SOLVING TRIANGLES USING THE SINE AND COSINE RULES


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1 Mthemtics Revision Guides  Solving Generl Tringles  Sine nd Cosine Rules Pge 1 of 17 M.K. HOME TUITION Mthemtics Revision Guides Level: GCSE Higher Tier SOLVING TRIANGLES USING THE SINE AND COSINE RULES Version: 3.1 Dte:
2 Mthemtics Revision Guides  Solving Generl Tringles  Sine nd Cosine Rules Pge of 17 SOLUTION OF GENERAL TRIANGLES  THE SINE AND COSINE RULES. The sine nd cosine rules re used for finding missing sides or ngles in ll tringles, not just rightngled exmples. The lelling is importnt here; uppercse letters re used for ngles nd lowercse ones used for sides. Also, lettered sides re opposite the corresponding lettered ngles. Are of tringle. One formul for finding the re of tringle is ½ (se) (height). This cn e dpted s follows: By drwing perpendiculr from A, its length cn e deduced y relising tht it is opposite to ngle C, nd tht the hypotenuse is of length. The length of the perpendiculr, nd thus the height of the tringle, is sin C. The re of the tringle is therefore ½ sin C. Since ny side cn e used s the se, the formul cn e rotted to give ½c sin B or ½c sin A. This formul holds true for cute nd otusengled tringles.
3 Mthemtics Revision Guides  Solving Generl Tringles  Sine nd Cosine Rules Pge 3 of 17 The sine rule. The sides nd ngles of tringle re relted y this importnt formul: sin A sin B sin C c or sin A sin B c sin C The formul is normlly used in the rerrnged forms sin B sin A when finding n unknown ngle, or sin A when finding n unknown side. sin B (The corresponding letterpirs re interchngele, thus exmples of other eqully vlid forms.) Bsin C sin B nd c c sin C sin A re Note tht n eqution of the form sin A = x hs two solutions in the rnge 0 to 180. Thus 30 is not the only ngle with sine of is nother one. Any ngle A will hve the sme sine s (180 A). This is importnt when solving certin cses, ut this is outside the scope of GCSE exms.
4 Mthemtics Revision Guides  Solving Generl Tringles  Sine nd Cosine Rules Pge 4 of 17 The cosine rule. This is nother formul relting the sides nd ngles of tringle, slightly hrder to pply thn the sine rule. c ccos A Here, A is the included ngle etween the sides c nd. It is used in this form when finding n unknown side, ut rerrnged s cos A c c when used to find n unknown ngle. This formul cn lso e rotted etween different sides nd ngles: thus c c c ccos A c cos B cos C ll hve the sme effect. The formule for missing ngles cn e similrly rotted: cos A c cos B cos C c c c c
5 Mthemtics Revision Guides  Solving Generl Tringles  Sine nd Cosine Rules Pge 5 of 17 Which rules should we use? This depends on the informtion given. i) Given two ngles nd one side  use the sine rule. (If the side is not opposite one of the ngles, you cn work out the third ngle simply y sutrcting the sum of the other two from 180 ). ii) Given two sides nd n ngle opposite one of them  use the sine rule. (Some cses cn give rise to two possile solutions, ut there is no need to know this t GCSE). This is the ngle not included nd two sides cse questions will usully specify tht the tringle is cutengled. The section on Congruent Tringles lso discusses this cse. iii) Given two sides nd the included ngle  use the cosine rule to find the third side nd then continue with the three sides nd one ngle cse elow. iv) Given three sides nd no ngles  use the cosine rule to find the ngle opposite the longest side, followed y the sine rule for either of the others. The third ngle cn e found y sutrction.
6 Mthemtics Revision Guides  Solving Generl Tringles  Sine nd Cosine Rules Pge 6 of 17 Exmple (1): Find the ngles mrked A nd the sides mrked in the tringles elow. Note tht tringles S nd T re cutengled. Tringle P. Two ngles nd side re known. The known side is not opposite either of the known ngles, ut the opposite ngle (cll it B) cn esily e worked out y sutrcting the other two ngles from 180. This mkes B = 111 nd = 6 units. We will lso lel the 3 ngle A s it is opposite side. We therefore use the sine rule in the form Tringle Q. sin A sin B, giving 6sin 3 sin111, or 3.41 units to d.p. All three sides re known here ut we re required to find ngle A. Lelling side s the opposite side (length 4 units), we will cll the side of length 5 side nd the side of length 6 side c. This time we use the cosine rule in the form Sustituting for, nd c gives cos A c c cos A nd hence A = 41.4 to 1 d.p. 60 Tringle R. Here we hve two sides plus the included ngle given. Lel the ngle of 34 s A, the side of length 8 s, nd the side of length 1 s c. We must therefore sustitute the vlues of A, nd c into the cosine formul c ccos A This gives cos 34, nd hence or 6.99 units to d.p.
7 Mthemtics Revision Guides  Solving Generl Tringles  Sine nd Cosine Rules Pge 7 of 17 Tringle S. Here we hve two sides given, plus n ngle not included. Lel the ngle opposite s A, the 75 ngle s B, the side of length 10 s, the side of length 9 s c, nd the ngle opposite c s C. To find we need to pply the sine rule twice. First we find ngle C using csin B 9sin 75 sin C, hence sin C 10 The vlue of sin C is to 4 dp, so in this cutengled cse ngle C is therefore To find side, we must find ngle A. The ngle cn e worked out s ( ) degrees, or Then we use the sine rule gin: Tringle T. sin A sin B or 10sin 44.6 sin 75, giving = 7.7 units to d.p. Agin we hve two sides given, plus n ngle not included. We use the sine rule gin, this time to find ngle A. Lel the side of length 8 s, the ngle of 1 s B, nd the side of length 3 s. Applying the sine formul in the form sin B sin A we get sin 8sin 1 3 A, or sin A = to 4 d.p. This gives ngle A = 7.9 nd, y sutrction, ngle C = 86.1.
8 Mthemtics Revision Guides  Solving Generl Tringles  Sine nd Cosine Rules Pge 8 of 17 Exmple (): Solve tringle Q from exmple 1 y finding ll three missing ngles, s well s its re. After lelling s ove, the first step would e to find ngle C, opposite the longest side. This uses the cosine formul. We use the form c cos C. Sustituting for, nd c gives cos C nd hence A = 8.8 to 1 d.p. (keep more ccurcy, 8.8, for future working) 40 We now hve enough informtion to work out the re of the tringle, s we hve found the included ngle C. The re of the tringle is thus ½ sin C, or 10 sin 8.8 = 9.85 sq.units. To find the other two ngles, we use the sine rule to find one of them nd then sutrct the sum of the other two ngles from 180 to find the third. The reson for using the longest side first is to prevent miguous results when using the sine rule. No tringle cn hve more thn one otuse ngle, nd the longest side is lwys opposite the lrgest ngle. The cosine rule would tke cre of the otuse ngle if there ws one, leving no possiility of confusion when using the sine rule to work out the other two. In fct, ngle C is cute in this cse, so we hve n cutengled tringle. We cn choose either remining side to work out the other ngles  here we'll find B first using the sine rule. sin C sin B, giving 5sin8.8 sin B nd sin B = c 6 This gives B = 55.8 to 1 d.p. (only the cute ngle is vlid here) To find C, we sutrct the sum of A nd B from 180, hence C = 41.4 to 1 d.p.
9 Mthemtics Revision Guides  Solving Generl Tringles  Sine nd Cosine Rules Pge 9 of 17 Exmple (3): Solve tringle R from exmple 1 y finding the two missing ngles nd the missing side. Find out its re s well. We cn work out the re t once s ½c sin A. This gives 48 sin 34 or 6.84 sq.units. First, we find the missing side y sustituting the vlues of A, nd c into the cosine formul c ccos A This gives cos 34 or 48.8, nd hence or 6.99 units to d.p. (Keep greter ccurcy for future clcultion ). After finding, the next step is to find one of the two missing ngles. Both methods re shown here for illustrtive purposes  choose the one you're hppier with. Using Cosine Rule. Choose ngle C s the next ngle, since it is opposite the longer side, here c. (This will tke cre of potentil otuse ngle solution.) cos C c, or cos C, or This gives C = 106. to 1 d.p. (note tht otuse ngles hve negtive cosine). Angle B cn e found simply y sutrcting the sum of A nd C from 180. It is thus ( ) or Using Sine Rule. We hve one known ngle, A, of 34, so we know tht one of the remining ones must e cute since ll tringles hve t lest two cute ngles. We therefore use the sine rule to find the ngle opposite the shorter of the remining sides, nmely side. Applying the sine formul in the form or sin B = to 4 d.p. Angle B is hence 39.8 sin A sin B we get sin 8sin B, Angle C cn e found y sutrction, eing equl to (180  ( )), or 106..
10 Mthemtics Revision Guides  Solving Generl Tringles  Sine nd Cosine Rules Pge 10 of 17 Rellife pplictions of Sine nd Cosine Rules. The sine nd cosine rules cn e used to solve rellife trigonometry prolems. Mny everydy prolems in trigonometry involve such terms s erings, ngles of elevtion, nd ngles of depression. Berings (revision). A ering of point B from point A is its compss direction generlly quoted to the nerest degree, nd stted s numer from 000 (North) to 359. Berings re mesured clockwise from the northline. Exmple(4): Express the eight points of the compss shown in the digrm s erings from north. N 000 ; NE 045 ; E 090 ; SE 135 S 180 ; SW 5 ; W 70 ; NW 315 Rellife prolems cn e solved y pplying trigonometric rules, nd often in differing wys. See the following exmples. Angle of elevtion nd depression. The ngle of elevtion of n oject is its ngulr height (in degrees) ove reference line. In the digrm on the right, the ngle A is the ngle of elevtion of the se of the lighthouse from the ycht. The ngle of depression of n oject is its ngulr depth (in degrees) elow reference line. In the digrm on the right, the ot mkes n ngle of depression A with the oserver s horizon t the cliff top.
11 Mthemtics Revision Guides  Solving Generl Tringles  Sine nd Cosine Rules Pge 11 of 17 Exmple (5): Two ships leve port t 10:00 nd ech one continue on strightline course. Ship A trvels on ering of 060 t speed of 3 km/h, nd ship B trvels on ering of 115 t speed of 8 km/h. How fr wy re the ships from ech other t 1:00, ssuming no chnges of course nd speed? Since two hours elpse etween 10:00 nd 1:00, ship A will hve trvelled 46 km nd ship B will hve trvelled 56 km from port. The sitution t 1:00 is represented y tringle where ship A is 46 km from port P t ering of 060 nd ship B 56 km from port t ering of 115. We hve lso included the northline t N. The ngle etween the ships erings is 55 ecuse ngle NPA = 60 nd ngle NPB = 115. We hve two sides nd the included ngle given in the tringle PAB, nd so we cn find the distnce AB using the cosine rule: (AB) = (AP) + (PB) (AP)(PB) cos 55 This gives (AB) = cos 55 or 96.9, nd hence AB = 47.9 km to 3 s.f.
12 Mthemtics Revision Guides  Solving Generl Tringles  Sine nd Cosine Rules Pge 1 of 17 Exmple (6): A ychtsmn psses lighthouse t point P nd sils for 6 km on ering of 080 until he reches point Q. He then chnges direction to sil for 4 km on ering of 150. Work out the ychtsmn s distnce nd ering from the lighthouse t point R, fter the second stge of his siling. (Although this is n ccurte digrm, only sketch is required ). We cn find PQR y relising tht N 1 PQ nd PQN re supplementry, i.e. their sum is 180. Hence dd PQN = (180 80) = 100. Becuse ngles t point dd to 360, PQR = 360 ( ) = 110 We cn now find the distnce PR s eing the third side of tringle PQR we hve two sides (4 km nd 6 km) nd the included ngle of 110. We lel ech side s opposite the ngles, nd use the cosine rule to find side q (PR) first : q = r + p pr cos Q This gives q = cos 110 or 68.41, nd hence q = 8.7 km (Keep higher ccurcy for future clcultion ). We cn then use the sine rule to find ngle QPR (P) nd hence the ycht s finl ering from northline N 1. Applying the sine rule, we hve sin QPR sin , nd thus sin QPR 4sin or sin QPR = to 4 d.p. Angle QPR is hence 7 The ycht s finl ering from northline N 1 is (80 + 7), or 107, nd its distnce from the lighthouse t P is 8.7 km.
13 Mthemtics Revision Guides  Solving Generl Tringles  Sine nd Cosine Rules Pge 13 of 17 Exmple (7): A ruined cstle is fenced off for sfety resons, nd surveyor mesures the ngle of elevtion of the tower t 16. He then wlks nother 40 metres in the direction of the tower to point B, where the ngle of elevtion is 8. Find the height, h, of the cstle tower. We strt y looking t the tringle ABT, ecuse we cn work out its ngles ; ABT = 15 (180 in stright line) nd therefore ngle ATB = 1. We then use the sine rule to find the side lelled : 40 sin16 sin1 40sin16 sin m. Then we cn use rightngled methods to find h = sin 8 m = 4.9 m the cstle tower is 4.9 m high.
14 Mthemtics Revision Guides  Solving Generl Tringles  Sine nd Cosine Rules Pge 14 of 17 Exmple (8): A ychtsmn t A mesures the ngle of elevtion of the se of lighthouse t point C nd finds it to e 35. He then mesures the ngle of elevtion of the top of the lighthouse t point D nd finds it to e 41. Given tht the lighthouse is on the top of verticl sewll 5m high meeting the se t B, clculte the height h of the lighthouse. Since the tringle ABD is rightngled, ngle ADB = 49. Since ngle BAC = 35, the smll ngle DAC = 6. We cn then use tringle ABC to find side AC (lso lelled d). Here d = 5 sin 35 m = 90.66m. We cn then use the sine rule to find h: h d sin 6 sin sin 6 h 1. 6 sin 49 Hence the height of the lighthouse, h, = 1.6 m. h m.
15 Mthemtics Revision Guides  Solving Generl Tringles  Sine nd Cosine Rules Pge 15 of 17 Exmple (9): Peel Tower is 1 km from Winter Hill, on ering of 083, wheres Juilee Tower is 7 km from Winter Hill, on ering of 016. Find the distnce nd ering of Juilee Tower from Peel Tower. First, we sketch the positions of the northline nd the three lndmrks in question. We lso lel sides opposite corresponding ngles with lowercse letters. Note tht ngle A = (83 16) = 67. Next we find the length of the side of the tringle, nd to do so, we use the cosine rule. = + c ccos A This gives = cos 67 or 17.35, nd hence = km (Keep higher ccurcy for future clcultions). To find the ering of Juilee Tower from Peel Tower, we drw southern continution of the northline t S nd use lternte ngles to find ACS = 83 The ering required is therefore ( ) + ngle C (to e determined). Angle C cn e found y the sine rule: sin C 7 sin sin 67 sin C 11.85, nd thus or sin C = to 3 d.p. Angle C is hence 35, so the ering of Juilee Tower from Peel Tower is ( ) = 98. Juilee Tower is 11.3 km from Peel Tower on ering of 98.
16 Mthemtics Revision Guides  Solving Generl Tringles  Sine nd Cosine Rules Pge 16 of 17 Exmple (10) : Two lmpposts A nd B re 40 metres prt on the sme side of stright rod. The points A nd B mke ngles of 41 nd 5 respectively with lmppost C on the opposite side of the rod. Clculte the width of the rod, w, to 3 significnt figures.. The first step is to find ngle ACB, which works out s 87 (sum of ngles of tringle = 180 ), nd from there we use the sine rule to find the distnce AC. Using the convention of lelling opposite sides nd ngles, AC =. By the sine rule, 40 sin 5 sin 87 hence = 40sin 5 sin87 = m. Finlly we drw perpendiculr cross the rod from C t the point X, nd use rightngled tringle methods to find the width of the rod, CX. Side AC is the hypotenuse of the tringle AXC, nd CX is the opposite, so CX = sin 41 = 0.71m. The width of the rod is 0.7m to 3 significnt figures. (We could eqully well hve chosen to find the length of BC, or. 40 sin 41 sin 87, hence = 40sin 41 sin87 = 6.8 m. From there the width of the rod would e clculted s 6.8 sin 5, or 0.7m )
17 Mthemtics Revision Guides  Solving Generl Tringles  Sine nd Cosine Rules Pge 17 of 17 Exmple (11): (Nonclcultor) ABDE is trpezium whose se length AE is 1 cm, nd dditionlly BC = CD = BD = 8 cm. In ddition, ngle CDE = 90. Clculte the perimeter of the trpezium, giving your result in the form + c where, nd c re integers. From the given dt, the tringle BCD is equilterl, so CBD = CDB = 60, nd ecuse AE nd BD re prllel, ngles ACB nd DCE equl 60 y lternte ngles. Tringle CDE is rightngled, so DE = 8 tn 60 = 83 cm, nd CE = 8 cos 60 = 16 cm. (Note tht cos 60 = ½ nd tn 60 = 3). By sutrction, AC = (1 16) cm = 5 cm, which leves us with side AB. The length of tht side cn e worked out using the cosine rule. (AB) = (AC) + (BC) (AC)(BC) cos 60 This gives (AB) = cos 60 or 89 40, or 49. Hence AB = 7cm. The perimeter of the trpezium is ( ) cm, or cm.
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