into orrespondene with the other is lled the ngle etween them. Angles re usully mesured in degrees (denoted ), rdins (denoted rd, or without unit), or sometimes grdins (denoted grd). II y I β nd re oterminl ngles euse they hve the sme initil nd sme terminl side. - β Vertex Terminl side + Initil side Rys re hlf-lines, tht is, portion of line tht strts vertex point nd extends indefinitely in one diretion. x III IV In retngulr oordinte system, ngles re defined s the rottion from one ry to nother. The ry t the strting position is lled the initil side nd the ry t then ending position is the terminl side. Rottion in ounterlokwise diretion is positive nd rottion in lokwise diretion is negtive. An ngle is in stndrd position if its vertex is t the origin nd its initil side oinides with the positive x-xis.
II Qudrntl Angles One full rottion in these three mesures orresponds to 360, π rdins, or 400 grdins. Hlf full rottion is 180 nd is lled stright ngle, nd qurter of full rottion is 90 nd is lled right ngle. An ngle less thn right ngle is lled n ute ngle, nd n ngle greter thn right ngle is lled n otuse ngle. y α 90, 40, 810, -70, et. I β 180, 40,900, et. β φ α δ 0, 360,70, et. III δ IV 360 is one full revolution. A right ngle, 90, is 90/360 1/4 revolution. Φ 70, 40, 810, - 90, et. Converting from Degrees, Minutes, Seonds (D, M,S ) to Deiml Form The use of degrees to mesure ngles hrks k to the Bylonins, whose sexgesiml numer system ws sed on the numer 60. likely rises from the Bylonin yer, whih ws omposed of 360 dys (1 months of 30 dys eh). The degree is further divided into 60 r minutes (denoted ), nd n r minute into 60 r seonds(denoted ). A more nturl mesure of n ngle is the rdin. It hs the property tht the r length round irle is simply given y the rdin ngle mesure times the irle rdius. The rdin is lso the most useful ngle mesure in lulus. Grdins re sometimes used in surveying (they hve the nie property tht right ngleis extly 100 grdins), ut re enountered infrequently, if t ll, in mthemtis. 1 ounterlokwise revolution 360 1 60 60 minutes 1 minute 1 60 60 seonds Exmple on p.117 Convert 0 6 1 to deiml in degrees. 1 60 whih mens 1 1 /60 1 60 so 1 1 /60 (1 /60)/60 1 /3600 0 6 1 0 + 6 *1 /60 + 1 *1 /3600 0.10833 Convert 1.6 to (D, M,S ) in degrees. Strt with the deiml prt,.6..6.6*60 / 1 1.36 Tke the deiml prt of 1.36 nd onvert to seonds..36 * 60 /1 1.6 Thus, 1.6 1 +1 +1.6 1 1
Converting from Degrees, Minutes, Seonds (D, M,S ) to Deiml Form Using TI-83 or TI-83 plus Convert 0 6 1 to deiml in degrees. 0 nd [ANGLE] 1 6 nd [ANGLE] 1 ALPHA ENTER Selets Degree nottion Selets Minute nottion Selets Seond nottion To onvert from deiml to DMS, just type the numer in deiml form, then press nd [ANGLE] 4 to see it in DMS form. Now you do #3 & #9 on p.1
Centrl Angles nd Ar Length A entrl ngle is n ngle whose vertex is t the enter of irle. The rys of entrl ngle sutend (interset) n r on the irle. [The entrl ngle is etween 0 nd 360 degrees]. 1 rdin is defined s entrl ngle of irle with rdius r tht hs n r length of r. If the irumferene of irle is πr, how mny rdins re there in 1 full revolution of the irle? Terminl Side Vertex r 3 r r 1 1 Ar length r 1 1 Ar length r 3 1 rdin Initil side r Ar length s r Ar length s 1 1 From geometry, we know tht the rtio of the mesures of the ngles equls the rtio of the orresponding lengths of the rs sutended y these ngles. s 1 s 1 s 1 r If 1 1 rdin, then the r length s 1 r, so nd ross-multiplyinggivess r for ny in theirle. ARC LENGTH For irle of rdius r, entrl ngle of rdins sutends n r whose length s is s r
How do you find r length if the ngle is given in degrees? Rell tht the irumferene of irle is πr. The r length of full revolution ( full ) is πr r full Therefore full π rdins 1 full revolution 360 π rdins ½ revolution 180 π rdins ¼revolution 90 π/ rdins Exmple 4 () on p. 10 Convert 60 to rdins. 60 * π rdins/180 60 π /180 rdins π/3 rdins Now you do #3 on p.1 Exmple () on p.10 Convert -3π/4 rdins to degrees. (-3π/4)*180 / π -13 Now you do #47 on p.1 Ltitude is G N s AG is the r length of the entrl ngle, AG, t point G. sutended y the ry from Erth s enter to Glsgow nd the ry from Erth s enter to Aluquerque. Tht is the distne from Glsgow to Aluquerque. G Step 1: AG G A 48 9-3 13 4 Step : Ltitude is 0 Convert from DMS to deiml degreest Equtor 13 4 13 + 4 /(1 /60 ) 13.0667 Step 3: Convert from deiml degrees to rdins. 13.0667 *(π rdins/180) 0.8 rdins. Step 4: Clulte r length s AG. s AG R Erth * AGrdins 3960* 0.8 903 miles s AG A G We use this ft to onvert from degrees to rdins. 1 rdin 180 / π 1 degree π rdins/180 Exmple 6 on p.11 Ltitude of lotion is the ngle formed y ry drwn from the enter of Erth to the Equtor nd ry drwn from the enter of Erth to the lotion. Glsgow, Montn is due north Aluquerque, New Mexio. Glsgow s ltitude is 48 9 N nd Aluquerque s ltitude is 3 N. Find the distne etween these ities. A Rdius 3960 miles Now do #101 on p.16
Common Degree to Rdin Conversions Degrees 0 30 4 60 90 10 13 10 180 10 40 70 300 31 330 360 Rdins 0 π/6 π/4 π/3 π/ π/3 3π/4 π/6 π 7π/6 π/4 4π/3 3π/ π/3 7π/4 11π/6 π * * * * * Memorize these
Are of Setor From geometry, we know tht the rtio of the mesures of the ngles equls the rtio of the orresponding res of the setors formed y these ngles. Tht is, A 1 A 1 If 1 π rdins, then A 1 re of irle. π A πr A A π πr r r Multiply oth sides y to solve for A. Are of setor formed y entrlngle of A 1 r rdins is Now you do #4 on p.109
Liner nd Angulr Speed Liner speed distne trveled round irle (s) divided y the elpsed time of trvel (t). veloity v s/t Angulr speed the entrl ngle swept out in time (), divided y the elpsed time, (t). Angulr speed ω /t where ngulr speed is in rdins per unit time. Exmple: An engine is revving t 900 RPM. If you re given RPM insted ofngulr speed, you n onvert to ngulr speed y using tht ft tht 1 revolution π rdins revolutions revolutions π rdins 900 900 1800π minute minute 1 revolution rdins minute Liner speed v s/t (r)/t r(/t) rω where ω is mesured in rdins per unit time. Exmple 8 on p.13 A hild is spinning rok t the end of -foot rope t rte of 180 revolutions per minute. Find the lier speed of the rok when it releses. revolutions revolutions π rdins 180 180 360π minute minute 1revolution rdins minute Liner speed v rw (ft) * (360π rdins/minute) 70π feet/min 6 feet/min feet 1mile 60 min 6 * *.7 mph min 80 feet 1hour Now you do # 97 on p.16
. Right Tringle Trigonometry is n ute ngle euse it is less thn 90 degrees. Hypotenuse Adjent to 90 - Opposite to From geometry, we know tht the rtios of the sides of similr tringles re equl so: ', ' ', ' ', ' ', ' These rtios re the sme for ny right tringle with ute ngle. They re lled the trigonometri funtions of ute ngles. ' ' ' ' FUNCTION NAME ABBREV. VALUE Notie these funtions re the reiprols of sine, osine, & tngent, respetively. Sine of Cosine of Tngent of Cosent of Sent of Cotngent of sin() /opposite/hypotenuse os() /djent/hypotenuse tn() /opposite/djent s() /hypotenuse/opposite se() /hypotenuse/djent ot() /djent/opposite Rememer SOH-CAH-TOA! In other words: s()1/sin(), se()1/os(), ot()1/tn() Do #11 on p.137
Finding Trig Funtions Exmple on p.130 Finding the Vlue of the Remining Trig Funtions, given sin nd os. Given sin( ), nd os( ) sin( ) 1 tn( ) os( ) se( ) 1/ os( ) s( ) 1/ sin( ), find thereminingtrig funtionsof NOW YOU DO #1 on p.138
Fundmentl Identities of Trigonometry We n use the Pythgoren Theorem to derive the fundmentl identities of trigonometry. We know +, right? Let s rerrnge some terms nd divide eh side y ( sin( )) + ( os( )) whih n lso e written s follows : sin Now we n get nother identity y dividing oth sides of sin ( ) os + os ( ) os tn sin sin + 1+ ot ( ) + os ( ) + 1 se ( ) os + ( ) sin 1 1 ( ) 1 ( ) ( ) ( ) Similrly, you hd divided eh side of + + ( ) s ( ) 1 1 os ( ) ( ) 1 ( ) sin ( ) Exmple 3 on p.11 tht eqution y sin Fundmentl Identities Hypotenuse 90 - Adjent to Opposite to this eqution yos ( ) : ( ) tn() sin()/ os() ot() os()/ sin() s() 1/sin() se() 1/ os() ot() os()/ sin() sin () + os () 1 tn () + 1 se () 1 + ot () s ()
Finding the ext vlue of the trig funtions, given one. Exmple 4 on p. 13 Given tht sin() 1/3 nd is n ute ngle, find the ext vlue of eh of the remining five trigonometri funtions of. Rememer sin() 1/3 opposite/hypotenuse so let s mke right tringle with the opposite of to e 1 nd the hypotenuse to e 3. We n use the Pythgoren Theorem to find the djent side,. 3? 1 3-1 9-1 8 8 4 Now tht you know ll three sides of the tringle, just use the definitions to find the ext vlues of the trig funtions. Trig Funtion sin() os() tn() s() se() ot() Definition /opposite/hypotenuse /djent/hypotenuse /opposite/djent /hypotenuse/opposite /hypotenuse/djent /djent/opposite Ext Vlue /1/3 / / /3/1 3 / / 3 3 1 1 4 3 4
Complementry Angles; Cofuntions α 90- β β Hypotenuse α Opposite to β Adjent to α α nd β re omplementry ngle euse their sum is right ngle, 90. Adjent to β Opposite to α Complementry Angle Theorem Cofuntions of omplementry ngles re equl. Cofuntions re trigonometri funtions tht shre the sme ngles of right tringle. sin(β)opposite to β / hypotenuse djent to α/hypotenuse os(α)os(90- β) os(β)djent to β / hypotenuse opposite to α/hypotenuse sin(α)sin(90- β) tn(β)opposite to β / djent to α djent to α/ opposite to β ot(α) ot(90- β) Likewise, the reiprol of these properties is true. s(β) hypotenuse / opposite to β / hypotenuse /djent to α se(α)se(90- β) se(β) hypotenuse /djent to β hypotenuse/opposite to α s(α) s(90- β) ot(β) djent to α /opposite to β opposite to β / djent to α tn(90- β)
HOMEWORK Homework p. 1-16 #3 103 EOO** p. 137-138 #3-60 ETP*, #67 * Every Third Prolem ** Every Other Odd prolem