TRIANGLE. The sides of a triangle (any type of triangle) are proportional to the sines of the angle opposite to them in triangle.

Transcription

1 19. SOLUTIONS OF TRINGLE 1. INTRODUTION In ny tringle, the side, opposite to the ngle is denoted by ; the side nd, opposite to the ngles nd respetively re denoted by b nd respetively. The semi-perimeter of the tringle is denoted by s nd its re by or S. In this hpter, we shll disuss vrious reltions between the sides, b, nd the ngles,, of. b. SINE RULE Figure 19.1 The sides of tringle (ny type of tringle) re proportionl to the sines of the ngle opposite to them in tringle b, sin sin sin sin sin sin Note: (i) The bove rule n lso be written s b (ii) The sine rule is very useful tool to express the sides of tringle in terms of sines of the ngle nd vie-vers b in the following mnner: k ( Let) ; k sin, b k sin, k sin sin sin sin sin sin sin Similrly, λ ( Let) ; sin λ, sin λ b, sin λ b 3. OSINE RULE b + + b + b In ny, os ; os ; os b b Note: In prtiulr 60 b + b 60 + b 60 + b b 4. PROJETION FORMULE Figure 19. If ny :(i) bos + os (ii) b os + os (iii) os + bos i.e. ny side of tringle is equl to the sum of the projetion of the other two sides on it.

2 19. Solutions of Tringle se I: When is n ute ngled tringle, D os D.os D.os nd D os D.os D bos then, D+D os + b os se II: When is n obtuse ngled tringle, D os D.os D D b.os nd os( 180 ) D.os then, nd D D bos + os Figure 19.3 Figure 19.4 Illustrtion 1: If 75, 45, then wht is the vlue of b+? (JEE MIN) Sol: Here, Therefore by using sine rule, we n solve the bove problem. Use sine rule b K sin75 sin45 sin60 k sin 75 b k sin 45 k sin 60 o o o 0 onsider, ( b + ) k ( sin45 + sin60 ) Illustrtion : In, if Sol: Here, by using osine rule k k k sin75 k sin 30 nd 3b, then find the vlue of. (JEE MIN) + b os we n esily solve the bove problem. + b 3 3b + b We hve os 3b b 0 OR b 0 ; 3b + b 0 ( b)( b) 0 Either b 30 or b 4b b Illustrtion 3: Prove tht sin( ) + bsin( ) + sin( ) 0 (JEE MIN) Sol: y sine rule i.e. b k we n simply prove the bove eqution. sin sin sin In eqution sin( ) + bsin( ) + sin( ) 0, putting k sin, b k sin, k sin k ( sin sin( ) sinsin( ) sinsin( ) ) (expnding ll terms gets nelled) (Using sin ( α β ) sin α. os β sin βos α) Illustrtion 4: Prove tht b sin sin (JEE MIN)

3 Mthemtis 19.3 ( ) b Sol: Given, sin( ) sin Tkin L.H.S., sin( ) (sinos ossin) sin( ) b sin Now using sine rule, sin sin sin k (sy) nd osine rule, b b b os +, nd os + b + b + b + b + b kb k k sin b RHS. b Illustrtion 5: The ngles of tringle re in 4:1:1 rtio. Find the rtio between its greter side nd perimeter? (JEE DVNED) Sol: Here, the ngles re 10, 30, 30. Therefore, by using sine rule, we will get the required rtio. ngles re 10, 30, 30. If the sides opposite to these ngles re, b nd respetively, will be the gretest side. b b Now from sine formul, ; sin10 sin30 sin30 3/ 1/ 1/ ; b k (sy) 3k 3 then 3k, perimeter ( + 3k ) ; Required rtio. + 3k + 3 Illustrtion 6: Solve bos + os in term of k where k is perimeter of the. (JEE DVNED) 1 + os θ Sol: We n solve the given problem simply by os θ b Here, bos + os ( 1 + os ) + (1 + os) [using projetion formul] b+ 1 + b+ + ; bos + os k [where k+b+, given] tn b Illustrtion 7: In ny tringle, show tht (JEE DVNED) + b + tn Sol: We n derive the vlues of, b nd using sine rule nd putting it to L.H.S. we n prove the bove problem. b We know tht, k sin sin sin k sin, b ksin, k sin (i) On putting the vlues of nd b from (1) on L.H.S., we get + os sin b k sin k sin sin sin L.H.S. + b k sin + k sin sin + sin + sin os tn + ot tn R.H.S. + tn

4 19.4 Solutions of Tringle 5. NPIER S NLOGY (LW OF TNGENTS) In ny, (i) b tn ot b+ (ii) b tn ot + b (iii) tn ot + Proof:(i) R.H.S. b ot k sin k sin ot [y the sine rule] b+ k sin + k sin + sin os.ot [y & D formule] + sin os sin sin ot sin + sin + tn.ot ot tn.tn.ot [y ondition identities] tn LHS Similrly, (ii) nd (iii) n be proved. Illustrtion 8: In ny tringle, if 30, b3 nd 3 3, then find nd. (JEE MIN) b Sol: y using formul, tn ot, we n esily obtin the vlues of nd. + b Here (i) Sine > b > nd (ii) b b + tn ot tn b b tn tn ; 3( 3 + 1) b π b tn ot tn + b + b 3( 3 1) ( 3 1) tn45 + tn30 [Using (1) ] tn tn( ) 3( 3 + 1) ( ) 1 tn45 tn30 1 ( ) tn ( ) ; ; 3 90 (iii) Solving (ii) nd (iii), we get 30 nd TRIGONOMETRI RTIOS OF HLF NGLES Sine, osine nd tngent of hlf the ngles of ny tringle re relted to their sides s given below. Note tht the perimeter of will be denoted by s i.e. s+b+ nd the re denoted by.

5 Mthemtis 19.5 Formule for (i) sin sin, sin, sin for ny ( s b)( s ) b (ii) sin ( s )( s ) (iii) sin ( s )( s b) b Formule for (i) os os, os, os for ny ss ( ) b (ii) os ss ( b) (iii) os ss ( ) b Formule for tn, tn, tn for ny s b s (i) tn ss s s (ii) tn ss b (iii) ( s )( s b) ( ) tn ss Illustrtion 9: In tringle, if s s b s, then find the vlue of s b s Sol: s we know, tn ss s s b s 3s + b + s ; Now tn tn (/ ). (JEE MIN). Therefore, by using this formul, we n solve the bove problem. ( s b)( s ) ss ( ) Illustrtion 10: In tringle, prove tht (+ b + ) tn + tn ot. (JEE MIN) s b s Sol: Here by using tn ss L.H.S. (+ b + ) tn + tn s s nd tn ss b s s b+ s s s s + b + b s s s b s s b lternte: L.H.S. s + s ( s ) s ( s b ) we n prove the bove problem. s b s s s s s b s s + s + ss ss b s s s b ( ) ( s )( s b) ss ot R.H.S. tn 1 1 s b s s + s s b s s b s s b.ot R.H.S. Illustrtion 11: In, if ot, ot, ot re in P, then prove tht the sides of re in.p. (JEE MIN) Sol: Here by using trigonometri rtios of hlf ngles formul, we n prove the bove illustrtion. Given ot, ot, ot re in.p.

6 19.6 Solutions of Tringle ss ( ) ss ( b ) ss ( ),, re in.p. ( s ), ( s b ), ( s ) re in.p., b, re in.p. Proved Illustrtion 1: In, the sides, b nd re in.p. Then wht is the vlue of tn + tn : ot? (JEE DVNED) Sol: Simply by using formul of tn, tn nd ot we n esily get the required result. tn + tn : ot s ( + ) : s ; s b s s s b s s b + : s s s s s s + b+ b: ; b : + b +.b. re in.p. ( ) + ( ) s s : s s b:+b+ :3 + b+ Illustrtion 13: In ny tringle, show tht ot + ot + ot ot. (JEE DVNED) + b Sol: Similr to the bove problem, by putting the vlues of problem. ot, ss ss b ss ot + ot + ot + + s b s s s s s b L.H.S. 1 ( )( )( ) s s s b s 1 ( )( )( ) s s s b s ( ) + ( ) + ( ) s s s s b s s { ( + + )} s 3s b 1 ( )( )( ) s s s b s ot nd 1 ( )( )( ) s s s b s ( s) s 3s ot we n prove the bove ( + + ) s s s b s s ot + ot + ot s s s b s ( )( )( ) (i) Now, R.H.S. L.H.S R.H.S + b + s s s ot ot ot ot + b + b + s s ss ( ) s s s s s b s s s b s + b+ s ; ot + b s s s b s (ii) From (i) nd (ii), we hve + b+ ot + ot + ot ot. + b Proved

7 Mthemtis RE OF TRINGLE If be the re of tringle, then bsin sin bsin Proof: Let be tringle. Then the following ses rise. se I: When D is n ute ngled tringle, sin 1 1 D sin ; D sin ; reof ; ( )( D ) ; sin D se-ii: When is n obtuse ngled tringle, sin( 180 ) ; D sin D sin re of ; () ( D ) ; sin ; So in eh se, sin (ii) Heron s formul s( s )( s b)( s ) Proof: 1 b sin os bsin.os ( s b)( s ) s( s ) b [y hlf ngle formul] s( s )( s b)( s ) b b Figure 19.5 Figure 19.6 (iii) 1 sinsin 1 b sinsin 1 sin sin sin sin sin ( + ) ( + ) ( + ) From the bove results, we obtin the following vlues of sin, sin nd sin b b (iv) sin s( s )( s b)( s ) (v) sin s( s )( s b)( s ) b b (vi) sin s( s )( s b)( s ) Further with the help of (iv), (v), (vi) we obtin sin sin sin b b Illustrtion 14: In ny tringle, prove tht Sol: We n prove the bove problem by using formul of re of tringle i.e. 1 os b + L.H.S. 4 ot 4. bsin. b os b. sin b 4 ot b +. (JEE MIN) 1 bsin nd b + R.H.S. b + os. b Illustrtion 15: In ny tringle, prove tht b sin sin. sin ( ) Sol: y putting k sin nd b k sin we n prove the bove illustrtion. b sin sin k sin k sin sin sin L.H.S.. sin sin( ) ( ) ( ) k sin sin sin sin sin( ) (JEE MIN)

8 19.8 Solutions of Tringle k sin + sin sinsin k [using sine formul k sin et.].sin ( + ) sinsin sin( ) 1 k sin k sin sin π ; + π ; 1 bsin R.H.S. Illustrtion 16: tree stnds vertilly on hill side whih mke n ngle of 15 o with the horizontl. From point on the ground 35m down the hill from the bse of the tree, the ngle of elevtion of the top of the tree is 60 o. Find the height of the tree. (JEE MIN) Sol: We n simply obtin the height of the tree from the given figure. P In QR, QR Qsin15 35sin15 ; R Qos15 35os15 In PR, PR tn60 ; PR R. 3 ; PQ + QR 3R R h h + 35sin os h 35 3 os15 sin ( 4) 35 m Hene, the height of the tree 35 m 60 o 15 o 35 m Figure 19.7 Q R Illustrtion 17: In ny tringle, prove tht 8 os + bos + os sinsin. (JEE DVNED) b Sol: we n solve this illustrtion by substituting s k sin, b k sin, nd k sin. k sin, b k sin, k sin (i) L.H.S. os + bos + os k sinos + k sinos + k sinos [using sine formul] k k sin os sinos + + { sin + sin} + sin k sin os ( ) + os ( π + ) k sinos sinos + k sin os os + k sin sinsin ( k sin ). sinsin sinsin 8 R.H.S. b ; b { } 1 sin sin ndsin b Illustrtion 18: The ngle of elevtion of the top point P of the vertil tower PQ of height h from point is 45 o nd from point, the ngle of elevtion is 60 o, where is point t distne d from the point mesured long the line whih mkes n ngle 30 with Q. Prove tht h d( 3 1) (JEE DVNED) Sol: y using sine rule in P, we n prove tht h d( 3 1) In the figure, PQ represents tower of height h. The ngle of elevtion of the point P from the point on the ground is

9 Mthemtis 19.9 PQ 45 ; P + Q 45 ; P P [Given Q 30 ] P (i) (Given) PH 45 ; P PH 45 + (given) P P 15 (ii) From (i) nd (ii), we hve P P So Pd; P d [Given d] gin PQ 45, Q 90 PQ 45 In PQ, PQ PQ Q PQ h P PQ + Q h + h P h P h pplying sine formul in P, we get d h sin15 h 3 1 d sin150 1 sin15 P sin150 d ( 3 1h ) d 45 o 30 o Figure 19.8 Illustrtion 19: lmp post is situted t the middle point M of the side of tringulr plot with 7m, 1 8m nd 9m. This lmp post subtends n ngle tn ( 3) t the point. Determine the height of the lmp post. (JEE DVNED) Sol: Here in post. MP PM tn PM, therefore by obtining the vlue of M we n find out the height of lmp M Here, is tringulr plot. lmp post PM is situted t the mid-point M of the side. Here PM subtends n 1 tn 3 t the point. 7m, b8m nd 9m ngle In, + b + os or os b os M M M 65 M In M, os ; os.m M 65 M [Using(i)] M M 7m In MP PM tn PM M Hene, the height of the lmp post 1m. 1 PM PM tn tn 3 3 PM 1m M 7 15 o P P h H Q (i) 9 M b8-1 tn (3) 7 Figure PROPERTIES OF TRINGLE 8.1 irumirle irle pssing through the verties of tringle is lled irumirle of the tringle. The entre of the irumirle is lled the irumentre of the tringle nd it is the point of intersetion of the perpendiulr bisetors of the sides of the tringle. The rdius of the irumirle is lled the irumrdius of the tringle nd is usully denoted by R nd is given by the following Figure 19.10

10 19.10 Solutions of Tringle b b formule: R Where is re of tringle nd sin sin sin 4 S 8. Inirle + b+ s. The irle whih n be insribed within the tringle so s to touh ll the three sides of the tringle is lled the inirle of the tringle. The entre of the inirle is lled the inentre of the tringle nd it is the point of intersetion of the internl bisetors of the ngles of the tringle. The rdius of the irle is lled the inrdius of the tringle nd is usully denoted by rin-rdius: The rdius r of the insribed irle of tringle is given by () (b) () r (ii) r ( s ) tn, r ( s b) tn s nd r ( s ) tn sin sin bsin sin sin sin r, r nd r os os os r 4R sin.sin.sin Figure entroid In, the mid-points of the sides, nd re D,E nd F respetively. The lines D, E nd F re lled medins of the tringle. The points of onurreny of three medins is lled the entroid. Generlly it is represented by G. lso, G D, G E nd G F Length of medins from Figure 9.1 b + D b + b 4 b b + 1 D D b Similrly, E + b nd F + b F D Figure 19.1 G E 8.4 pollonius Theorem + D + D Proof: D + D b + + b F E 8.5 Orthoentre O The point of intersetion of perpendiulrs drwn from the verties on the opposite sides of tringle is lled its orthoentre.let the perpendiulr D,E nd F from the verties, nd on the opposite sides, nd D of, respetively meet t O. Figure Then O is the orthoentre of the. The tringle DEF is lled the pedl Tringle of the.

11 Mthemtis entroid (G) of tringle is situted on the line joining its irumentre (O) nd orthoenter (H) show tht the line divides joining its irumentre (O) nd orthoenter (H) in the rtio 1:. Proof: Let L be perpendiulr from on, then H lies on L. If OD is perpendiulr from O on, then D is mid-point of. D is medin of. Let the line HO meet the medin D t G. Now, we shll prove tht G is the entroid of the. Obviously, OGD nd HG re similr tringles. OG GD OD R os 1 L D HG G H R os Figure GD G G is entroid of nd OG : HG 1 : The distnes of the orthoenter from the verties nd the sides: If O is the orthoenter nd DEF the pedl tringle of the, where D, E, F re the perpendiulrs drwn from,, on the opposite sides,, respetively, then (i) O R os, O R os nd O R os (ii) OD R osos, OER osos nd OF R osos, where R is irumrdius. H G O (iii) The irumrdius of the pedl tringle R (iv) The re of pedl tringle ososos. (v) The sides of the pedl tringle re os, bos nd os nd its ngles re π, π nd π. (vi) irumrdii of the tringles O, O, O nd re equl. MSTERJEE ONEPTS The irumentre, entroid nd orthoentre re olliner. In ny right ngled tringle, the orthoentre oinides with the vertex ontining the right ngle. The mid-point of the hypotenuse of right ngled tringle is equidistnt from the three verties of tringle. The mid-point of the hypotenuse of right ngled tringle is the irumentreof the tringle. The entroid of the tringle lies on the line joining the irumentre to the orthoentre nd divides it in the rtio 1: Vibhv Krishnn (JEE 009,IR ) 9. PEDL TRINGLE The tringle formed by the feet of the ltitudes on the side of tringle is lled pedl tringle. In n ute ngled tringle, orthoentre of is the in-entre of the pedl tringle DEF. F H E Proof: Points F,H,D nd re onyli π FDH FH E Similrly, points D, H, E nd re onyli D Figure 19.15

12 19.1 Solutions of Tringle π HDE HE F Thus, FDH HDE D is the ngle bisetor of FDE. Hene, ltitudes of re internl ngle bisetors of the pedl tringle. Thus, the orthoentre of is the inentre of the pedl tringle DEF. Sides of pedl tringle in ute ngled tringle In EF,F bos,e os y osine rule, EF E + F E Fos( EF) 3 EF b os + os bos EF os b + bos os EF os irumrdius of pedl tringle EF os os Let irumrdius be R' R ' R sin EDF sin π sinos sin R' R / MSTERJEE ONEPTS The irle irumsribing the pedl tringle of given tringle bisets the sides of the given tringle nd lso the lines joining the verties of the given tringle to the orthoenter of the given tringle. This irle is known s Nine point irle. The irumentre of the pedl tringle of given tringle bisets the line joining the irumentre of the tringle to the orthoentre. It lso psses through midpoint of the line segment from eh vertex to the orthoenter. Orthoenter of tringle is in entre of pedl tringle. Shriknt Ngori (JEE 009, IR 30) 10. ESRIED IRLES OF THE TRINGLE The irle whih touhes the sides nd two sides nd produed of tringle is lled the esribed irle opposite to the ngle. Its rdius is denoted byr 1. Similrly, r ndr 3 denote the rdii of the esribed irles opposite to the ngles nd respetively. The entres of the esribed irles re lled the ex-entres. The entre of the esribed irle opposite to the ngle is the point of intersetion of the externl bisetor of ngles nd. The internl bisetor lso psses through the sme point. This entre is generlly denoted by I 1. Formule for r 1,r,r 3 In ny, we hve (i) r 1,r,r s s b 3 s (ii) r 1 stn, r stn, r 3 stn os os os os os os (iii) r 1,r b,r 3 os os os (iv) r 1 4R sin os os, r 4R os sin os, r 3 4R os os sin Figure 19.16

13 Mthemtis MSTERJEE ONEPTS If I 1 is the entre of the esribed irle opposite to the ngle, then OI1 R 1 + 8sin.os.os ; OI R 1 + 8os.sin.os ; OI3 R 1 + 8os.os.sin Where R is irum rdius The Sum of the opposite ngles of yli qudrilterl is 180 o. In yli qudrilterl, the sum of the produts of the opposites is equl to the produt of digonls. This is known s Ptolemy s theorem. If the sum of the opposite sides of qudrilterl is equl, then nd only then irle n be insribed in the qudrilterl. If l 1, l nd l 3 re the entres of esribed irles whih re opposite to, nd respetively nd l is the entre of the inirle, then tringle is the pedl tringle of the tringle lll 13 nd l is the orthoenter of tringle lll 13. The irle irumsribing the pedl tringle of given tringle bisets the sides of the given tringle nd lso the lines joining the verties of the given tringle to the orthoenter of the given tringle. This irle is lso known s nine point irle. 1 + bd d + b b + d The irumrdius of yli qudrilterl, R 4 s s b s s d Nitish Jhwr (JEE 009, IR 7) 11. LENGTH OF NGLE ISETOR ND MEDINS If m nd β re the lengths of medin nd n ngle bisetor from the ngle then, b os 1 m b + nd β b+ 3. Note tht m + mb + m ( + b + ) 4 Illustrtion 0: The rtio of the irumrdius nd in rdius of n equilterl tringle is (JEE MIN) Sol: Here, s we know, ll ngles of n equilterl tringle re In rdius we n obtin the required rtio. 0 60, therefore by using formul of irumrdius nd r os + bos + os + b + os In n equilterl tringle, 60 R + b+ + b+ 0 Illustrtion 1: In, 18 nd b4m nd 30m then find the vlue of r 1, r nd r 3. (JEE MIN) Sol: s we know, r 1 s, r s b nd r 3 s. Hene, we n solve the bove problem by using this formul. 18m, b4m,30m; s b 7m + + ; s36m ut, s( s )( s b)( s ) sq. units Then, r 1 1m ; r s 18 18m s b 1 So, r 1,r, r 3 re 1m, 18m, nd 36m respetively. 16 ; r 3 36m s 6

14 19.14 Solutions of Tringle Illustrtion : If the exrdii of tringle re in HP, the orresponding sides re in (JEE MIN) Sol: Here, in this problem, r 1, r nd r 3 re in H.P s s b s,, re in.p.,, re in.p. s,s b,s re in.p. r r r 1 3, b, re in.p., b, re in.p. Illustrtion 3: Find the vlue of Sol: y using r 1 b b + +. (JEE DVNED) r r r 1 3 s, r s b nd r, we n solve the bove problem. 3 s ( b ) ( ) ( b) + + ( b ) + ( ) + ( b) r r r 1 3 ( s )( b ) + ( s b)( ) + ( s )( b) s s b s s b + + b b + b b + b 0 0 Illustrtion 4: Find the vlue of the r ot ot. (JEE DVNED) Sol: Here, in this problem, r4rsin/.sin/.sin/. y putting this vlue, we n solve the bove problem. rot/.ot/ os / os / 4R sin / sin /.sin /.. [s r4rsin/.sin/.sin/] sin / sin / 4R.sin/.os/.os/ r { s,r 4R sin/.os/.os/ } rot /.ot / r EXENTRL TRINGLE I 3 I The tringle formed by joining the three exentres I 1,I nd I 3 of is lled the exentrl or exentri tringle. Note tht: (i) The inentre I of is the orthoentre of the exentrl I13 II. (ii) is the pedl tringle of the III I 90 o (iii) The sides of the exentrl tringle re 4R os, 4R os nd 4R os nd π π π Its ngles re, nd. I 1 Figure (iv) Distne between the inentre nd exentre I I 1 4R sin ; I I 4R sin ; I I3 4R sin. m D n Figure 19.18

15 Mthemtis M-N THEOREM (RTIO FORMUL) If D be point on the side of () ( m n) ot mot not + θ α β (b) Proof: () Given tht, D m nd D θ D n D 180 θ ; D α nd D β suh tht D:D m:n nd D θ, αnd D β. m + n otθ not mot D 180 α+ 180 θ θ α nd D 180 ( θ+β) From D D D, sin α sin ( θ α) D D From D, or sin β sin 180 ( θ+β) dividing (i) by (ii), then ( θ+β) Dsinβ sin Dsinα sin θ α D D sinβ sin ( θ+β) m sinβ sin θ.osβ+ os θ.sinβ or n sinα sin θ.osα osθsinα or msinθsinβosα mos θ.sin α.sinβ nsinαsinθosβ+ nsinαosθsinβ motα motθ notβ+ notθ [dividing both sides by sin sin sin nd α β θ] or (b) Given D m nd D θ; D 180 θ; D nd D D n D θ+ θ ; D 180 ( θ+ ) nd now from nd from D D D. sin sin ( θ ) D D D, or sin 180 θ+ sin dividing (i) by (ii), then D. D ( ) sin( θ+ ) sin( θ ) D sin ( θ+ ) D sin sin m sinθ os + osθsin sin or, sin n sinθos osθsin sin or, msinθ ossin + mosθ sinsin nsinθsinos nosθsinsin or, mot + motθ not notθ [dividing both sides by sinsinsinθ ] or, ( m + n) otθ not mot m + n otθ motα notβ... (i)... (ii) (i)... (ii) Illustrtion 5: In tringle, if ot ot, ot ot nd ot ot b, then find the vlue of + +. (JEE MIN) s s b s Sol: Here, by using trigonometri rtios of hlf ngle, we n solve bove problem. ( ) ( ) ss ss b ot ot ; s b s s s Similrly So tht 1 1 b nd s s s b s b + s + + s s b s s s s 1 s s s

16 19.16 Solutions of Tringle 14. SOLUTION OF DIFFERENT TYPES OF TRINGLE In tringle, there re six elements- three sides nd three ngles. In plne geometry, we hve done tht if three of the elements re given, t lest one of whih must be side, then the other three elements n be uniquely determined. The proedure of determining unknown elements from the known elements is lled solving tringle. Solution of right ngled tringle se I: When two sides re given: Let the tringle be right ngled t. Then we n determine the remining elements s given in the tble. (i) (ii) Given, b, Required tn, 90, b sin sin, b os, 90 se II: When side nd n ute ngle re given: In this se, we n determine the remining elements s given in the tble. (i) Given, Required 90, b ot, sin (ii), 90, sin, b os Solution of tringle in generl se I: When three sides, b, re given: In this se, the remining elements re determined by using the following formule. s( s )( s b)( s ), where s + b + sin, sin, sin. OR tn, tn, tn b b ss ss b ss se II: When two sides, b nd the inluded ngle re given: In this se, we use the following formule: 1 b bsin, tn ot + b ; + 90 nd sin sin se III: When one side nd two ngle nd re given: In this se, we use the following formule to determine the remining elements sin 1 ; 180 ( + ) nd ; sin sin se IV: When two sides, b nd the opposite to one side is given: In this se, we use the following formule. b sin sin... (i) sin 180 ( + ), sin From (i), the following possibilities will rise: When is n ute ngle nd < bsin.

17 Mthemtis b In this se, the reltion sin sin gives tht sin > 1, whih is impossible. Hene no tringle is possible. When is n ute ngle nd bsin. In this se, only one tringle is possible whih is right ngled t. When is n ute ngle nd > bsin bsin In this se, there re two vlues of given by sin sy 1nd suh tht nd side n sin be obtined by using sin Some useful results: Solution of oblique tringles: The tringle whih re not right ngled re known s oblique tringles. The problems on solving n oblique tringle lie in the following tegories: () When three sides re given (b) When two side nd inluded ngle re given () When one side nd two ngles re given (d) When ll the three ngles re given (e) mbiguous se in solution of tringle When the three sides re given: When three sides, b, of tringle re given, then to solve it, we hve to find its three ngles,,. For this osine rule n be used. When two sides nd inluded ngle re given: Problem bsed on finding the ngles when ny two sides nd the ngles between them or ny two sides nd the differene of the opposite ngles to them re given, Npier s nlogy n be used. When one side nd two ngles re given: Problems bsed on finding the sides nd ngles when ny two nd side opposite to one of them re given, then sine rule n be used. When ll the three ngles re given: In this se unique solution of tringle is not possible. In this se only the rtio of the sides n be determined. b For this the formul, n be used sin sin sin mbiguous se in solution of tringles: When ny two sides nd one of the orresponding ngles re given, under ertin dditionl onditions, two tringles re possible. The se when two tringles re possible is lled the mbiguous se. In ft, when ny two sides nd the ngle opposite to one of them re given either no tringle is possible or only one tringle is possible or two tringles re possible. Now, we will disuss the se when two tringles re possible. Illustrtion 6: Solve the tringle, if b 7.95, 8.31, 4 47' (JEE MIN) Sol: y using sine rule i.e. sin sin, we n solve the given tringle. b sin sin sin 8.31 sin4 47' (i) To find sin b b sin '1'' nd 19 58'48''

18 19.18 Solutions of Tringle 50 1'1" I solution ' '1" 87 11'48" To find b sin sin bsin sin sin87 11'48" sin4 7' 19 58'48" II solution 180 (4 47' '48") 7 14'1" To find b sin sin bsin sin sin7 14'1" sin4 7' Two solutions re '1" '48" '48" Geometrilly, we drw the tringle with given dt,b nd ngle. () If N( sin) b (extly). The tringle is right ngled tringle. (b) If N( sin) > b, the tringle nnot be drwn. () If N( sin) < b <, two tringles re possible. (d) b>, only one tringle is possible. 7 14'1" N b Figure b Illustrtion 7: In tringle, b16m, 5m, nd 33 15'. Find the ngle. (JEE MIN) Sol: Simply by using sine rule, we n find out the ngle. We know tht, sin sin [Here, b16m, 5m, 33 15' ] b 5sin33 15' 1 sin sin ; sin ( ) 58 57' ; ' ; ' 11 3' b 16 PROLEM SOLVING TTIS In the pplition of sine rule, the following points re to be noted. We re given one side nd some other side x is to be found. oth these re in different tringles. We hoose ommon side y of these tringles. Then pply sine rule for nd y in one tringle nd for x nd y for the other tringle nd eliminte y. Thus, we will get the unknown side x in terms of. In the djoining figure, is the known side of nd x is the unknown side of tringle D. The ommon side of these tringles is y(sy). Now, pply sine rule. y y x y (i) nd. (ii) sinα sinβ sinθ sinγ x D Dividing (ii) by (i) we get, xsin α sin β sinβsinθ ; x sinθ sinγ sinαsinγ Figure 19.0 In se of generlized tringle problems, option verifition is very useful using equilterl, isoseles or right ngle tringle properties. So, it is dvised to remember properties of these tringles.

19 Mthemtis FORMULE SHEET () In, + + π () sin( + ) sin( π ) sin (b) os( + ) os ( π ) os () + π sin sin os + π (d) os os sin b (b) Sine rule: In, R Where R irumrdius nd, b, re sides of tringle. sin sin sin () b + + b + b osine rule: os, os, os b b (d) Trigonometri rtios of hlf ngles: () sin (e) re of tringle : ( s b)( s ) where s + b + ; (b) b bsin sin bsin (f) Heron s formul : s( s )( s b)( s ), where (g) irumirle Rdius : b b R sin sin sin 4 os + b+ s. ( ) ss ; () b ( s b)( s ) ( ) tn ss (h) Inirle Rdius : () r s (i) Rdius of the Esribed irle : () 1 3 r s tn ; (b) r ( s ) tn, r ( s b) tn nd r, r,r s s b s r stn, r stn, r stn os os os os os os r,r b,r os os os r 4R sin os os, r 4R os sin os, r 4R os os sin (b) 1 3 () 1 3 (d) 1 3 (j) Length of ngle bisetor nd Medin: b os 1 m b + nd β b+ m - length of medin, β - length of bisetor.