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gINE AND COSINE FORMULAE, TEZIR APPLICATION, , INTRODUCTION, , triangle the three sides and the three angles are generall, e. A triangle which does not contain a right a ee ee, , jn any 5, wang ns : ingle is called an oblique triangle., , jangle , the measures of the angles ZBAC, /C, jnany triang : " & , £CBA and ZACB are denoted by the, ttt A, Band C respectively. The sides BC, CA and AB opposite to the angles A, B and Cc, respectively are denoted by a,b and c. These six elements of a triangle are not independent and, aeconnected by the relations: (i) A+B+C =n (ii) a+b>c;b+c>a;c+a>b. Inadditionto, , shese relations, the elements of a triangle are connected by some trigonometric relations. We, intend to discuss those relations in the sections to follow of this chapter., {92 THE LAW OF SINES OR SINE RULE, , THEOREM The sides of a triangle are proportional to the sines of the angles opposite to them i.e. ina, , ABC, <2 =$——=—, AA Gn A sinB sinc’, , , , , , , , , , poor The following cases arise:, case] When A ABC is an acute angled triangle:, Draw AD perpendicular from A to the opposite side BC meeting it in the point D., Inthe triangle ABD, we have, sinB = 12 = sing - 22 = ap = csinB i), AB c, Inthe triangle ACD, we have, snc = 42 5 sinc = AP = ap = bsinc fii), AC b, From (i) and (ii), we get A, : @, csnB =bsnC > — = A, sin sinC, hha similar manner, by drawing a perpendicular from Bon AC, c b, We obtain, & c, sinA sinC, Hence, 4 _ b _i¢ B D a e, Sn A ~ Gin B ~ sin€ Fig. 10, ws ,, “SEU When A ABC is an obtuse angled triangle. ., . AD Perpendicular from A on CB produced meeting asin)., ADB, we h:, ” We have AD E, ‘ AD apse =csi, sin ZABD = 42 = sin (180 -B) =" sin B= 3 = AD=csin B (i), , , , A, , Scanned with CamScanner

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10.2 ATHE a,, , In A ACD, we have, AD, , sinc = 42 = sinc = = AD = bsinC, , From (i) and (ii), we obtain, , , , b G, csin B=b sinC > ——=, " mm Sin B sin C, , Similarly, by drawing perpendicular from B on AC, we obtain, , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , a _ oc, sin A sinC, b c, Hence, = = ij,, sinA sinB_ sinC, cast ti When A ABC is a right angled triangle:, In A ABC, we have, sin C =sin = =1, sin a= 2E = © and, aa p-2Gt 4, Zz AB c AB c, a b, = =c, sinA sin B, a b e y fl, = - =— ==, sin A sinB 1, a b 2, > 5 =——= = : sinC =sin = =1|, sinA sin B sinC 2, Hence, in all the cases, we obtain 7 — c, Fig. 10.3, ab ee, sinA sinB sinC, Q, in A si ‘, REMARK | The above rule may also be expressed as snd, sind, Se, , a b c, des of a triangle in terms of the sines of angles, , , , , , REMARK — The sine rule isa very useful tool to express, and vice-versa in the following manner., a b c, , , , , , , , , , , , , , , , , , , , , , Let — =— =— =k (say) Then, a=k sin A, b=k sin B, c=k sinC., sinA sin B sinC, Similarly,, sin = = B a8 c =A(say) = sin A=2a, sin B=7.b and sinC =4¢., a h c |, ILLUSTRATIVE EXAMPLES |, LEVEL-1, , Type 1 PROBLEMS BASED ON SINE RULE, , RESULTS Ina AABC, we have, sin (B4C) = sin A, sin(C + A) = sin B, sin(A + B) =sinC, cos(B+C) = —cos A, cos(C + A) =—cos B, cos (A + B) =-cosC, tan (B+C) = ~tan A, tan (C + A) =~ tan B, tan (A + B) =—tanC, , EXAMPLE 1! Ina A ABC, ifa=2,b=3and sin A = 5 find 2B., , SOLUTION We have,, : c, , ig a, sinA sinB_ sinC, , Scanned with CamScanner

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LEVEL-1, , , , , , iIfinaAABC, ZA = 45°, ZB = 60°, and ZC = 75°; find the ratio of its sides,, , 2. Ifin any AABC, ZC = 105°, ZB = 45°, a=2, then find b., 3.In A ABC, ifa=18,b =24andC = 30 and ZC =90°, find sin A, sin Band sin C., , In any triangle ABC, prove the following: (4-24), , tan | ———, a-b _ 9, , ‘ath tan 422), 2, , =

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O_O, a ANO COSINE FORMULAE AND THEIR APPLICATIONS, , a ei 10.13, je Ds esin( 422), " (NCERT], A B, aes tan| —, wn(3)* (3), 3am G, a tan 2 2, , "a+b V4 tan 5 tan 3], 2 2., cos(458), ‘ a+b kee, c ane. [NCERT], 2, . ot) - -c oat, 9. sin 2 a 2, , @-c? _ sin(A-C), , 0 Tr © Sin(A+C), , i. bsinB-csinC = asin(B-C), , 2. a? sin(B-C) = (0? -c?) sin A, , 5, Sin A = sin B _ a+b -2Vab, , “sin A + [sin B a-b, , 14. a(sin B-sin C) +b (sin C -sin A) + ¢ (sin A -sin B) =0, , 2 pp 2 _ 2 as i., , 3, 2 sin (B C) 1b sin (C A) ¢ sin (A - B) =0, sin A sin B sin C, , . a (cos? B- cos? C) +b? (cos? C - cos? A) + c? (cos? A - cos? B) = 0, , 1. beos B+ ccosC = acos(B-C), cos 2A cos 2B, , , , 19, c0s* B-cos*C cos? C - cos? A cos? A -cos® B 7, . m =, bec c+a a+b, , y 5 - C-A «C ..fA-8, 2 ssin Fin BC), b sin 2 sin 5 )+esin $ sin Jo., , 0, , , , , , , , >, , yy, bsecB+esecC csecC+asecA — asecA+bsecB, / sm 2b —$—<$<—S$= 5 ———————_., tan B+ tan C tan C + tan A tan A + tan B, ~ 4800 A+bcosB+ccosC = WsinAsinC = 2csinAsinB, * 4(cos B cos C + cos A) =b (cos C cos A + cos B) =¢ (cos A cos B + cos C)., , ' 8(€08 C ~ cos B) =2(b—c) cos? 4. (NCERT], , Be, , - InaABc prove that, if @ be any angle, thenb cos 8 = ccos(A -@)+acos(C + 8)., "Wa ABC, if sin? A+ sin2 B = sin? C, show that the triangle is right angled., , , a, , ne, , Scanned with CamScanner

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10.14 MaMa, , : LEVEL-2 y, , b2, c? are in A.P., prove that cot A, cot Band cotC are also in, , 27. Inany AABC, if a’, a oe, ‘ akes an angle o with D, Th er part of a tree broken over by the win make: the, 28. and RG dictenice from the root to the point where the top of the tree toug Boy, ground is 15 m. Using sine rule, find the height of the tree. te, he elevation of its summit is 45°; after ascending 1009 1, , . At the foot of a mountain t is 45", the mountain up a slope of 30°inclination, the elevation is found to be 60° Fing th, , of the mountain. ., 30. A person observes the angle of elevation, cmetres along a slope inclined at the ang,, , the hill to be y. Show that the height of the peak above the ground is, , , , hei, , of the peak of a hill froma station tobe o, ty,, , le Band finds the angle of elevation of the, c sin a sin, , Wa, ( meat’, ——\I=f), (sin y~ a), 2B, , ‘ vio A os .2C, 31. If the sides a,b,c ofa AABC are in H.P., prove that sin a sin y sin 3 are ingp, , ANS Weng, , , , 4, , - 3, 4, 2:V6:¥3 +1 2, 242 3. sin A==, sin Bo, , 28, 15V3 m 29, 500 (3 +1) metres, , 10.3 THE LAW OF COSINES, THEOREM In any AABC, we have:, 2 2 0? + c = a, (i) a = b*+c*-2becosA or, cosA = —————, , a +c? -b?, , c2 +a” —2ac cos B or, cos B =, e 2ac, 2, , 2,722, (iii) c2 = a2 +b? —2ab cosC or, cosC = lesa, mt 2ab, , PROOF The following cases may arise:, CASEI When AABC is an acute angled triangle:, , Draw perpendicular AD from A on BC. A, In AABD, we have, , cos B = mo BD =c cosB (i), , In AACD, we have c, cosC & Os CD =b cosC, , (ii) b?, , In AACD, using Pythagoras theorem, we have, , AC? = AD* +cD?2 Ba, AC? = AD? +(BC - BD)?, , AC? = AD? + BC? +BD2-2BC.BD, , AC? = BC? +(AD? + BD?) —2BC.BD, , AC? = BC? +.AB2-2BC.BD [= AB? =BD, , 2. 2, b a* + c* —2ac cos B, , UY uy, “ou, , You, H, , I, , Scanned with CamScanner