hpter 12 Trigonometry ontents: Using scle digrms Lelling tringles The trigonometric rtios Trigonometric prolem solving erings 3-dimensionl prolem solving
268 TRIGOOMTRY (hpter 12) OPIG PROLM indi s office uilding hs stellite dish on the roof. It is ngled t 41 o to horizontl. indi is concerned tht new uilding cross the rod my ostruct the pth of the stellite connection. The new uilding is 26 m higher thn the stellite dish, nd is 30 m from it. n you nswer indi s concern? 30 m 26 m Prolems like the Opening Prolem cn e solved y either scle digrm or y right ngled tringle trigonometry. Trigonometry is the study of the reltionship etween lengths nd ngles of geometricl figures. Trigonometry uses other rnches of mthemtics such s lger, rithmetic nd geometry to find unknown lengths nd ngles of tringles. USIG SL IGRMS To represent physicl sitution we cn often drw scle digrm. ll lengths on the digrm re in proportion to those in relity ccording to the digrm s scle. The ngles in scle digrm will equl the corresponding ngles in relity. Scle digrms cn e used to find the lengths of sides nd ngles of geometricl figures. xmple 1 rom the top of n emnkment, ion mesured the ngle etween the horizontl ground nd the top of tower to e 47 o. The emnkment is 35:4 m from the centre of the se of the tower. How high is the tower? 47 35.4 m emnkment irst we choose suitle scle, in this cse 1 cm 10 m or 1 mm 1 m. We drw horizontl line segment [] which is 35:4 mm. t the left end we drw verticl line using set squre. We use protrctor to drw 47 o ngle t. The point where the two lines meet is, which is the top of the tower. We mesure in mm nd use the scle to convert ck to metres. ¼ 38 mm. ) the tower is pproximtely 38 metres high. 35.4 mm 47
TRIGOOMTRY (hpter 12) 269 ISUSSIO Wht fctors might cuse errors when using scle digrms? omment on the ccurcy of nswers otined using scle digrms. XRIS 12 1 onvert this rough sketch into n ccurte scle digrm using scle of 1 cm 1 m. Use the scle digrm to find s ccurtely s you cn the length of: [QR] [PR]. P 34 6m R Q 2 Use scle digrm to find the height of the flgpole. 33 79 m 3 The tringulr grden XYZ hs XY =12m, YZ =10:2 m, nd XZ =8:4 m. Use compss nd ruler to drw n ccurte scle digrm of the grden nd hence find the mesures of the grden s ngles. 4 mountin rises in the distnce from horizontl plne. surveyor mkes two ngle mesurements from the plne to the top of the mountin. The results re shown in the digrm longside. Use scle digrm to find the height of the mountin. 33.7 41.6 400 m 5 Try to solve the Opening Prolem on pge 268 using scle digrm. o you think scle digrm is sufficient to nswer indi s concerns in this cse? LLLIG TRIGLS Trigonometry enles us to find lengths nd ngles to greter ccurcy thn is possile using scle digrms. Loosely trnslted, trigonometry mens tringle mesure. hypotenuse In this section we will consider only right ngled tringles. The hypotenuse is the longest side of right ngled tringle, nd is opposite the right ngle. or given ngle µ in tringle, the opposite side is opposite the ngle µ. The third side is next to the ngle µ nd so is clled the djcent side.
270 TRIGOOMTRY (hpter 12) or exmple: djcent opposite opposite djcent djcent opposite opposite djcent xmple 2 In the digrm longside, find the: hypotenuse side opposite ngle c side djcent to ngle d side opposite ngle e side djcent to ngle. The hypotenuse is []. [] is the side opposite ngle. c [] is the side djcent to ngle. d [] is the side opposite ngle. e [] is the side djcent to ngle. XRIS 12 1 In the digrms elow, nme the: i hypotenuse ii side opposite ngle µ iii side djcent to ngle µ T c R S 2 The right ngled tringle longside hs hypotenuse of length units nd other sides of length units nd c units. µ nd Á re the two cute ngles. ind the length of the side: opposite µ opposite Á c djcent to µ d djcent to Á c TH TRIGOOMTRI RTIOS onsider the right ngled tringle opposite. We use: HYP to indicte the length of the hypotenuse OPP to indicte the length of the side opposite µ J to indicte the length of the side djcent to µ. HYP J OPP
TRIGOOMTRY (hpter 12) 271 We use these revitions to define the three sic trigonometric rtios: The sine of ngle µ is The cosine of ngle µ is The tngent of ngle µ is sin µ = OPP HYP. cos µ = J HYP. tn µ = OPP J. These rtios re the tools we use for finding the lengths of unknown sides nd ngles of right ngled tringles. IIG TRIGOOMTRI RTIOS xmple 3 or the tringle shown, find: sin µ cos Á c tn µ q r p sin µ = OPP HYP = p r cos Á = J HYP = p r c tn µ = OPP J = p q XRIS 12.1 1 or ech of the following tringles find: i sin µ ii cos µ iii tn µ iv sin Á v cos Á vi tn Á c c k l 3 4 m d e f 2 3 6 4 5 5 IIG SI LGTHS In right ngled tringle, if we re given nother ngle nd side length we cn find: ² the third ngle using the ngle sum of tringle is 180 o ² the other side lengths using trigonometry.
272 TRIGOOMTRY (hpter 12) The method: Step 1: Redrw the figure nd mrk on it HYP, OPP, nd J reltive to the given ngle. Step 2: or the given ngle choose the correct trigonometric rtio which cn e used to set up n eqution. Step 3: Set up the eqution. Step 4: Solve to find the unknown. When we solve equtions involving trigonometry, we evlute the trigonometric rtios for ngles using our clcultor. xmple 4 ind, to 2 deciml plces, the unknown length in the following tringles: 58 10 cm 8m x m 39 x cm sin 58 o = x f sin µ = OPP 10 HYP g ) sin 58 o 10 = x fmultiplying oth sides y 10g 58 10 cm J HYP ) x ¼ 8:48 f sin 58 ) 10 TR g OPP 8m x cm OPP HYP J x m 39 The length of the side is out 8:48 cm. tn 39 o = 8 x f tn µ = OPP J g ) x tn 39 o =8 fmultiplying oth sides y xg ) x = 8 tn 39 o fdividing oth sides y tn 39 o g ) x ¼ 9:88 f8 tn 39 ) TR g The length of the side is out 9:88 m. XRIS 12.2 1 Set up trigonometric eqution connecting the ngle nd the sides given: x x c 32 65 c 56 x d e f d 49 e 37 x x f 73 x
TRIGOOMTRY (hpter 12) 273 g h i x 54 27 h x g i 59 x 2 ind, to 2 deciml plces, the unknown length in: x cm x cm c 27 54 4.82 m 68 19 cm 10 cm d e f 12.96 km 31 39 8.67 m x m x km g h i 28 x m x km x mm 52.4 m 79.5 km 59 j k l 5.64 cm 64 66 x cm 3 ind, to one deciml plce, ll unknown ngles nd sides of: c cm 14.6 m 28 cm m 14.2 cm m 63 IIG GLS In the right ngled tringle shown, cos µ = 3 5 : So, to find µ we need n ngle with cosine of 3 5. If cos 1 (::::::) reds the ngle with cosine of..., we cn write µ = cos 1 3 5. We could lso sy tht µ is the inverse cosine of 3 5. In similr wy we cn define the inverse sine nd inverse tngent of n ngle. These inverses cn lso e found on clcultor. or grphics clcultor instructions, see pge 14. To find µ in this cse, press: 2nd cos 3 5 ) TR. The nswer is µ ¼ 53:1 o : With n ordinry scientific clcultor you my need to press IV, 2nd or SHIT nd then press cos ( 3 5 ) =. x m 26.8 m x km 43.2 cm HYP 5cm 38 49 x m 82.7 cm cm 74 34 11 mm 23.9 km J 3cm x cm cm
274 TRIGOOMTRY (hpter 12) xmple 5 ind, to one deciml plce, the mesure of the ngle mrked µ in: 8cm 5cm 5.88 m 3.17 m tn µ = 5 8 f tn µ = OPP J g ) µ = tn 1 5 HYP OPP 8 5cm ) µ ¼ 32:0 o f 2nd tn 5 8 ) TR g 8cm So, the ngle mesure is out 32:0 o. J cos µ = 3:17 5:88 fs cos µ = J HYP g ) µ = cos 1 3:17 5:88 3.17 m 5.88 m ) µ ¼ 57:4 o f 2nd cos 3:17 5:88 ) J TR g HYP So, the ngle mesure is out 57:4 o. XRIS 12.3 1 ind, to one deciml plce, the mesure of the ngle mrked µ in: 5cm c 4cm 6cm 4cm 4.8 cm 3.7 cm d e f 3.2 m 5.2 m 3.1 km 2.1 km 3.1 m 4.2 m g h i 5.1 km 4.4 m 6.1 m 7.5 mm 4.1 km 12.5 mm
TRIGOOMTRY (hpter 12) 275 2 ind, to 1 deciml plce, ll unknown sides nd ngles in the following: x cm c x m 9.45 km 5cm 3.5 m 8cm 6.1 m 12.62 km 3 heck your nswers for x in question 2 using Pythgors theorem. 4 ind µ in the following using trigonometry. Wht conclusions cn you drw? c 8.12 m 11.7 km x km 17.9 mm 6.45 m 15.6 mm 11.7 km TRIGOOMTRI PROLM SOLVIG The trigonometric rtios cn e used to solve wide vriety of prolems involving right ngled tringles. When solving these prolems it is importnt to follow the steps elow: Step 1: Step 2: Step 3: Step 4: Step 5: Step 6: Step 7: Red the question crefully. rw digrm, not necessrily to scle, with the given informtion clerly mrked. If necessry, lel the vertices of tringles in the figure. Stte clerly ny ssumptions you mke which will enle you to use right ngled tringles or properties of other geometric figures. hoose n pproprite trigonometric rtio nd sustitute into trigonometric eqution connecting the quntities. On some occsions more thn one eqution my e needed. Solve the eqution(s) to find the unknown. nswer the question in sentence. GLS O LVTIO PRSSIO Suppose you re stnding t the top of cliff. s you look out you cn see helicopter in the distnce, high overhed. The ngle etween horizontl nd your line of sight to the helicopter is clled the ngle of elevtion. Looking down, there is ot out to se. The ngle etween the horizontl nd your line of sight to the ot is clled the ngle of depression. ngle of elevtion ngle of depression
276 TRIGOOMTRY (hpter 12) If the ngle of elevtion from to is µ, then the ngle of depression from to is lso µ. or exmple: ngle of depression When using trigonometry to solve prolems we often use: ² true erings ² ngles of elevtion nd depression ² the properties of isosceles nd equilterl tringles ² the properties of circles nd tngents. ngle of elevtion pex rdius rdius tngent chord se xmple 6 12 m n -frme house hs the shpe of n isosceles tringle with se ngles of 70 o. The olique wlls of the uilding re 12 m long. How wide is the uilding t ground level? 12 m 70 x m 70 x m cos 70 o = x 12 ) 12 cos 70 o = x ) x ¼ 4:1042 ) 2x ¼ 8:21 f cos µ = J HYP g t ground level, the uilding is out 8:21 m wide. xmple 7 stircse 36 : m in length is needed to ccess pltform which is 29 : m ove ground level. ind: the ngle the stircse mkes with the ground the distnce from the foot of the stircse to the post supporting the pltform. 3.6 m 2.9 m
TRIGOOMTRY (hpter 12) 277 3.6 m x m sin µ = OPP HYP = 2:9 3:6 2.9 m ) µ = sin 1 µ 2:9 3:6 ) µ ¼ 53:7 o So, the stircse mkes n ngle of out 53: 7 o with the ground. cos µ = J HYP ) cos 53:7 o = x 3:6 ) 3:6 cos 53:7 o = x ) 2:13 ¼ x So, the foot of the stircse is out 213 : m from the post. XRIS 12 In this exercise, round your nswers correct to 3 significnt figures. 1 n -frme house hs the shpe of n isosceles tringle with se ngles 67 o. The olique wlls re 13:2 m long. ind: how wide the uilding is t ground level 13.2 m the height of the pex ove the ground. 67 2 or the tringulr roof truss illustrted, find: the length of rfter if the em is 13:8 m nd the pitch is 20 o the pitch of the roof if the rfter is 8:85 m long nd the em is 13:2 m long. pitch rfter em rfter 3 onsider ldder lening ginst verticl wll. ind: how fr up the wll ldder of length 4:2 m reches if its ngle µ with the ground is 70 o ldder wll the length of ldder if it mkes n ngle of 67 o with the ground nd its feet re 0:9 m from the wll c the ngle etween the ground nd 4:8 m long foot of ldder ldder tht reches 4:1 m up the wll. 4 The lmp in lighthouse is 64 m ove se level. The ngle of depression from the light to fishing ot is 11 o. How fr horizontlly is the ot from the lighthouse? 5 ind the height of verticl cliff if the ngle of elevtion from point 318 m from the se of the cliff is 23 o. 6 rmp for wheel-chir ccess is illustrted. ind: the length needed to rise 0:66 m t n ngle of 8 o its ngle if the rmp rises 11 : m over length of 10: 8 m. length ngle rise
278 TRIGOOMTRY (hpter 12) 7 cyclist in rnce rides up long incline with n verge rise of 6 o. If he rides for 6:2 km, how fr hs he climed verticlly? 8 5cm rcket is mde from flt steel to the dimensions 50 cm shown. ind the length of the olique strut []. 58 ind the length of metl needed to mke one rcket. 60 cm c If the metl costs $5:60 per metre, find the cost of the metl needed to mke 200 rckets. 9 rom horizontl ground, the ngle of elevtion from point to the top of tower is 32 o. Given tht is 586 m from t the se of the tower, find the tower s height. 10 130 3m 32 Illustrted is the end wll of grge. ind: how high is ove ground level the re of this wll. 10 m 11 housing lock is 42 my18 m. ind: µ correct to 4 significnt figures the length of [] using trigonometry c the length of [] using Pythgors theorem. 42 m 18 m 12 circle hs rdius 8 cm nd chord [] of length 13 cm. If the circle s centre is O, find the mesure of O. 13 The ngle etween the tngent from point P to circle nd the line from P to the centre of the circle is 23 o. etermine the length of the line from P to the centre of the circle if the rdius is 4 cm. 14 tngent from point X to circle of rdius 5 cm is 12 cm long. ind: the distnce of X from the centre of the circle the size of the ngle etween the tngent nd the line joining X to the centre of the circle. 15 rhomus hs sides of length 12 cm. The ngle etween two djcent sides is 68 o. ind the length of the shorter digonl of the rhomus. 16 flg pole csts shdow of length 12:8 m when the sun is t n elevtion of 68 o. How tll is the flg pole?
TRIGOOMTRY (hpter 12) 279 17 hord [] is drwn in circle with centre O. ind: the length of the rdius if chord [] is 8 cm long nd mkes n ngle of 87 o t the centre O the length of the chord if the rdius is 10 cm nd ngle O mesures 108 o. rdius chord 18 The hypotenuse [] of tringle is three times longer thn side []. ind the mesure of. 19 n eroplne tkes off t n ngle of 18 o. Its verge speed in the first 20 seconds of flight is 240 km h 1. Wht is the ltitude of the plne t the end of this time? 20 n oserver notices tht n eroplne flies directly overhed. Two minutes lter the eroplne is t n ngle of elevtion of 25 o. If the speed of the eroplne is 350 ms 1, find the ltitude of the eroplne. 21 nswer the Opening Prolem on pge 268. RIGS OMPSS RIGS West W orth 45 45 45 st SW S South W S TRU RIGS We cn mesure direction y compring it with the true north direction. We cll this true ering. Mesurements re lwys tken in the clockwise direction.
280 TRIGOOMTRY (hpter 12) Imgine you re stnding t point, fcing north. You turn clockwise through n ngle until you fce. The ering of from is the ngle through which you hve turned. So, the ering of from is the clockwise mesure of the ngle etween [] nd the north line through. In the digrm longside, the ering of from is 120 o from true north. We write this s 120 o Tor120 o. To find the true ering of from, we plce ourselves t point nd fce north. We then mesure the clockwise ngle through which we hve to turn so tht we fce. The true ering of from is 300 o. ote: ² true ering is lwys written using three digits. or exmple, we write 070 o rther thn 70 o. ² The erings of from nd from lwys differ y 180 o. You should e le to explin this using ngle pir properties for prllel lines. True north lines re prllel, so we cn use ngle pir properties to find unknown ngles in ering prolems. XRIS 12 1 rw digrms to represent erings from O of: 055 o 140 o c 330 o d 255 o 2 ind the ering of Q from P if the ering of P from Q is: 124 o 068 o c 244 o d 321 o 3, nd re checkpoints in n orienteering course. or ech of the following, find the ering of: i from ii from iii from iv from v from vi from. north 120 T 120 T north 300 T 40 130 25 150 270 285 xmple 8 cyclist rides 21:3 km due west nd then 13:8 km due north. ind, to the nerest degree, the ering of the finishing point from the strting point.
TRIGOOMTRY (hpter 12) 281 13.8 km 21.3 km S tn µ = 13:8 21:3 ) µ = tn 1 µ 13:8 21:3 ) µ ¼ 33 o nd so 270 o + µ ¼ 303 o ¼ 32:9 o So, the ering of from S is 303 o. 4 hiker wlks 900 m est nd then 500 m south. ind the ering of his finishing position from his strting point. 5 Two runners meet t n intersection, then leve it t the sme time. Runner P runs t 12 km h 1 due north, while runner Q runs t 14 km h 1 due est. ind the distnce nd ering of runner Q from runner P fter 30 minutes. 6 helicopter pilot flies in the direction 147 o nd lnds when she is 12 km south of her strting point. How fr did she fly? 7 ship sils for 180 km on the ering 058 o. How fr is the ship north of its strting point? 8 n eroplne trvels on the ering 315o until it is 650 km west of its strting point. How fr hs it trvelled on this ering? xmple 9 rmen deprts from point nd wlks in the direction 055 o for 8:91 km to X. She then chnges direction nd wlks for 7:29 km on the course 145 o to point. rw fully lelled sketch of the sitution. ind the distnce of from, correct to 4 significnt figures. c ind the ering of from. d In wht direction must rmen trvel to return from to? X = 145 o 55 o =90 o 8.91 km ) 2 = 891 : 2 +7:29 2 145 X fpythgorsg ) = p 55 8:91 2 +7:29 7.29 km 2 ) ¼ 11:51 km c tn µ = 7:29 d 8:91 µ 7:29 ) µ = tn 1 8:91 ) µ ¼ 39:3 o ) the ering of from is 55 o +39:3 o = 094:3 o 94:3 o + 180 o = 274:3 o ) the ering of from is 274:3 o. rmen must trvel in this direction.
282 TRIGOOMTRY (hpter 12) 9 cyclist deprts point R nd rides on stright rod for 2:3 km in the direction 197 o. She then chnges direction nd rides for 1:8 km in the direction 107 o to point S. rw fully lelled sketch of this sitution. ind the distnce etween R nd S. c ind the ering from R to S. 10 fishing trwler sils from port in the direction 083 o for 14 km. nother ot rdios messge tht they hve found lrge schools of fish, so the trwler cptin chnges course nd heds in the direction 173 o for 20 km. rw fully lelled sketch of this sitution. ind the distnce of the trwler from the port. c ind the ering of the trwler from the port. d ind the direction the trwler needs to trvel in order to return to the port. 3-IMSIOL PROLM SOLVIG Right ngled tringles occur frequently in 3-dimensionl figures. trigonometry to find unknown ngles nd lengths. We cn therefore use TH GL TW TWO LIS To find the ngle etween two lines, we consider tringle in which two sides re these lines. xmple 10 cue hs sides of length 12 cm. ind the ngle etween the digonl [] nd one of the edges t. The ngle etween [] nd ny of the edges t is the sme. ) the required ngle is. x cm y Pythgors: 12 cm x 2 =12 2 +12 2 ) x 2 = 288 ) x = p 288 fx >0g tn µ = 12 cm p 288 12 ) µ = tn 1 Ãp 288 12 ) µ ¼ 54:7 o ) the required ngle is out 54:7 o.! ~`2`8`8 12
TH GL TW LI PL To find the ngle etween line nd plne, we consider the projection or shdow cst onto the plne y light shining perpendiculrly ove the line. The required ngle is the ngle etween the line nd its shdow. TRIGOOMTRY (hpter 12) 283 lick on the icon to otin demonstrtion which shows how to locte the ngle etween line nd plne. MO XRIS 12 1 The figure longside is cue with sides of length 10 cm. ind: G G H G 2 The figure longside is rectngulr prism. X nd Y re the midpoints of the edges [] nd 4cm [G] respectively. ind: H G HX the size of XH 6cm 8cm X Y c HY d the size of YH 3 In the tringulr prism longside, find: the size of. 6cm 3cm 4cm 4 very edge of the illustrted pyrmid hs length 20 m. ind: the length of digonl [] the ngle tht [] mkes with []. M 5 ind the shdow or projection of ech of the following in the se plne if light is shone from directly ove the figure: [UP] [WP] c [VP] d [XP] S W P T X R V Q U 6 ind the projection on the se plne of: [] [] c [] d [X] X
284 TRIGOOMTRY (hpter 12) 7 In the squre-sed pyrmid shown, find the projection on the se plne of: [P] [P] P M xmple 11 The given figure shows squre-sed pyrmid with pex directly ove the centre of its se. 14 m The se is 10 my10 m nd its slnt edges re 14 m long. ind: the length of [M] the ngle tht [] mkes with the se. M 10 m 10 m M 10 m x m M x m 10 m 14 m ~`5`0 m M is right ngled nd isosceles. Let M = M = x m ) x 2 + x 2 =10 2 fpythgorsg ) 2x 2 = 100 ) x 2 =50 ) x = p 50 fs x>0g So, [M] is out 7:07 m long. The shdow of [] cst onto plne is [M], so the required ngle is M, mrked µ. p 50 ow cos µ = 14 Ãp! 50 ) µ = cos 1 ¼ 59:66 o 14 So, [] mkes n ngle of out 59: 7 o with. 8 ind the size of the ngle etween the following lines nd the se plne GH: [G] [H] H 5cm G 3cm 4cm
TRIGOOMTRY (hpter 12) 285 9 The digrm longside shows cue with side H G length 2 cm. ind the size of the ngle etween se nd: [] [G]. 10 ind the size of the ngle etween se PQRS nd: [PU] [PM]. P 4cm T S Q M U 1cm R 3cm 11 The given figure is n equilterl tringulr prism with ll edges of length 10 cm. ind the size of the ngle etween se nd [M]. M 12 In the pyrmid longside, the pex is directly ove the middle of the rectngulr se. ind the size of the ngle tht [] mkes with 18 m the se. M 20 m 16 m RVIW ST 12 1 ind the exct vlue of x. x cm ind s frction: i cos µ ii sin µ iii tn µ 3cm 2 ind the vlue of in: c 2cm 38 12 cm cm m 63 21.7 m cm 53 58.2 cm 3 ind the vlue of µ in: c 10.1 m 5m 4m 11.3 m 22 km 14 km
286 TRIGOOMTRY (hpter 12) 4 Using compss, drw circle of rdius 2:6 cm. refully drw ny chord of the circle such tht the chord hs length 4 cm. Using the digrm only, find the shortest distnce from the chord to the circle s centre. Use trigonometry to find more ccurte nswer to. 5 rom the top of 56:2 m high uilding, the ngle of depression to the se of shed is 58 o. How fr is it from the se of the uilding to the se of the shed? 6 ot sils 50 km on the ering 054 o. How fr is the ship north of its strting point? 58 56.2 m 7 30 m high tower is supported y 4 ropes. The ropes re tied 5 m from the top of the tower to four points on the ground. ch rope mkes n ngle of 60 o with the ground. ssuming the ropes do not sg, find the length of ech one. If 3% extr rope is needed to llow for sgging nd tying, how much rope is needed to tie down the tower? 8 ind the ngle tht: [G] mkes with [G] 7cm [G] mkes with the se plne GH. 5cm H 10 cm G 5cm 9 ind: the size of ngle the length of [] c the ngle [] mkes with the se plne. 3cm 2cm RVIW ST 12 1 ind the exct vlue of y. ind s frction: i cos µ ii sin µ iii tn µ 2 ind the vlue of in: c 5m 51 km 38 m 2.37 km 4cm y cm 2cm 11.6 m m 53
TRIGOOMTRY (hpter 12) 287 3 ind the vlue of Á in: c 4.2 m 317 cm 5.2 m 424 cm 11 m 8m 4 motorcyclist trvels 87 km west nd then 63 km north. How fr is she from her strting point? Wht is her ering from her strting point? c Wht is her ering to her strting point? 5 Use compss to construct scle digrm of tringulr field with sides 100 m, 100 m nd 60 m. Use scle of 1 cm 20 m. Use protrctor to mesure the smllest ngle of the tringle to the nerest degree. c Use trigonometry to find more ccurte nswer to, correct to 1 deciml plce. 6 The ngle of elevtion to the top of verticl cliff is 13:8 o. This mesurement ws tken 1:234 km from the se of the cliff. How high is the cliff, to the nerest metre? 7 ship leves port nd trvels for 60 km in the direction 038 o. It then sils 40 km in the direction 128 o to n islnd port. How fr is from? If the ship wishes to sil ck directly to from, in wht direction must it sil? 8 The illustrted figure is cue with sides H 6 m. ind: G the lengths of [] nd [G] the ngle [G] mkes with plne. 9 M is the midpoint of [] on the tringulr prism. ind the ngle tht [M] mkes with se plne. M 6m 4m 3m