Mthemtics SKE: STRND I UNIT I Trigonometric Prolems: Text STRND I: Geometry nd Trigonometry I Trigonometric Prolems Text ontents Section * * * I. Mixed Prolems Using Trigonometry I. Sine nd osine Rules I.3 ppliction: re of ny Tringle I.4 Heron's Formul IMT, Plymouth University
Mthemtics SKE: STRND I UNIT I Trigonometric Prolems: Text I Trigonometric Prolems I. Mixed Prolems Using Trigonometry When you look up t something, such s n eroplne, the ngle etween your line of sight nd the horizontl is clled the ngle of elevtion. ngle of elevtion liffs ngle of depression Similrly, if you look down t something, then the ngle etween your line of sight nd the horizontl is clled the ngle of depression. Worked Exmple mn looks out to se from cliff top t height of metres. He sees ot tht is 50 metres from the cliffs. Wht is the ngle of depression? Solution The sitution cn e represented y the tringle shown in the digrm, where is the ngle of depression. m 50 m In this tringle, opposite = m djcent = 50 m Using tn = opposite djcent gives tn = 50 = 008. Using clcultor gives = 46. (to d.p.) IMT, Plymouth University
I. Mthemtics SKE: STRND I UNIT I Trigonometric Prolems: Text Worked Exmple person wlking on open lnd cn see the top of rdio mst. The person is 00 metres from the mst. The ngle of elevtion of the top of the mst is 3. Wht is the height of the mst? Solution The tringle illustrtes the sitution descried. In this tringle, opposite = x djcent = 00 m x Using tn = opposite djcent gives tn3 = x 00 Multiplying oth sides y 00 gives x = 00 tn3 = 0. 5 metres (to d.p.) 00 m 3 Worked Exmple 3 ldder is 3.5 metres long. It is plced ginst verticl wll so tht its foot is on horizontl ground nd it mkes n ngle of 48 with the ground. () (c) Drw digrm which represents the informtion given. Lel the digrm showing the ldder, the wll nd the ground nd insert ll mesurements given. lculte, to two significnt figures, (i) the height the ldder reches up the wll (ii) the distnce the foot of the ldder is from the wll. The top of the ldder is lowered so tht it reches.75 m up the wll, still touching the wll. lculte the ngle tht the ldder now mkes with the horizontl. Solution () 3.5 m 48 IMT, Plymouth University
I. Mthemtics SKE: STRND I UNIT I Trigonometric Prolems: Text (i) height = 35. sin 48. 60 (ii) distnce = 35. cos 48 = 6. m to significnt figures. 34 (c) = 3. m to significnt figures 3.5 m.75 m sin = 75. 35. = 05. = 30 Exercises. In order to find the height of tree, some students wlk 50 metres from the se of the tree nd mesure the ngle of elevtion s 0. Find the height of the tree. 0 50 m. From distnce of 0 metres from its se, the ngle of elevtion of the top of pylon is 3. Find the height of the pylon. 3. The height of church tower is 5 metres. mn looks t the tower from distnce of 0 metres. Wht is the ngle of elevtion of the top of the tower from the mn? 4. costgurd looks out from n oservtion tower of height 9 metres nd sees ot in distress t distnce of 500 metres from the tower. Wht is the ngle of depression of the ot from the tower? 5. lighthouse is 0 metres high. life-rft is drifting nd one of its occupnts estimtes the ngle of elevtion of the top of the lighthouse s 3. () Use the estimted ngle to find the distnce of the life-rft from the lighthouse. If the life-rft is in fct 600 metres from the lighthouse, find the correct ngle of elevtion. 6. rdio mst is supported y two cles s shown. Find the distnce etween the two points nd. 30 m 60 50 IMT, Plymouth University 3
I. Mthemtics SKE: STRND I UNIT I Trigonometric Prolems: Text 7. mn stnds t distnce of 8 metres from lmppost. When stnding s shown, he mesures the ngle of elevtion s 34. Find the height of the lmppost. 34.8 m 8 m 8. Find the unknown length (x) in ech digrm. () 40 5 m x x 7 m 3.5 m 8 m 4 m (c) (d) x 0 m 6 m 60 3 m 4 m 4 m 8 x 9. From his hotel window tourist hs cler view of clock tower. The window is 5 metres ove ground level. The ngle of depression of the ottom of the tower is 5 nd the ngle of elevtion of the top of the tower is 7. () How fr is the hotel from the tower? Wht is the height of the tower? 5 m 7 5 0. rdr opertor notes tht n eroplne is t distnce of 000 metres nd t height of 800 metres. Find the ngle of elevtion. little while lter the distnce hs reduced to 00 metres, ut the height remins 800 metres. How fr hs the eroplne moved? 000 m 800 m IMT, Plymouth University 4
I. Mthemtics SKE: STRND I UNIT I Trigonometric Prolems: Text. The digrm represents tringulr roof frme with window frme EF. D nd EF re horizontl nd D nd F re verticl. () lculte the height D. 4.4 m.8 m E F (c) lculte the size of the ngle mrked x in the digrm. lculte F. x D 75.4 m. Two ships nd re oth due est of point t the se of verticl cliff. The cliff is 30 metres high. The ship t is 350 metres from the ottom of the cliff. 30 m () (i) lculte the distnce from the top of the cliff to the ship t. (ii) lculte the ngle of depression from the top of the cliff to the ship t. 33 350 m The ngle of elevtion of the top of the cliff from the ship t is 33. lculte the distnce. I. Sine nd osine Rules In the tringle, the side opposite ngle hs length, the side opposite ngle hs length nd the side opposite ngle hs length c. The sine rule sttes c sin sin sin = = c Proof of Sine Rule c If you construct the perpendiculr from vertex to meet side t N, then N = csin (from Δ N) = sin (from Δ N) N similrly for sin. Hence sin csin = sin = sin c IMT, Plymouth University 5
I. Mthemtics SKE: STRND I UNIT I Trigonometric Prolems: Text The cosine rule sttes = + c ccos = c + ccos c = + cos Proof of osine Rule If N = x, then N = x nd c ( ) = c = N + x when x N = ( ) + ( ) = sin cos, since x cos ( ) x (x 0) x N = sin + cos cos + ( ) + = sin + cos cos i.e. c = + cos, since sin + cos = Worked Exmple Find the unknown ngles nd side length of the tringle shown. Solution. cm Using the sine rule, sin sin 70 sin = =. 35. 70 From the first equlity,. sin 70 sin = = 0. 5638 35. = 34. 3 Since ngles in tringle dd up to 80, = 80 70 = 75. 68 3.5 cm From the sine rule, sin 70 sin = 35. = 35. sin sin70 3. 5 sin 75. 68 = sin 70 = 36. cm IMT, Plymouth University 6
I. Mthemtics SKE: STRND I UNIT I Trigonometric Prolems: Text Worked Exmple Find two solutions for the unknown ngles nd side of the tringle shown. Solution Using the sine rule, sin sin sin 4 = = 6 5 From the second equlity, 6 sin 4 sin = = 0. 8030 5 5 cm grph of sin x shows tht etween 0 nd 80 there re two solutions for. 6 cm 4 0.8030 y 0 0 80 x These solutions re = 53. 4 nd, y symmetry, = 80 53. 4 Solving for ngle we hve From the sine rule, = 80 4 = 6. 59 when = 53. 4, = 84. 59 when = 6. 59, =. 4 = 6 sin sin For = 84. 59, = 53. 4, = 7. 44 cm For =. 4, = 6. 59, =. 48 cm Worked Exmple 3 Find the unknown side nd ngles of the tringle shown. Solution To find, use the cosine rule: 3.7 = 49. + 37. 49. 37. cos65 =. 3759 = 473. (to d.p.) 65 4.9 IMT, Plymouth University 7
40 I. Mthemtics SKE: STRND I UNIT I Trigonometric Prolems: Text To find the ngles, use the sine rule: sin 65 sin sin = = 4.9 3.7 49. sin 65 49. sin 65 sin = = 4.73 = 69. 86 = 0. 9389 37. sin 65 37. sin 65 sin = = = 0. 7090 4.73 = 45. 5 (lterntively, use + + = 80 to find ) hecking, + + = 65 + 69. 86 + 45. 5 = 80. 0. The three ngles should dd to 80 ; the extr 00. is due to rounding errors. Worked Exmple 4 The digrm elow, not drwn to scle, shows the journey of ship which siled from Port to Port nd then to Port. Port is locted 3 km due West of Port nd Port is 45 km from Port on ering of 040. 45 km lculte, giving your nswers correct to 3 significnt figures () the distnce the ering of Port from Port. Solution () Using the OSINE rule, = + cos ˆ s ngle = 90 40, then = 3 + 45 3 45 cos50 97. 77 34.6088 3 km = 34.6 to 3 significnt figures IMT, Plymouth University 8
I. Mthemtics SKE: STRND I UNIT I Trigonometric Prolems: Text The ering of from is 70 + ngle 45 Using the SINE rule, sin 45 = sin 50 34. 6088 50 3 45 sin 50 sin = 34. 6088 0.9960 So ngle 84. 903 = 84. 9 to 3 significnt figures Hence the ering of Port from Port is 70 + 84. 9 = 354. 9 = 355 to 3 significnt figures Some importnt vlues of sin, cos nd tn re shown in this tle. sin cos tn 0 0 0 30 3 3 45 60 3 90 0 infinite α 3 The grphs of sin nd cos for ny ngle re shown in the following digrms. y = sin 360 70 80 90 90 80 70 360 450 540 630 70 y = cos 360 70 80 90 90 80 70 360 450 540 630 70 IMT, Plymouth University 9
I. Mthemtics SKE: STRND I UNIT I Trigonometric Prolems: Text The grphs re exmples of periodic functions. Ech sic pttern repets itself every 360. We sy tht the period is 360. Worked Exmple 5 n oil tnker leves Town X, nd trvels on ering of 050 to Town Z, 50 km wy. The tnker then trvels to Town Y, 70 km wy, on ering of 0. () Drw crefully lelled digrm of the tnker's journey, clerly showing the North line. lculte the distnce of Y from X, giving your nswer to 3 significnt figures. (c) (i) On your digrm, mrk the ngle tht shows the ering of X from Y. (ii) lculte the ering of X from Y, giving your nswer to the nerest degree. Solution () N N Z 0 50 60 50 km 70 km Not to scle N X 50 60 Y ering of X from Y In tringle XZY, ngle XZY = 50 + 60 = 0 so, using the cosine rule, XY = 50 + 70 50 70 cos0 = 500 + 4900 7000 cos0 ( ) 7400 7000 0. 340 9794.4 XY 98.965 = 99.0 to 3 significnt figures. (c) (i) s mrked on digrm. (ii) We need to find the ngle ZYX to determine the ering of X from Y. IMT, Plymouth University 0
I. Mthemtics SKE: STRND I UNIT I Trigonometric Prolems: Text In tringle ZYX, using the sine rule, sin ZYX 50 sin0 99. 0 sin ZYX 0.4748 ngle ZYX 8. 34 = 8 to the nerest degree Hence the ering of X from Y = 360 ( 60 + 8 ) = 7 Worked Exmple 6 Find the shded ngle in the tringle shown. Solution Using the cosine rule, 5 = 6 + 3 6 3cos x 6 cm Z 5 cm Rerrnging, 6 3cos x = 6 + 3 5 X x 46 cos x = 00 cos x = 5 5 x cos 5 5 = 8. 7 to deciml plce 3 cm Y Note using the SHIFT nd OS uttons on clcultor. ( ) s cos x is negtive, the ngle will e otuse 90 < x < 80. Exercises. For ech of the tringles, find the unknown ngle mrked. () 85 8. 3.6.9 0.3 79 IMT, Plymouth University
I. Mthemtics SKE: STRND I UNIT I Trigonometric Prolems: Text (c) (d) 4. 5. 4.8 66 (e) (f) 64 5 6.7 99 05 3.9 9.4.. For ech tringle, find the unknown side mrked, or c. () (c) 50 80 5 6.5 44 60 70 45 55 68 4.5 68 (d) (e) (f) 75 5 c 7. 95 0 8.3 55 60 45 37 7.9 48 3. For ech of the tringles, find the unknown ngles nd sides. () 85 3. 4.9 0 4.6 5.4 (c) (d) 50 79 5. cm 3.9 57 c 6. cm IMT, Plymouth University
I. Mthemtics SKE: STRND I UNIT I Trigonometric Prolems: Text 4. Which of the following tringles could hve two solutions? () 30 36.4 35.8 8.7 7.9 (c) (d) 6.4 8 0 5. Find the remining ngles nd sides of the tringle if = 67, = 5 nd c = 00. 6. Find the remining ngles nd sides of the tringle if = 8, = nd c =. 7. For ech of the following tringles, find the unknown ngles nd sides. () 3.5 60 3 3 50 5 (c) (d). 3. 5 7.8 4.3. (e) 8.9 (f) 4.7 c 30 6. 55 9.3 IMT, Plymouth University 3
I. Mthemtics SKE: STRND I UNIT I Trigonometric Prolems: Text 8. To clculte the height of church tower, surveyor mesures the ngle of elevtion of the top of the tower from two points 50 metres prt. The ngles re shown in the digrm. () lculte the distnce. Hence clculte the height of the tower D. 47 40 D 50 m 9. The ngles of elevtion of hot ir lloon from two points, nd, on level ground, re 4. nd 46. 8, respectively. The points nd re 8.4 miles prt, nd the lloon is etween the points in the sme verticl plne. Find the height of the lloon ove the ground. 8.4 miles 5.7 m 0. The digrm shows crne working on whrf. is verticl. () Find the size of ngle..4 m Find the height of point ove the whrf. 7.6 m. The rectngulr ox shown in the digrm hs dimensions 0 cm y 8 cm y 6 cm. Find the ngle formed y digonl of the se nd digonl of the 8 cm y 6 cm side. 6 cm 0 cm 8 cm IMT, Plymouth University 4
I. Mthemtics SKE: STRND I UNIT I Trigonometric Prolems: Text. () lculte the length K. lculte the size of the ngle NK. 4 cm K N 8 cm 3. Stewrt Town is 35 km due north of Willimsfield Edenford is 40 km from Willimsfield nd 45 km from Stewrt Town. lculte the ering of Edenford from Willimsfield. 45 km Stewrt Town N Ipswich, Edenford Sint Elizeth 40 km 35 km 4. In tringle, =. 6 cm, =. cm nd ngle = 54. The lengths nd re correct to the nerest millimetre nd ngle is correct to the nerest degree. Use the sine rule sin sin = to clculte the smllest possile vlue of ngle. Willimsfield, Mnchester.6 cm. cm 54 5. The nks of river re stright nd prllel. To find the width of the river, two points, nd, re chosen 50 metres prt. The ngles mde with tree t on the opposite nk re mesured s ngle = 56, ngle = 40. lculte the width of the river. 56 50 m 40 River IMT, Plymouth University 5
Mthemtics SKE: STRND I UNIT I Trigonometric Prolems: Text I.3 ppliction: re of ny Tringle n importnt ppliction of trigonometry is tht of finding the re of tringle with side lengths nd nd included ngle. The re () is given y sin. = sin Proof If you construct the perpendiculr from the vertex to, then its length, p, is given y p = sin Thus the re of is given y re = se height c = p = ( sin) = sin s required. Worked Exmple The digrm shows circle of rdius 64 cm. The length of the chord is 00 cm. () Find the ngle, to d.p. Find the re of tringle O. O 64 cm 00 cm Solution () If = 00 cm then, y symmetry, = 50 cm. sin = 50 64 = 5. 38 O 64 cm 50 cm The re of the tringle O is 64 sin = 997 cm IMT, Plymouth University 6
I.3 Mthemtics SKE: STRND I UNIT I Trigonometric Prolems: Text Worked Exmple Given tht sin = 3, 0 90 () Express in frctionl or surd form the vlue of cos. Show tht the re of tringle DE is 50 3 squre units, where D = 30 units nd DE = 0 units. 30 D 0 E (c) lculte the length of the side E. Solution so sin + cos = cos = 3 = 3 4 = 4 Hence cos = [n lterntive pproch is to consider the right ngled tringle, s shown. If the lengths = 3, = then clerly sin = 3. We cn find the length : = (Pythgors) 3 = 4 3 = So = nd cos = = ] IMT, Plymouth University 7
I.3 Mthemtics SKE: STRND I UNIT I Trigonometric Prolems: Text In the given tringle, re = D DE sin 30 = 30 0 = 50 3 3 D 0 (c) We cn find the length E using the cosine rule, nmely E = DE + D DE D cos = 0 + 30 0 30 = 400 + 900 600 = 700 E 6.5 E Exercises. Find the re of the shded region in ech of the following figures. () 6 cm 4 cm 5 cm cm 5. Find the re of the shded region. () 30 9 cm 7 cm 3 cm O 3 cm 3. Find the re of the shded region. () 4 cm O 60 0 00 O 7 cm 5 cm = IMT, Plymouth University 8
I.3 Mthemtics SKE: STRND I UNIT I Trigonometric Prolems: Text 4. S 5 cm T 5 9 cm W In the digrm ove, not drwn to scle, ST = 5 cm, TW = 9 cm nd STW = 5. lculte () the length of SW the re of Δ STW. 5. X P Q Y Z In the digrm ove, not drwn to scle, P nd Q re midpoints of the sides XY nd XZ of tringle XYZ. Given tht XP = 7.5 cm, XQ = 45. cm nd the re of tringle XPQ = 3. 5 cm, clculte () the size of ngle PXQ, expressing your nswer correct to the nerest degree. the re of tringle YXZ. 6. On the digrm elow, not drwn to scle, TU = 8 m, TW = 0 m, VW = m, ngle UTW = 60 nd ngle WUV = 40. U 40 8 m V T lculte () the length of UW 60 0 m W m the size of the ngle UVW (c) the re of tringle TUW. IMT, Plymouth University 9
Mthemtics SKE: STRND I UNIT I Trigonometric Prolems: Text I.4 Heron's Formul You hve lredy met the formul for the re of tringle when the lengths of two sides nd the included ngle re known, = sin c We will use this result to find the formul for the re of tringle when the lengths of ll three sides re known. The formul is credited to Heron (or Hero) of lexndri, nd proof cn e found in his ook, Metric, written in out D 60. The formul, known s Heron's formul, is given y where ( )( )( ) = s s s s c + + c s = The proof is given elow. lthough strightforwrd, it does involve detiled lgeric mnipultion. Proof You strt with the formul = sin c y definition, in right ngled tringle, sin = opp hyp ( sin ) = ( opp ) hyp ( ) = ( hyp ) ( dj ) ( hyp) hyp opp dj (using Pythgors' theorem) = ( ) dj ( hyp) = ( cos) IMT, Plymouth University 0
I.4 Mthemtics SKE: STRND I UNIT I Trigonometric Prolems: Text Thus = ( cos) = ( cos) + cos ut, from the cosine rule (Unit 34), c = + cos + c i.e. cos = ( ) Sustituting into the formul for, = + c + c + ( ) ( + + ) =. + c c ( ) ( + ) = c ( ) ( ) c 4 = ( c ( ) ) ( c + ( ) ) (( + ) c) ( + ) + c 4 ( ) = ( c + ) ( c + ) ( + c) + + c 4 ut s = + + c nd ( ) s = ( + + c) = + c s = + c ( ) ( ) ( ) giving s c = + c ( ) = ( ) ( ) ( ) 4 s s s c s s required. ( ) ( ) ( ) = ss s s c IMT, Plymouth University
I.4 Mthemtics SKE: STRND I UNIT I Trigonometric Prolems: Text Worked Exmple For the tringle shown, find () the re of the tringle, ngle. Solution 4 cm 7 cm 5 cm () Using Heron's formul ( )( )( ) = s s s s c where s = ( + + ) = 4 5 7 8 cm = 8 4 3 = 96 = 980. cm Using the formul = sin 9. 8 sin = = 4 5 = 098. = 78. 5 Exercises. lculte the res of the tringles shown. () 5 cm 5 cm 5 cm 0 cm 6 cm 6 cm (c) (d) 5 cm 6 cm 3 cm cm 4 cm 3 cm IMT, Plymouth University
I.4 Mthemtics SKE: STRND I UNIT I Trigonometric Prolems: Text. For ech of the tringles shown find () the re of the tringle, the ngle shown y. (i) 8 mm 7 mm (ii) 4 cm 8 cm 0 mm 6 cm (iii) (iv).7 cm.6 cm 58 5 cm 3.3 cm 7 cm IMT, Plymouth University 3