SOLUTION OF TRIANGLE GENERAL NOTATION : 1. In a triangle ABC angles at vertices are usually denoted by A, B, C

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GENERL NOTTION : SOLUTION OF TRINGLE In tringle BC ngles t vertices re usully denoted by, B, C PGE # sides opposite to these vertices re denoted by, b, c respectively

1 GENERL NOTTION : SOLUTION OF TRINGLE In tringle BC ngles t vertices re usully denoted by, B, C PGE # sides opposite to these vertices re denoted by, b, c respectively Circumrdius is denoted by R 3 Inrdius is denoted by r 4 Ex-rdii re denoted by r, r, r 3 opposite to vertices, B, C respectively 5 s + b + c s c + b c + b c (s c) + c b (s b)

2 PGE # b c b c b c 4s(s c) b c b c b c 4s s b s, s b, s c will be lwys +ve 6 or S is used for re of tringle 7 p p perpediculr r of opp side 8 B C D BD c b D 4 D b c length of medin, D b c In tringle the ngle opp to gretest side will be gretest 3 In centroid nd incenter will lwys lie within it wheres orthocentre nd circumcentre my lie outside the 4 The rtio of the re of the tringles mde on sme bse (ltitudes) will be equls to rtio of ltitudes (bse) r BD BD r DC DC

3 PGE # 3 r BC p r DBC p 4 In BC (i) sin, sin B, sin C > 0 (ii) () sin( + B) sin C (b) cos( + B) cos C (c) tn( + B) tn C (iii) () B C sin cos (b) B C cos sin B C (c) tn cot 5 # Sine Rule : In BC b c R sin sinb sinc In BD & CD : PROOF : D D c sinb bsinc In BD CD b sinb sin O is circumcentre In BOM BM sin R sin we hve b c R sin sinb sinc

4 PTTERN IDENTIFICTION : (i) sin (ii) b PROOF : b sin B sin B sincsin B 4R sin sin B 4R sin B sin B 4R sin B sinc (ie nd sin() re mutully convertible) PGE # 4 b sin sinb (iii) c sinc b R sin sinb B B 4R sin cos C B 4R sin sin (iv) cosb sin (v) tn cos 6 To find side indentify the in which it lies preferbly right ngle or the tringle one side of which nd two of the ngles re known 7 If in the ques two of the ngles sy & B nd one of the corresponding sides sy is given this my implies sine rute NPIER S NLOGY : b c sinb sinc b c sinb sinc B C B C sin cos B C B C sin cos B C B C tn cot B C tn tn B C b c tn cot b c

5 which gives the result : PGE # 5 B C b c tn tn b c Similrly C c B tn tn c & B tn b tn C b NOTE : If trigonometric rtio of difference of two of the ngles (sy B) nd the corresponding sides ( & b) re given in the ques this my imply Npier s nlogy PROJECTION FORMUL : BC BD + DC c cosb b cosc c cosb b cosc Similrly b cosc c cos & c cosb bcos COSINE RULE : In BD (refer bove figure) B D D C b cosc b sin C b cos C bcosc b sin C C b bcosc b c c cosb b c bc cos cosc b c b Similrly b c cos bc c b cosb c GENERL NOTE : (i) for three quntities, b, c (i) b 0 (ii) b (iii) b c 0

6 b c (iv) (v) b b b PGE # 6 NOTE: If two of the sides (sy & b) nd the third ngle (C) is given in the ques then this my imply ppliction of cosine rule bc cos b c 3 cot cot sin b c bc sin b c 4 c 4 4 b c 4 4 If BCD is cyclic qudrilterl : BD BD d dcos b c bc cos cos d bc b b c d b c cos d bc b c d cosb b cd cos C bc d b c d d bc b cd c bd BD Similrly b cd c bd d bc C

7 PGE # 7 BD C c bd BDC c + bd BDC BCD + DBC (Ptolemy s Theorem) NOTE : If then cos 3 cos 3 b c bc b c bc b c bc Q (i) If 5, b 4 & cos ( B) 3 3 find C Q PT b c cos Q3 PT bcos c c cosb b c Q4 If 3, b & c 60, Find other sides & ngles Q5 If C, 3, b 4 nd D is on B such tht BCD 6 Find CD 6 If,, re legnth of ltitudes of BC 7 PT cot TPT sin B b csin 8 If b c c b 3 TPT cos cosb cosc 7 5 b c sin 0 9 PT 0 PT sec tnbtnc cosb cosc

8 PGE # 8 B cos B tn cos B B cos B tn cos B Using Npier s 3/ 3 3/ B b c tn cot b c cot c 7 tn 3 7 / 9 cosc 7 / 9 cosc 8 c b bcosc c 6 Q LHS bcos c cos + cosb c cos B + cosc bcos C c + b + Q3 bcosc c cosb b c c b b c Q4 cosine Rule : b c c b bcosc 3 4 3

9 PGE # 9 c 6 Q5 BCD : CD BC sinb sin B 6 CD 3sinB sin B Q6 b c 4 b c 4 Q7 consider cot b c sinb sinc sin B C B C sin cos sin cos B C cos sin B C (s cos sin )

10 PGE # 0 C B cos sin B B cos sin b csin cos B sin B Q8 b c c b 3 b c 36 (Using C & D) c b c 5 b 6 7 pply cosine rule b c cos bc etc Q9 b c sin b c sin cos b c R cos b c b c Rbc b c 4 b c 0 Rbc 0 cosbcosc cos cosbcosc cos

11 PGE # cos cosbcosc R sin cos R 4sin sinbsinc cos cosbcosc (using sin 4 sin sec tnb tnc ) HLF NGLE FORMULE : cos b c bc b c cos bc b c cos bc b c bc bc b c bc s s bc s s cos bc Similrly B cos s s b c, C s s c cos b gin cos sin b c bc b c sin bc b c bc s b s c sin bc Similrly B sin s s c c lso, s b s c tn s s, s s c B tn s s b RE OF TRINGLE : ht bse c sinb

12 PGE # c sinb bc sin [using sin B b sin ] RsinB R sinc sin R lso C sin bc sin bcsin cos s bs c ss bc bc Ss s b s c bc bsinc R sin Ss s b s c * s b s c tn S s s b s c ss * Ss b B cot s s c s s c s s b * sin bc * If ques contins hlf s nd sides, then use bove formul for mnipultion COT - m-n THEOREM : If bse BC is divided by pt D in rtio m : n, then

13 () (m + n) cot m cot n cot () (m + n) cot n cot B m cot C Proof : in BD : PGE # 3 D sin In CD : BD sin () D DC sin sin () () () sin sin sin sin BD sin DC sin m sin n sin nsin sincos cossin msin sin cos cos sin Dividing by sin sin sin, we get (m + n) cot m cot n cot * If in question, one of the side is bisected/trisected etc then this my imply ppliction of cot m n theorem Q PT cot cot Q If b c, then find tn Q3 Let C, 75, If D is on C 3 r (BD) 3 r (BCD) Find BD Q4 If medin D B, in, PT tn + tn B 0 Q5 If medins D & BE intersects t, then PT b 5c Q6 Find, if ( + b + c) (b + c ) 3bc Q7 If, TPT 3 4D b bc c, where D is medin Q8 If, B, C re in P, nd if 3c b, then find ngles of

14 PGE # 4 Q9 In, PT B c tn tn b c Q0 If medin D is 'r to C, TPTs c cos cosc 3c ns cot ss ss 3s s lso lt : b c cot cot cot 4 cot 3 4 cot s s s b s c b c b c 4s b s c s s s s b s c 4 tn 4 s b s c tn s s s b s c tn tn tn cot m n th : 3 75 cot 75 3 cot 60

15 PGE # 5 4 cot B cot cot tnb tn tn tnb 0 5 Dist of frm centroid G D 3 b c 3 E G GE (where is mid pt of C) 6 b b c 9 s s 3 s s 3bc bc 4 + c b 36 cos cos 3 lt : b c 3bc b c bc b c cos bc b c bc lso

16 D b c 4D b c b c b c bc PGE # 6 b c bc 8, B, C P B 3 3c b c sinc b 3 sinb 3 sinc sin C B tn tn s b s c s s c s s s s b s c c c c s b c 0 b c b D b 4 4b b c c 3b b c cos bc b c cosc b

17 PGE # 7 b 3b b c 4b cos cosc 3bc b c 3c Q PT B C cos b c + B C sin b c Q BCD is trpezium such tht B & DC re & BC is 'r to then If DB, BC p & CD q, Q3 If TPT p q sin B bcos qsin b c tn sin, b TPT c b sec Q4 PT c c c b cos b sin Q5 PT cosbcosc cos c cos cosb cosc bcos cos C cosb Q6 cos B b cos bcos B c Q7 Q8 Q9 Q0 bc cos s b c cos 0 cos tn b c Sol LHS B C B C cos sin b c b c

18 PGE # 8 B C cos 4R sin B sin C + B C sin 4R sin B sin C 4R B C B C 4sin 4cos B C B C sin cos 4R B C B C 4sin cos 4R sin B C Sol In BD B sin BD sin B sin BD sin cos cos sin sin p q q p sin p q p q cos p q sin qsin pcos Sol3 sec tn C 4b sin b b b cosc b b bcos C b C b

19 PGE # 9 C C b cos b sin Sol4 C C b cos sin C C b cos sin b bcosc C Sol5 cos Bcos C cos cosbcosc cos B C cos B cos C cos BcosC sinbsin C sinbsinc Rsin sinb sinc Using symt ll terms re equls to R sin Sol6 cosb b cos bcos B cos B sin B + b cos sin + b cos cos B sin sinb cos B b cos b cos cos B sin B b sin b sin sinb cos B bcos sinb b sin c 0 c Sol7 bccos ss bc s s bc s(s) s Sol8 b c cos b c ss s b c s bc bc s s b c b c 0 bc

20 DIFFERENT CENTRES OF TRINGLE : PGE # 0 Circum centre : bc () R 4 () OM R cos Dist of BC frm cc B Incentre : Dist of side frm cc R cos, R cos B, R cos C () BD c DC b () () BD ly DC c b c b b c (3) In BD : I ID B BD c c b c b c (4) IB (5) (i) In BD : B c D sinb BD sin D sinb C [using sin B b sin ] sin / b c bc cos b c length of r bisector similrly, c B BE cos c (ii) r BC r IBC r (IC) + r(ib) rs r s r br r

21 PGE # (iii) r s s s r s s tn B s b tn C s c tn (iv) B C 4R sin sin sin s b s c s s c 4R bc c s s b b 4R s s b s c bc ss 4R bc s r s s r 4R sin (i) r s s tn 4R sin (iv) In BPI BI B r cosec C 4R sin sin Dis of vertex B from incenter Dis of vertices from incenter rcsc, rcsc B nd rcsc C

22 In BPI PGE # B r tn BP BP r s b B tn length of tngent from B to incircle BP BR s b CP CQ s c Q R s (vi) IO B B C B B C I O R In IO B C 4R sin sin IO I O OIcos IO B C IO R 6R sin sin B C B C R 4R sin sin cos B C B C B C R 8R sin sin sin sin cos R 8R sin IO R 8 sin R (i) R r (ii) Rr sin 8 Orhocentre : (i) BD ccosb (ii) DC bcos C

23 (iii) BD DC (iv) BD c cos B tn C bcos C tn B BH ccosb cosec C PGE # 3 c sinc Rcos B cos B HD c cosb cotc cos C ccos B sin C Rcos B cosc Hence Dis of verteces from orthocenter will be R cos () Rcos B, Rcos C nd the distnce of the sides from orthocenter will be R cos(b) cos(c), Rcos Ccos nd Rcos cosb HO B C H R cos() O R HC C OM B OM B HO C B B C OH O H OHcos B C R 4R cos RRcos cosb C R 4cos cos cosb C R 4cos cos B C cos B C R 8R cos OH R 8 cos (iii) EX-CIRCLE : (i) r s

24 (ii) r s b PGE # 4 (iii) r 3 s c B C r 4R sin cos cos B C r 4R cos sin cos 3 () r s tn B (b) r s tn C (c) r3 s tn 4 B r r s tn tn r r s 5 Proof: r r s s 3s s s r 6 r r r3 r Proof: s s b s c s s s Q Prove tht r cos R

25 Sol cos 4sin PGE # 5 4R sin R + r R Q Prove tht if r, r, r 3 in HP then BC re in P Sol,, r r r P 3 s s b s c,, r P s, s b, s c P, b, c P, b, c P Q3 Let dis of orthocenter from vertices is p, q, r then prove tht Proof: qr bc p Rcos q Rcos B r Rcos C LHS qr Rsin RcosB Rcos C 3 8R sin cosb cos C 3 8R sin bc LT : r BC qrsin B C qr 4R 3 bc qr bpr cpr 4R 4R 4R 4R

26 PGE # 6 bc qr Q If r r r 3 r PT is right ngled Q If B / PT r b c Q3 If ltitudes from, B, C re produced to meet circum centre, if length of produced prts is,, then prove tht centre, if length of produced prts is,, then prove tht tn Q4 If,, re length of internl bisectors TPT cos Q5 Q6 b Rr b c 0 r Q7 If,, re the distnce of the vertex of from the corresponding pts of contect with the in-circle TPT r s s b s c s s s s b s c s s s s s b c s b s c s s s b c s bc s b c bc b c b c bc b c bc bc b c B r s b tn 4 s b tn 4 s b

27 s b c b PGE # 7 5 b c bc c bc s bc 4 s bc R r b c s 6 b c s b c b c 0 b c r b c s s b c b c 0 7 s s b s c s s s s s s s r r 4 bc cos b c cos b c bc b c cos b c

28 PGE # 8 b c 3 s BP BC BP c lso, In BDP BDcot C cos C ccos B sin C Rcos BcosC Rcos B cos C sin cos B cos C C sin B C B C C tnb tnc q t B t C t Q If p, p, p 3 re length of (i) (ii) p r cos p R 'r from vertices to sides then PT (iii) bp c + cp + cp3 b b c R Q PT pro I bc B C tn tn tn Q3 If x, y, z re respectively distnce of the vertex from its orthocentre then prove tht x R r Q4 If x, y, z respectively re the 'r from the circum center to the sides of the tringle BC then PT bc x 4xyz Q5 (i) PT in tringle tn

29 PGE # 9 (ii) (iii) (iv) cos 3 4 s 3 3 sin 4 ns p p (iii) cos p s r cos R R sin 4 sin R sin R b (iii) bp c bp b bc pb p c bc R TP I tn LHS B C 4R sin sin 3 B C B C 64R s s s s s s Q3 x Rcos 3 s s B s C B C 64R t 8 bc tn

30 Rsin x Rcos tn PGE # 30 x tn tn x Rcos x R 4 sin R 4R sin R r Q4 x R cos x Rsin Rcos tn() tn x tn x 4 x Q5 pply M GM on tn, B tn B B tn tn tn tn similrly dding tn (s B tn tn ) (ii) cos b c cos cos

31 R sin 4R sin PGE # 3 4 sin 4 4 cos 8 sin (Using sin ) 8 / 3 s s b s c s 3 (iii) 3 s s 3 4 s s s 3 s 3 3 s mx ****-----

SOLUTION OF TRIANGLES

SOLUTION OF TRIANGLES SOLUTION OF TIANGLES DPP by VK Sir B.TEH., IIT DELHI VK lsses, -9-40, Indr Vihr, Kot. Mob. No. 989060 . If cos A + cosb + cos = then the sides of the AB re in A.P. G.P H.P. none. If in tringle sin A :