ME Jmi: STR I UIT 32 ngles, irles n Tngents: Stuent Tet ontents STR I: Geometry n Trigonometry Unit 32 ngles, irles n Tngents Stuent Tet ontents Setion 32.1 ompss erings 32.2 ngles n irles 1 32.3 ngles n irles 2 32.4 irles n Tngents IMT n e-lerning Jmi
ME Jmi: STR I UIT 32 ngles, irles n Tngents: Stuent Tet 32 ngles, irles n Tngents 32.1 ompss erings When esriing iretion, the points of ompss n e useful, e.g. S (south) or SW (south west). W E W E SW S SE ering n lso e use, often in nvigtion n y people wlking on rough or open moorln or hills. ote erings re lwys mesure lokwise from north n use 3 igits. The ering of from is 050. The ering of from is 210. Worke Emple 1 n mp of Keny, fin the erings of () () Wjir from iroi Mkinu from Moms. IMT n e-lerning Jmi 1
32.1 ME Jmi: STR I UIT 32 ngles, irles n Tngents: Stuent Tet Solution Mp of Keny () First rw in north line t iroi n nother line from iroi to Wjir. Then mesure the ngle lokwise from north to the seon line. In this se the ngle is 47 so the ering is 0 47. () rw north line t Momss n line from Momss to Mkinu. The ering n then e mesure s 312. Worke Emple 2 ot sils for 500 miles on ering of 070 n then sils further 700 nutil miles on ering of 200. Fin the istne of the ot from its strting point n the ering tht woul hve tken it stright there. Solution To fin the solution use sle rwing. 1. rw north rrow t the strting point. 2. Mesure n ngle of 70 from orth. 3. rw line 5 m long. (1 m represents 100 nutil miles.) 70 5 m IMT n e-lerning Jmi 2
32.1 ME Jmi: STR I UIT 32 ngles, irles n Tngents: Stuent Tet 4. rw seon north rrow n mesure n ngle of 200. 5. rw line 7 m long. 70 5 m 200 7m 6. Join the finl point to the strting point n mesure the istne. It is 5.4 m, whih represents 540 nutil miles. 5 m 200 70 7. The ering n lso e mesure s 155. 155 7m 5.4m Worke Emple 3 ship leves port n sils to port Q on ering of 124. From Q, the ship trvels to port R on ering of 320. Given tht the ering of R from is 025 : () rw refully lelle igrm to represent the journey of the ship. () etermine the ering of from R. Solution () R 025 ot to sle 124 () ering of from R is 180 + 025 = 205 Q IMT n e-lerning Jmi 3 320
32.1 ME Jmi: STR I UIT 32 ngles, irles n Tngents: Stuent Tet erings re use gin in setion 4 of Unit 34, when you lulte istnes etween points. Eerises 1. The igrm shows the positions of 8 friens. Tevin Rih Kneil W E Riky Krist Shmoy Shvon Joel S () () () () (e) (f) (g) Who is iretly south of Rih? If Kneil wlks SE, whom will she meet? If Rih wlks SW, whom will she meet? Who is iretly west of Krist? Who is W of Krist? Who will Shvon meet if he wlks W? In wht iretion shoul Shmoy wlk to fin Rih? 2. The mp shows some towns n ities in the UK. ernrfon Rhyl ngor Mount Snowon Snowon Llner Llngollen erystwyth Lmpeter Milfor Hven Swnse Rhon ewport riff IMT n e-lerning Jmi 4
32.1 ME Jmi: STR I UIT 32 ngles, irles n Tngents: Stuent Tet Write own the ering of eh of the following ples from Mount Snowon. () Llngollen () ewport () Swnse () ngor (e) Milfor Hven (f) erystwyth 3. In orer to voi n re of ngerous roks yht sils s shown in the igrm. RKS Roks 1m r (1 m represents 100 m) () () Fin the ering of the yht s it sils from (i) to (ii) to (iii) to. How muh further oes the yht trvel to voi the roks? 4. roughly skethe mp of the US is shown elow. SETTLE S FRIS ETRIT EW YRK LS GELES MEMHIS HUST MIMI Fin the erings of: () Mimi from Memphis. () etroit from ew York () Los ngeles from etroit. () Settle from Houston. (e) Memphis from Sn Frniso. IMT n e-lerning Jmi 5
32.1 ME Jmi: STR I UIT 32 ngles, irles n Tngents: Stuent Tet 5. Use sle rwing to fin the nswers to eh of the following prolems. () () () () (e) If mn wlks 700 m on ering of 040, how fr north n how fr est is he from his strting point? og runs 50 m on ering of 230 n then runs north until he is west of his strting point. How fr is the og from its strting point? heliopter flies 80 km north n then 20 km SW. Wht ering woul hve tken the heliopter iretly to its finl position? How fr is the heliopter from its strting point? ot trvels 500 m E n then 500 m south. Wht ering woul tke the ot k to its strting point? plne flies 300 km west n then further 200 km SW. It then returns iretly to its strting point. n wht ering shoul it fly n how fr oes it hve to trvel? 6. Use sle rwing to illustrte eh of the following journeys. esrie the return journey in eh se using single ering n istne. () 120 m on 090 followe y 120 m on 180. () 500 km on 045 followe y 200 km on 270. () 300 km on 220 followe y 300 km on 170. () 25 km on 330 followe y 30 km on 170. (e) 10 km on 160 followe y 2 km on 300. (f) 30 km on 120 followe y 30 km on 270. (g) 1000 m on 050 followe y 1200 m on 310. 7. ship sils from point to nother point, 8000 m ue est of. It then sils in nother iretion n rrives t point, 10 000 m SE of. n wht ering i the ship sil on the seon stge of the journey n how fr i it trvel? 8. The position of ship from is 5 km on ering of 040. The position of ship from is 10 km on ering fo 040. 0 10 20 30 40 50 60 70 80 90 2.5 5 7.5 10 12.5 km Wht is the position of ship from ship? IMT n e-lerning Jmi 6
32.1 ME Jmi: STR I UIT 32 ngles, irles n Tngents: Stuent Tet 9. orth orth orth y 80 z 32 The igrm is not rwn to sle. The igrm shows the position of three ples,, n. is the sme length s. () (i) lulte the size of the ngle mrke. (ii) Eplin why the ngle mrke y is equl to 32. (iii) lulte the size of the ngle mrke z. () Use your nswers to () to lulte the ering of (i) from (ii) from (iii) from. 10. The figure elow, not rwn to sle, represents the journey of n irrft flying from Y to X n then from X to Z. X Y Z The ering of X from Y is 035. The ering of Z from X is 125. Z is ue est of Y. () opy n omplete the igrm, showing LERLY the erings 035 n 125. () etermine the size of ngle YXZ. () etermine the size of the ngle XZY. (X) IMT n e-lerning Jmi 7
32.1 ME Jmi: STR I UIT 32 ngles, irles n Tngents: Stuent Tet 11. The igrm shows y in whih yhts re moore. The igrm hs een rwn to sle of 5 m to 1 km. White Rok orth y View () The yht res is moore t. Mesure the ering of this yht from y View. () The yht Wet-n-Winy is moore 1.2 km from White Rok on ering of 210. Tre the igrm n mrk with ross the position of this yht on the igrm. Investigtion rw retngle of ny size. Use your ruler to lote the mi-points of the sies. Join these mi-points to form new qurilterl. Wht is the nme of the qurilterl you hve otine? Repet the ove y rwing () trpezium () prllelogrm () kite () rhomus (e) qurilterl of 4 unequl lengths. Wht onlusion n you rw from these? IMT n e-lerning Jmi 8
ME Jmi: STR I UIT 32 ngles, irles n Tngents: Stuent Tet 32.2 ngles n irles 1 The following results re true in ny irle. When tringle is rwn in semi-irle s shown, the ngle on the perimeter is lwys right ngle. roof Join the entre,, to the point,, on the perimeter. Sine then = ngle = ngle (equl rii) (=, sy) y Similrly, tringle is lso isoseles n ngle = ngle (= y, sy). y In tringle, the sum of the ngles must e 180. Then y + + ( + y) = 180 2 + 2y = 180 (olleting like terms) + y = 90 ( 2) ut ngle = + y, n this is right ngle. tngent is line tht touhes only one point on the irumferene of irle. This point is known s the point of tngeny. tngent is lwys perpeniulr to the rius of the irle. hor is line joining ny two points on the irle. The perpeniulr isetor is seon line tht ivies the first line in hlf n is t right ngles to it. The perpeniulr isetor of hor is lwys rius of the irle. Rius hor oint of tngeny Tngent Rius hor When the ens of hor re joine to the entre of irle, n isoseles tringle is forme, so the two ngles mrke re equl. IMT n e-lerning Jmi 9
32.2 ME Jmi: STR I UIT 32 ngles, irles n Tngents: Stuent Tet Worke Emple 1 Fin the ngles mrke with letters in the igrm, if is the entre of the irle. Solution s oth tringles re in semi-irles, ngles n must eh e 90. S 40 70 R Q The other ngles n e foun euse the sum of the ngles in eh tringle is 180. For the tringle QS, For the tringle QRS, 40 + 90 + = 180 = 180 130 = 50 70 + 90 + = 180 = 180 160 = 20 Worke Emple 2 20 Fin the ngles, n, if is tngent n is the entre of the irle. Solution First onsier the tringle. s is rius n is tngent, the ngle etween them is 90. So 90 + 20 + = 180 = 180 110 90 20 = 70 Then onsier the tringle. s n re oth rii of the irle, it is n isoseles tringle with =. So 2 + 70 = 180 2 = 110 = 55 70 n = 55. IMT n e-lerning Jmi 10
32.2 ME Jmi: STR I UIT 32 ngles, irles n Tngents: Stuent Tet Worke Emple 3 Fin the ngles mrke in the igrm, where is the entre of the irle. Solution First onsier the tringle. 100 s the sies n re oth rii, the tringle must e isoseles with =. So + + 100 = 180 ut s =, 2 + 100 = 180 n = 40. ow onsier the tringle. 2 = 80 = 40 100 s the line is imeter of the irle, the ngle must e 90. So + = 90 or 40 + = 90 = 50 The ngles in the tringle must totl 180, so 40 40 40 + 90 + = 180 = 50 Eerises 1. Fin the ngles mrke with letter in eh of the following igrms. In eh se the entre of the irle is mrke. () () 25 18 IMT n e-lerning Jmi 11
32.2 ME Jmi: STR I UIT 32 ngles, irles n Tngents: Stuent Tet () () 14 16 32 58 (e) (f) g 41 f 40 e 60 e (g) (h) e 92 f 20 e 2. opy the igrm opposite, n mrk every right ngle, if is the entre of the irle. 3. Fin the ngles mrke with letters in eh igrm elow, if is the entre of the irle. () 25 () 106 is tngent n re tngents IMT n e-lerning Jmi 12
32.2 ME Jmi: STR I UIT 32 ngles, irles n Tngents: Stuent Tet () 40 15 is tngent () 10 n re tngents 4. Fin the ngles mrke with letters in eh of the following igrms, if is the entre of the irle. () () 20 130 () () 80 30 IMT n e-lerning Jmi 13
32.2 ME Jmi: STR I UIT 32 ngles, irles n Tngents: Stuent Tet (e) (f) 48 e e 30 (g) e (h) 70 42 28 e 5. Fin eh of the mrke ngles if is the entre of the irle. () 80 () 70 30 40 () 60 e 50 f g () 55 50 6. Fin the imeter of eh irle elow, if is the entre of the irle. () 4 3 () 5 6 IMT n e-lerning Jmi 14
32.2 ME Jmi: STR I UIT 32 ngles, irles n Tngents: Stuent Tet () 8 10 is tngent () 13 12 is tngent 32.3 ngles n irles 2 There re numer of importnt geometri results se on ngles in irles. (The first you hve met lrey.) ny ngle sutene t the irumferene from imeter is right ngle. The ngle sutene y n r, Q, t the entre is twie the ngle sutene t the irumferene. roof 2 = (equl rii), so ngle = ngle (=, sy). Q Similrly, ngle Q = ngle Q (= y, sy). ow, etening the line to, sy, note tht ngle = + y n, similrly, = 2 ngle Q = y + y y = 2y Q IMT n e-lerning Jmi 15
32.3 ME Jmi: STR I UIT 32 ngles, irles n Tngents: Stuent Tet Hene, s require. ngle Q = 2 + 2y = 2 ( + y) = 2 ngle Q ngles sutene t the irumferene y hor (on the sme sie of the hor) re equl; tht is, in the igrm =. roof The ngle t the entre is 2 or 2 (oring to the first result). Thus 2 = 2 or =, s require. Q In yli qurilterls (qurilterls where ll 4 verties lie on irle), opposite ngles sum to 180 ; tht is + = 180 n + = 180 roof onstrut the igonls n, s elow. Then lel the ngles sutene y s w; tht is ngle = ngle (= w) Similrly for the other hors, the ngles eing mrke, y n z s shown. ow, in tringle, the sum of the ngles is 180, so w + z + ( + y) = 180 w z w y z y You n rerrnge this s ( + w) + ( y + z) = 180 whih shows tht ngle + ngle = 180 proving one of the results. The other result follows in similr wy. IMT n e-lerning Jmi 16
32.3 ME Jmi: STR I UIT 32 ngles, irles n Tngents: Stuent Tet Worke Emple 1 Fin the ngles mrke in the igrms. In eh se is the entre of the irle. () 35 () 25 Solution () s oth ngles re rwn on the sme hor, the ngles re equl, so = 35 () ngle n the 25 ngle re rwn on the sme hor, so = 25 ngle is rwn t the entre on the sme hor s the 25 ngle, so Worke Emple 2 = 2 25 = 50 Fin the ngles mrke in the igrms. is the entre of the irle. () () 110 210 20 80 e Solution () pposite ngles in yli qurilterl up to 180. So + 80 = 180 = 100 n + 110 = 180 = 70 IMT n e-lerning Jmi 17
32.3 ME Jmi: STR I UIT 32 ngles, irles n Tngents: Stuent Tet () onsier the ngles n 210. Sine the ngle t the entre is oule the ngle in segment rwn on the sme r, 2 = 210 = 105 ngles n e up to 180 euse they re opposite ngles in yli qurilterl. + e = 180 105 + e = 180 e = 180 105 = 75 onsier the qurilterl. The four ngles in ny qurilterl up to 360. So + e + 210 + 20 = 360 = 360 210 20 = 130 = 130 75 = 55 e Worke Emple 3 In the igrm the line is imeter n is the entre of the irle. Fin the ngles mrke. Solution onsier tringle. Sine n re rii, tringle is isoseles. So = 50 50 20 e The ngles in tringle to 180, so for tringle, + + 50 = 180 = 180 50 = 80 Sine is imeter of the irle, the ngle is right ngle, so + 20 + = 90 = 90 20 = 20 ngle n ngle re ngles in the sme segment, so = ngle = 50 IMT n e-lerning Jmi 18
32.3 ME Jmi: STR I UIT 32 ngles, irles n Tngents: Stuent Tet ngle e is rwn on the sme r s the ngle t the entre,, so e = 2e = 1 2 = 40 Worke Emple 4 In the igrm the hors n re prllel. rove tht the tringles E n E re isoseles. Solution E ngles n re ngles in the sme segment, so ngle = Sine n re prllel, ngles n re equl lternte ngles, ngle = = ngle Hene in tringle E, the se ngles t n re equl, so the tringle is isoseles. The ngle t, ngle, equls the ngle t, ngle, euse they re ngles in the sme segment: ngle = ngle = Hene tringle E is isoseles, sine the ngles t n re equl. Eerises 1. Fin ll the ngles mrke with letter in eh of the following igrms. In eh se the entre of the irle is mrke. Give resons for your nswers. () 30 () 120 () () 73 70 e IMT n e-lerning Jmi 19
32.3 ME Jmi: STR I UIT 32 ngles, irles n Tngents: Stuent Tet (e) (f) g f 35 y 47 (g) 50 (h) 27 2. In the igrm, is the entre of the irle n, F n G re imeters. F E () () Ientify the equl ngles. Ientify the right ngles. Give resons for your nswers. G 3. Fin ll of the ngles mrke with letter in eh of the following igrms. Give resons for your nswers. () 105 60 () 57 31 () () 110 80 (e) 105 42 (f) 75 100 IMT n e-lerning Jmi 20
32.3 ME Jmi: STR I UIT 32 ngles, irles n Tngents: Stuent Tet (g) (h) 85 115 80 4. Whih of the following points re onyli points? 108 115 72 H 100 G 80 100 F U 65 R 80 E V 56 28 T Q S 5. Fin ll the ngles mrke with letter in the following igrms. In eh se, the point is the entre of the irle. () 35 144 () e 65 15 () () 68 25 = IMT n e-lerning Jmi 21
32.3 ME Jmi: STR I UIT 32 ngles, irles n Tngents: Stuent Tet 6. Fin the vlue of in eh of the following igrms. () 2 + 30 () 2 + 70 6 20 () 2 10 + 30 7. In the igrm, is the entre of the irle, = n is stright line. rove tht isets the ngle. 8. In the igrm is the entre of the irle, = Q = QR R rove tht n QR re prllel lines. 9. is the entre of the irle. Q 47 is tngent to the irle t. y ot to sle 28 Work out the size of ngles, y n z. z IMT n e-lerning Jmi 22
32.3 ME Jmi: STR I UIT 32 ngles, irles n Tngents: Stuent Tet 10. E In the igrm, is the entre of the irle, is imeter n is tngent. ngle E =. Fin, in terms of, the size of: () ngle E () ngle E () ngle E () ngle E 11. 45 y The igrm ove, not rwn to sle, shows irle, entre. is prllel to n ˆ = 45. () lulte, giving resons, the vlues of n y. () Show tht is squre, giving the resons for your nswer. (X) IMT n e-lerning Jmi 23
32.3 ME Jmi: STR I UIT 32 ngles, irles n Tngents: Stuent Tet 12. M L 64 20 Q The igrm shows irle LQ, not rwn to sle, with entre, ngle QM = 20 n ngle ML = 64. lulte, in egrees, giving resons for your nswers, the size of ngles () () () () LQ Q L L (X) 32.4 irles n Tngents Some importnt results re stte elow. 1. If two tngents re rwn from point T to irle with entre, n n R re the points of ontt of the tngents with the irle, then, using symmetry, T () () R T = RT Tringles T n TR re ongruent. IMT n e-lerning Jmi 24
32.4 ME Jmi: STR I 2. The ngle etween tngent n hor equls n ngle t the irumferene sutene y the sme hor; e.g. = in the igrm. This is known s the Q lternte segment theorem n nees proof, s it is not oviously true! UIT 32 ngles, irles n Tngents: Stuent Tet R T roof onstrut the imeter S, s shown. We know tht Q S ngle SR = 90 sine S is imeter. R ow ngle SR = ngle QR =, sy, so ngle SR = 180 90 ut = 90 ngle RT = 90 ( ngle SR) = 90 ( 90 ) = = ngle QR n the result is prove. T 3. For ny two interseting hors, s shown, X X = X X The proof is se on similr tringles. X roof In tringles X n X, ngle = ngle (equl ngles sutene y hor ) n ngle = ngle (equl ngles sutene y hor ) s X n X re similr, s require. X X X = X.X = X X.X IMT n e-lerning Jmi 25
32.4 ME Jmi: STR I UIT 32 ngles, irles n Tngents: Stuent Tet This result will still e true even when the hors interset outsie the irle, s illustrte opposite. How n this e prove? X When the hor eomes tngent, n n oinie t the point, then X X = X X or X X = X 2 X Worke Emple 1 Fin the ngle in the igrm. Solution T 65 The tringles TR n T re ongruent, so ngle TR = 65 R Sine TR is tngent to the irle n R is rius, Hene ngle TR = 90. Worke Emple 2 = 180 90 65 = 25 Fin the ngles n y in the igrm. Solution The lternte ngle segment theorem gives = 62 The tngents T n T re equl in length, so the tringle T is isoseles. So ngle T = = 62 Hene y + 62 + 62 = 180 (the ngles in tringle T up to 180 ) y = 56 T y 62 IMT n e-lerning Jmi 26
32.4 ME Jmi: STR I UIT 32 ngles, irles n Tngents: Stuent Tet Worke Emple 3 Fin the unknown lengths in the igrm. Solution Sine T is tngent, Hene T 2 = T.T 36 = T 4 T = 9 y y + 8 = T = 9 2.5 m 4 m 6 m 4 m T y = 1m n re interseting hors, so. =. 25. = 1 4 = 4 25. = 16. m Eerises 1. Fin the ngles mrke in the igrms. In eh se is the entre of the irle. () () 40 35 () 20 IMT n e-lerning Jmi 27
32.4 ME Jmi: STR I UIT 32 ngles, irles n Tngents: Stuent Tet () 117 52 (e) (f) 35 52 59 2. Fin the unknown lengths in the following igrms. () () 7 m 5 m T X 4 2 3 () 3 5 T 5 2.5 y () 2 y 6 3 4 (e) 3 2 4 y 2 2 (f) 3 2 3 2 2 y 2 IMT n e-lerning Jmi 28
32.4 ME Jmi: STR I UIT 32 ngles, irles n Tngents: Stuent Tet 3. In the igrm, T is tngent to the irle. () rove tht =. () () Show tht tringles T n T re similr tringles. If = 5 m T = 4 m T lulte the length of the tngent T. 4. The tringle in the igrm is n equilterl tringle. The line isets ngle. () rove tht the line is imeter of the irle. () Hene fin the ngle T, where T is tngent to the irle. T 5. In the igrm, T n T re tngents. 36 74 Fin the ngles, n. Give resons for your nswers. T 6. T n T re tngents to the irle, entre. is point on the irumferene, s shown. y 65 z T ngle T = 65 lulte the size of () () y () z IMT n e-lerning Jmi 29
32.4 ME Jmi: STR I UIT 32 ngles, irles n Tngents: Stuent Tet 7. 40 T In the igrm ove, not rwn to sle, T is irle. n T re imeters. T, the tngent t T, meets proue t, so tht T = 40. lulte, giving resons for ll sttements, the size of () () () () T T T T (X) IMT n e-lerning Jmi 30