GEOMETRICAL PROPERTIES OF ANGLES AND CIRCLES, ANGLES PROPERTIES OF TRIANGLES, QUADRILATERALS AND POLYGONS:

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1 GEOMETRICL PROPERTIES OF NGLES ND CIRCLES, NGLES PROPERTIES OF TRINGLES, QUDRILTERLS ND POLYGONS: 1.1 TYPES OF NGLES: CUTE NGLE RIGHT NGLE OTUSE NGLE STRIGHT NGLE REFLEX NGLE ngles less thn ngles equl to ngles greter thn 90 0 but less thn ngles equl to ngles greter thn but less thn GEOMETRICL PROPERTIES OF NGLES:

2 1.COMPLEMENTRY NGLES: Two ngles re complementry if their sum dd up to b 90 0 b. SUPPLEMENTRY NGLES: Two ngles re supplementry if their sum dd up to b b 3. The sum of DJCENT NGLES on stright line is equl to b + c ( dj. s on str. line. ) b c

3 4. ngles t point dd up to b + c + d + e ( s, t point. ) e b d c 5. Verticlly opposite ngles re equl. c b d ( vert. opp. s.) b d C 6. ngles formed by prllel lines cut by trnsversl, l. ) CORRESPONDING NGLES re equl. l b ( corr. s, // CD ) b C D

4 b) LTERNTE NGLES re equl. l c d c ( lt. s, // CD ) d C D c) INTERIOR NGLES re supplementry. l e f e ( int. s, // CD ) f C D 1.3 NGLES PROPERTIES OF TRINGLES ND QUDRILTERLS: NGLE PROPERTIES OF TRINGLES: 1. The sum of the 3 ngles of tringle is equl to x + y + z x ( sum of Δ ) y z

5 . The exterior ngle of tringle is equl to the sum of the interior opposite ngles. x y + z y ( ext. of Δ ) z x 3. n Isosceles Tringle hs equl ngles opposite the equl sides. x y ( bse s of isos. Δ ) x y 4. n Equilterl Tringle hs 3 equl sides nd 3 equl ngles, ech equl to x + y + z 60 0 x ( of equi. Δ ) y z

6 TIPS FOR STUDENTS : Sclene Tringle hs no equl sides nd ll the ngles re different in size. 5. tringle cn be grouped ccording to the types of ngles it contins s shown in the tble below. CTUE NGLED TRINGLE RIGHT NGLED TRINGLE OTUSE NGLED TRINGLE ll Three ngles re cute. One Right ngle. One Obtuse ngle. NGLE PROPERTIES OF QUDRILTERLS: 1. The sum of ll the ngles in qudrilterl is

7 + b + c + d ( sum of qud. ) d b c. The properties of some specil qudrilterls re given in the tble below. NME DESCRIPTION EXMPLE TRPEZIUM One pir of prllel opposite sides. ISOSCELES TRPEZIUM One pir of prllel opposite sides Non prllel sides re equl in length.

8 PRLLELOGRM Two pirs of prllel opposite sides. Opposite sides re equl in length. Opposite ngles re equl. D Digonls bisects ech other. X XC X XD C NME DESCRIPTION EXMPLE RECTNGLE Two pirs of prllel opposite sides. Opposite sides re equl in length. ll four ngles re right ngles. (90 0 ) Digonls re equl in length. Digonls bisect ech other.

9 RHOMUS Two pirs of prllel opposite sides. Four equl sides. Opposite ngles re equl. Digonls bisect ech other t right ngles. D Digonls bisect the interior ngles. X XC X XD C NME DESCRIPTION EXMPLE SQURE Two pirs of prllel opposite sides. Four equl sides. ll four ngles re right ngles. (90 0 ) Digonls re equl in length. Digonls bisect ech other t right ngles. Digonls bisect the

10 interior ngles. KITE No prllel sides. Two pirs of equl djcent sides. One pir of equl opposite ngles. Digonls intersect t right ngles. D One digonls bisect the interior ngles. C DC C DC C TIPS FOR STUDENTS: rectngle with 4 equl sides is squre. Prllelogrm with 4 right ngles is rectngle. prllelogrm with 4 equl sides is rhombus. rhombus with 4 equl ngles is squre. EXMPLE1: In the digrm, is prllel to CD nd PT is prllel to QR. Given tht PST 5 0, PQR 10 0 nd RUD 60 0.

11 Find ) PTU, b) SPT, c) QRU. P Q C S T U D R SOLUTION: P Q X 60 0 Y 60 0 R C S T U D

12 ) PTU PQR ( corr, s, PT // QR ) 10 0 PTU PT ( lt, s, // CD ) 10 0 b) SPT + PST PTU ( ext, s of Δ ) SPT c) Drw the line XY through R which is prllel to. QRX + PQR 1800 ( int, s, // XY ) SPT QRX 60 0 XRU RUD ( lt, s, XY // CD ) 60 0 QRU QRX + XRU

13 EXMPLE : In the digrm, PQR is tringle. The point S is on PQ producer, the point U is on PR produced nd STU is stright line. SQ ST nd QR is prllel to SU. Given tht QRU 13 0 nd QST P CLCULTE z ) x, Q R b) y, y 13 0 c) z x S T U SOLUTION: ) x ( int, s, QR // SU ) x b) STQ ( bse, s of isos. Δ ) 6 0

14 y STQ (lt, s, QR // SU ) 6 0 c) z x ( sum of Δ ) z y NGLE OF PROPERTIES OF POLYGONS: 1. polygon is closed plne figure with three or more stright lines. e.g. 3 SIDES 4 SIDES 5 SIDES TRINGLE QUDRILTERL PENTGON. Polygons re nmed ccording to the number of sides they hve.

15 The tble below shows the nmes of some polygons. NUMER OF SIDES NME OF POLYGON 3 Tringle 4 Qudrilterl 5 Pentgon 6 Hexgon 7 Heptgon 8 Octgon 9 Nongon 10 Decgon 3. In regulr polygon, the sides re ll equl in length nd the interior ngles re equl. REGULR PENTGON REGULR HEXGON TIPS FOR STUDENTS: polygon with n sides is clled n n gon. e.g. Polygon with 1 sides is clled 1 gon.

16 INTERIOR NGLES OF POLYGON: For n sided polygon, Sum of the interior ngle (n ) X For regulr n sided polygon, Size of ech interior ngle ( ) INTERIOR NGLES EXMPLE3: CLCULTE ) The sum of the interior ngles of n octgon, b) The size of ech interior ngles in regulr 18 sides polygon. c) The size of ech interior ngle in regulr hexgon. SOLUTION:

17 ) Sum of interior ngles of n octgon ( 8 ) X X Sum of ech interior s of regulr polygon (n ) X n octgon hs 8 sides n 8 b) Size of ech interior ngle of regulr 18 sides polygon ( ) Sum of the interior s of polygon ( ) Hence, n 18. c) Size of ech interior ngle of regulr hexgon ( ) 10 0 EXMPLE4 :

18 CDEF is hexgon. Find the vlue of x. F x E D SOLUTION: x Sum of interior ngles of hexgon ( 6 ) X C x x x x 5 0 x 75 0 EXTERIOR NGLES OF POLYGON: Sum of ll the exterior ngles of ny polygon 360 0

19 For regulr n sided polygon, Size of ech exterior ngle EXTERIOR NGLES EXMPLE 5 : ) Clculte the size of ech exterior ngles of regulr pentgon. b) The exterior of regulr polygon is 4 0. How mny sides does it hve? SOLUTION: ) Size of ech exterior ngles of regulr pentgon 7 0 Sum of the exterior s of regulr polygon pentgon hs 5 sides. n 5

20 b) Number of sides of the regulr polygon 15 EXMPLE6: ) Ech interior ngles of regulr polygon is 10 0 greter thn ech exterior ngles of the polygon. Clculte the number of sides of the polygon. b) One of the exterior ngles of n octgon is 66 0 while the rest of the seven exterior ngles re ech equl to x 0. Find the vlue of x. SOLUTION : ) Let the size of ech exterior ngles be x 0. Size of ech interior ngle x x + ( x ) x 60 0 The sum of the int. nd ext. is equl to x 30 0 Size of ech exterior ngle 30 0.

21 Number of sides of polygon 1 b) Sum of exterior ngles of n octgon x x 94 0 The sum of the ext. s of ny polygon is x 4 0 EXMPLE7 : The digrm shows regulr pentgon CDE nd regulr hexgon PQRS which re drwn on opposite sides of the common line. CLCULTE E S ) E, D b) S, R c) ES. C Q P SOLUTION:

22 E S D R C Q P ) E ( ) Sum of the interior s of regulr polygon ( ) Hence, n 5. b) S ( ) 10 0 c) ES O ( s t point ) ES ( bse s of isos. Δ ) SYMMETRICL PROPERTIES OF CIRCLE:

23 1. The perpendiculr bisects of chord psses through the centre of the circle. Conversely, line drwn from the centre of circle to the midpoint of the chord is perpendiculr to chord. If OM, then M M. O Conversely, M If M M, then OX.. Ech chords of circle re equidistnt from the centre of the circle. Conversely, chords which re equidistnt from the centre of the circle re equl. Q If PQ RS, then OX OY. X O Conversely, P M

24 If OX OY, then PQ RS. EXMPLE8 : circle of rdius 8 cm hs centre O. chord XY is 1 cm long. Clculte the distnce from O to the midpoint of the chord. SOLUTION: Let M be the mid point of the chord XY. XM XY ( 1 ) 6 cm 8 O OMX 4 0 If XM MY, then X 6 M 6 Y OM XY. Using Pythgors Theorem on OXM, OM OM OM

25 5.9 cm ( correct to 3 sig. fig. ) The required distnce is 5.9 cm. 1.6 NGLE PROPERTIES OF CIRCLE: 1. n ngle t the centre of circle is twice the ngle t the circumference subtended by the sme rc. C x x x O x O O x x O C ( t centre t circumference )

26 . The ngle in semicircle is right ngle. C If is the dimeter of the circle, centre O, then C 90 0 ( rt. in semicircle ) O. The ngle in semicircle is right ngle. If is the dimeter of the circle, centre O, then C C 90 0 ( s in the sme segment ) x x x D 4. The ngle in opposite segments of circle re supplementry ( i.e. the sum of the ngles dd up to ) + c ( s in opp. Segments re supp.) b + c b d c

27 TIPS FOR STUDENTS: 1. cycle qudrilterl is qudrilterl drwn inside circle so tht ll its 4 vertices lie on the circumference of the circle.. The opposite ngles of cyclic qudrilterl dd up to ( Supplementry ngles ) EXMPLE9 : In the digrm, O is the center of the circle PQRST. POR is the dimeter of the circle, ROS 7 0 nd QPR 5 0. FIND P ) RPS, b) QES, c) PTS, Q O 5 0 d) PRQ. T 7 0 R S

28 SOLUTION : P Q 16 0 O 7 0 T 54 0 R S ) RPS 7 0 ( s t centre t circumference ) 36 0 b) QES ( bse s of isos. Δ) 54 0 c) PTS PRS ( s in opp. Segments re supp. ) d) PRQ 90 0 ( rt. s in Semicircle. ) ( s sum of Δ)

29 65 0 EXMPLE10 : The points,, C, D nd E lie on the circle. E 59 0, DE 3 0, DC 48 0 nd is prllel to DC. FIND ) E b) DE, 59 0 E c) D, 3 0 d) CD. D 48 0 C

30 SOLUTION : E D 48 0 C ) E DE ( s in the sme segment ) 3 0 b) D DC ( lt. s // DC ) c) D E ( s in the sme segment ) 59 0 PTS D - D ( s sum of Δ)

31 73 0 d) CD D ( s in opp. Segments re supp. ) EXMPLE11 : In the digrm,,, C nd D lie on the circle. DE nd CE re stright lines. C 41 0, D 48 0 nd DCE FIND ) DC, 41 0 D b) CD, c) DC, d) CED. E C

32 SOLUTION: D 41 0 E C ) DC C ( s in the sme segment ) 41 0 b) D ( ext. s of Δ) CD 34 0 c) DC C ( s in opp. Segments re supp. ) ( ) 98 0

33 d) CED ( ext. s of Δ) CED TNGENT THEOREMS: tngent to circle is line which touches the circle t only one point. tngent perpendiculr to the rdius t the point of contct. If C is tngent to the circle t, then O O OC 90 0 ( tn rd.) C TNGENTS FROM N EXTERNL POINT 1. Tngents drwn from n externl point to circle re equl. If P nd P re tngents to the circle, then

34 P P O b b P. The tngents subtend equl ngles t the centre. If P nd P re tngents to the circle center O, then PO PO 3. The lie joining the externl point to the centre of the circle bisects the bgles between the tngents. If P nd P re tngents to the circle, center O, then PO PO

35 EXMPLE11 : The tngents T nd TD re drwn from point T to the circle, center O. The dimeter D nd the tngent T when produced meet t E. Given tht DO 6 0. E FIND ) DT, b) TD, c) E, O d) CD. T C 6 0 D SOLUTION :

36 E O 5 0 T C D ) ODT 90 0 ( tn rd.) ODT b) T T ( tngents from n externl point) DT DT 64 0 ( bse s isos. Δ) DT ( s sum of Δ) 5 0 c) D 90 0 ( rt. s in semicircle. )

37 E ( dj. s on str. line) 6 0 d) D ( s of sum Δ) 64 0 CD D ( s in the sme segment ) 64 0 EXMPLE1 : In the digrm,,, C nd D lie on the circle, center. SCT is tngent to the circle t C, O is prllel to C nd CS 4 0. FIND ) OC, b) DC, c) OC, O D d) O. 4 0 S C T

38 SOLUTION : O D S C T ) OCS 90 0 ( tn rd.) OC OC ( int. s O // C ) 13 0 b) DC X OC ( t center t circumference. ) X

39 c) OC ( bse s of isos. Δ) 4 0 d) C ( s in opp. segment re supp.) O ( int. s O // C ) 66 0 EXMPLE13 : T nd T re tngents to circle, center O nd rdius r cm. Give tht T 1 cm nd CT 8 cm. r 8 1 O C T FIND ) n expression for OT in term of r,

40 b) The vlue of r, c) The re of qudrilterl OT, d) O un rdins, e) The re of the shded region. SOLUTION: r r 8 1 O C T ) OT r cm ( rdius of circle ) OT ( r + 8 ) cm b) OT OT 90 0 ( tn rd.) Using Pythgors Theorem on Δ OT,

41 OT O + T ( r + 8 ) r + 1 r + 16 r + 64 r r 80 r 5 cm c) re of qudrilterl OT, X re of Δ OT X ( X 1 X 5 ) 60 cm re of Δ X se X Height h X b X h b d) tn OT OT tn rd O X OT X rd ( correct to 3 sig. fig. )

42 e) re of shded region re of sector O r re of qudrilterl OT re of minor r sector O 60 - X 5 X cm ( correct to 3 sig. fig. ) O r 1.8 PERPENDICULR ISECTS ND NGLE ISECTORS: PERPENDICULR ISECTOR OF LINE 1. The Perpendiculr isector of line segment forms right ngle with the line segment nd divides the line segment into equl prts. e.g. P Q

43 PQ is clled perpendiculr bisector for line segment. PROPERTY OF PERPENDICULR ISECTOR ( isector ) ny point on the perpendiculr bisector of line segment is equidistnt to the two end points of the line segment.. To construct the perpendiculr bisector of the line : ) Set your compsses to more thn hlf the length of. b) With s center, mrk rcs bove nd below. ( Figure 1 ) c) Repet the process with s the centre. ( Figure 1 ) d) Drw stright line tht joins P nd Q ( where the two sets of rcs intersect ). ( Figure 1 ) PQ is the perpendiculr bisector of.

44 P P Figure 1 Q Figure Q Figure 3 NGLE ISECTOR: 1. n ngle bisector is ry tht divides n ngle into equl ngles. e.g. If X splits C into two ngles such tht X of C. C C, then X is the ngle bisector C

45 PROPERTY OF NGLE ISECTOR ( isector ) ny point on the ngle bisector is equidistnt from the two sides of the ngle.. To construct the ngle bisector of C : ) Use pir of compsses nd with S s centre, drw n rc to cut t X nd C t Y. ( Figure 1 ) b) Using sme rdius, with X nd Y s centers, drw two rcs tht intersect t Z. ( Figure ) c) Drw stright line from through Z. ( Figure 3 ) Z is the ngle bisector of C. X X X Z Y C Y C Y C Figure 1 Figure Figure

46 1.9 CONSTRUCTIONS OF SIMPLE GEOMETRIC FIGURES: Drw rough sketch nd write down the given informtion before constructing the ctul figure. EXMPLE14 : Drw tringle C in which 7.5 cm, C nd C 8 cm. ) Mesure nd write down the length of C. b) On the tringle, construct i) circle of rdius 3.8 cm with centre, ii) The ngle bisector of C. c) The ngle bisector cuts the circle t the point P, given tht P lies inside tringle C, Complete the sentence below. The point P is 3.8 cm from the point nd is equidistnt from the lines nd. SOLUTION :

47 ), b) C SKETCH: C cm 3 cm b) (ii) P b) (i) ) y mesurement, the length of C is 1.7 cm. b) The point P is 3.8 cm from the point nd is equidistnt from the lines nd C. EXMPLE15 : Construct the qudrilterl CD in which the bse 6 cm, C 108 0, C 5 cm, ) Mesure nd write down the size of DC, b) On the qudrilterl, construct i) The ngle bisector of CD,

48 ii) The perpendiculr bisector of. c) These two lines intersect t the point T. Mesure nd write down the length of DT. SOLUTION: ) y the mesurement, the size of DC is SKETCH : D 8 cm C D 7 cm 5 cm cm C b) (ii) T b) (i)

49 b) y the mesurement, the length of DT is 5.6 cm. PROPERTIES OF TRINGLES 1.1 RELTION RETWEEN SIDES ND NGLES OF TRINGLE: 1. tringle consists of three sides nd three ngles clled elements of the tringle. In ny tringle C,,, C denotes the ngles of the tringle t the vertices. + + C The sides of the tringle re denoted by, b, c opposite b c to the ngles, nd C respectively. C Fig (1) 3. + b + c s The perimeter of the tringle. 1. THE SINE RULE: In tringle C, prove tht

50 R Where, R is the circum rdius of the tringle. PROOF: Let S be the circumcentre of the tringle C. First prove tht R CSE (I) : Let S be n cute ngle. Let P be ny point on the circle. Join P, Which pss through S. Join CP, so tht CP 90 O. C PC (ngles in the sme segment). FROM Δ PC, sin C S P C sin, R Fig ( ) CSE (II) : Let be right ngle, ie., C R 90 0 ( Fig 3 ), Then C is the dimeter. sin S C R Fig ( 3)

51 CSE (III) : Let be n obtuse ngle ( Fig 4 ). join P, pssing through S. Join CP, so tht P Now C ( C ) ( Since PC is cyclic qudrilterl ) From Δ PC, sin ( C ) S C Fig ( 4) i.e. sin (180 0 ) sin R R is true for ll vlues of. Similrly, we cn prove, R, Thus, R. or r sin, b r sin, c r sinc.

52 1.3 THE COSINE RULE: In ny tringle C, prove tht b + c - bc cos b c + - c cos c + b - b cos C C C c b b b D C c D C ( Fig 5 ) ( Fig 6 ) ( Fig 7 ) PROOF: Cse ( I ) Let be n cute ngle ( Fig 5 ) Drw CD. From Δ DC

53 C D + DC ( D ) + DC -. D + D + DC C -. D + C ( Since D + DC C ) c C. D + b ut from Δ DC, Cos D cos b cos c c.b.cos + b Or b + c bc cos Cse ( II ) Let be right ngled, ie., 90 0 ( Fig 6 ) C + C ie., b + c ut b + c bc cos b + c bc cos 90 0 b + c, which is true for right ngled tringle. Cse ( III ) Let be obtuse ngle, ie., > 90 0 ( Fig 7 ).

54 Drw CD produced. From Δ DC C D + DC ( + D ) + DC C +. D + D + DC C +. D + C ( Since, D + DC C ) c.c. D + b ut from Δ DC, Cos ( D C ), ( D C ) Cos ( ) - cos, D b cos b + bc ( b cos ) + c b + c bc cos Similrly, we cn prove, b c + - c cos c + b - b cos C Tips for Students: The bove formule cn be written s: Cos b c bc, Cos c b c, Cos C b b, These results re useful in finding the cosines of the ngle when numericl vlues of the sides re given. Logrithmic computtion is not pplicble since

55 the formule involve sum nd difference of terms. However, logrithmic method cn be pplied t the end of simplifiction to find ngle 1.4 THE PROJECTION RULE: In this rule, we show how, one side of tringle cn be expressed in terms of other two sides. It is clled projections rule. b cos C + c cos, b c cos + cos C, c cos + b cos. PROOF: Let C be n cute ngle Drw D C produced. In Fig ( i ) C D + DC [ NOTE : D is clled projection of on C nd DC is the projection of C ] D + DC (1) From Δ D, cos D cos c cos From Δ CD, cos C CD C cos C b cos C c b From (1) c cos + b cos C b cos C + c cos D C (Fig 8)

56 CseII: When C is right ngle, ie., 90 0 ( Fig 9 ). cos, c cos () Since 90 0, cos cos , c We get, b cos c cos c cos (3) C ( Fig 9 ) Cse III : When c is obtuse ngle ( Fig (iii) ) From Δ D, C D - CD (4) From Δ D, cos, D c cos From Δ CD Cos ( C ) c b cos, CD - b cos C C From (4) C c cos ( - b cos C ) C D ie., c cos + b cos C b cos C + c cos ( Fig 10 ) Similrly, b c cos + cos C, c cos + b cos.

57 THE LW OF TNGENTS: In ny Δ C, Prove tht : 1. tn tn,. tn tn C C, 3. tn tn C C PROOF: Using sine rule,.cos cos.sin cos, cos.tn 1.5

58 tn tn u sin g cot 1 tn Similrly, other two results cn be proved by chnging sides nd ngles in cycle order. 1.6 EXPRESSIONS FOR HLF NGLES IN TERMS OF, b, c : In ny tringle C, prove tht 1. sin S bs C bc,. cos S S bc, 3. tn S bs c SS PROOF: (1) We know tht sin 1 cos sin 1 - b c bc ( Using cosine rule for )

59 sin bc b c bc b c bc bc b c b c bc b c bc b c b c bc S b S c bc Since + b + c s + b s - c + c s - c sin s bs c bc ( Divide by ) sin ± S b S c bc If is cute, then sin is lwys positive. sin S bs c bc

60 b. sin 1 + cos 1 + c bc ( Using cosine rule for ) sin bc b c bc b c bc sin b c b c bc s s bc Using + b + c s + b s c Dividing by, we get cos s s bc ± S S bc Since is cute, cos is lwys positive nd therefore, cos S S bc 3. tn sin cos s bs c bc ss bc s bs c ss Similrly, we cn show tht

61 sin s s c c, cos s s b c, tn s s c ss b sin s s b c, cos s s c c, tn s s b ss c WORKED EXMPLES 1. If 3, b 4, c 5, in tringle C, find the vlue of.) sin C b.) sin 4C + cos 4C SOLUTION: Since, c b + is stisfied by the given sides, they form right ngled tringle C 90 0, sin C 90 0 sin sin nd, sin 4C + cos 4C sin( 4 X 90 0 ) + cos( 4 X 90 0 ) sin cos

62 . Prove tht sin ( C ) + b sin ( C ) + C sin ( ) 0. SOLUTION: Now, sin( C ) R sin. sin( C ) ( Since, R sin ) R sin.sin( C ) Since + + C 180 0, + C sin( + C ) sin R sin ( + C ) sin ( C ) R sin [ sin sin C ] Similrly, b sin ( C ) R [ sin C sin ] C sin ( ) R [ sin sin ] L. H. S. sin ( C ) + b sin ( C ) + C sin ( ) R [ sin sin C ] + R [ sin C sin ] + R [ sin sin ] R [ sin sin C + sin C sin + sin sin ] 3. Prove tht, in ΔC, sin C sin C b c SOLUTION:

63 L. H. S. sin C sin C X sin C sin C sin sin sin c C Since sin ( + C ) sin ( - C ) sin sin C sin sin sin b C 4R 4R 4R b c 4R 4R C sin( + ) sin C in ΔC ( using sine rule ) b c R. H. S. 4. Prove tht (b cos C c cos ) b - c SOLUTION: L. H. S. ( b cos C c cos ) b cos C c cos

64 b b bc - c c b bc (Using cosine rule) b bc - c b bc b c b - c R. H. S. b c 5. Prove tht c sin + b b sin + c sinc 0 SOLUTION: b c Now b c sin X sin cos (Since, sin sin cos ) b c X R X b c bc (Using sine rule nd cosine rule) b c b c 4R. bc c Similrly, b c c b sin 4R. bc b c b b c sinc 4R. bc

65 L. H. S R bc c b c b. 4 + R bc b c c. 4 + R bc c b b. 4 4R.bc 1 [ b 4 c 4 ( b c ) + c 4 4 ( c ) b + 4 b 4 ( b ) c ] 4R.bc 1 [ 0 ] R. H. S 6. Find the gretest side of the tringle, whose sides re x + x + 1, x + 1, x 1. SOLUTION: Let x + x + 1, b x + 1, c x 1 Then, is the gretest side. Therefore is the gretest ngle. cos bc c b x x x x x x x x x x x x x x x x x x cos x x x x x x 1 - cos60 0 cos( ) 1 cos Therefore, the gretest ngle is 10 0

66 7. If sin + sin sin C in Δ C, Prove tht either 90 0 or SOLUTION: sin + sin sin C sin.cos sin C Using sin C + sin D C D sin C D. cos sin ( + ). cos ( ) sin C cos C sin ( + ) Sin( C ) sin C sin C cos ( ) sin C cos C Dividing by sin C both sides, we get, Cos( ) cos C lso, Cos( ) cos -C [ Since, cos( -C ) cos C ] ± C When C, + C ut, + + C 180 0, gives , i.e., 180 0, 90 0 When - C, + C + + C 180 0, gives , i.e., 180 0, 90 0 Therefore, tringle is right ngled tringle.

67 cos cos 1 8. Prove tht - b - 1 b SOLUTION: cos cos L. H. S - b (Using cos 1 sin ) 1 sin 1 sin b - 1 sin b sin b b - sin + sin 1-1 b - 1 R + 1 R Since, sin 1 R sin b 1 R 1-1 b R. H. S SUMMRY ND KEY POINTS 1.) ) cute ngle: ngles less thn b) Obtuse ngle: ngles greter thn 90 0 but less thn c) Stright ngle: ngles equl to

68 d) Reflex ngle: ngles greter thn but less thn ) COMPLEMENTRY NGLES: Two ngles re complementry if their sum dd up to ) SUPPLEMENTRY NGLES: Two ngles re supplementry if their sum dd up to ) The sum of DJCENT NGLES on stright line is equl to ) ngles t point dd up to e b d c + b + c + d + e ) Verticlly opposite ngles re equl. 7.) ngles formed by prllel lines cut by trnsversl Line, l. ) CORRESPONDING NGLES re equl. l b (corr. s, // CD ) b C D

69 b) LTERNTE NGLES re equl. l c d c (lt. s, // CD ) d C D c) INTERIOR NGLES re supplementry. l e f e ( int. s, // CD ) f C D 8.) NGLE PROPERTIES OF TRINGLES:.) The sum of the 3 ngles of tringle is equl to x + y + z x ( sum of Δ ) y z b.) The exterior ngle of tringle is equl to the sum of the interior opposite ngles.

70 x y + z y ( ext. of Δ ) z x c.)n Isosceles Tringle hs equl ngles opposite the equl sides. x y ( bse s of isos. Δ ) x y d.)n Equilterl Tringle hs 3 equl sides nd 3 equl ngles, ech equl to x + y + z 60 0 x ( of equi. Δ ) y z

71 9.) NGLE PROPERTIES OF QUDRILTERLS:.) The sum of ll the ngles in qudrilterl is b + c + d ( sum of qud. ) d b c b.) The properties of some specil qudrilterls re s follows: i) Trpezium: one pir of prllel opposite sides. ii) Isosceles Trpezium: one pir of prllel opposite sides. Non prllel sides re equl in length. iii)prllelogrm: Two pirs of prllel opposite sides. Opposite sides re equl in length. Opposite ngles re equl. iv)rectngle: Two pirs of prllel opposite sides. Opposite sides re equl in length. ll four ngles re right ngles(90 0 ). Digonls re equl in length. Digonls bisect ech other.

72 V) Rhombus: Two pirs of prllel opposite sides. Four equl sides. Opposite ngles re equl. Digonls bisect ech other t right ngles. Digonls bisect the interior ngles. vi)squre: Two pirs of prllel opposite sides. Four equl sides. ll four ngles re right ngles. (90 0 ) Digonls re equl in length. Digonls bisect ech other t right ngles. Digonls bisect the interior ngles. vii) Kite: No prllel sides. Two pirs of equl djcent sides. One pir of equl opposite ngles. Digonls intersect t right ngles. One digonls bisect the interior ngles. Key Points: 1.) rectngle with 4 equl sides is squre..) Prllelogrm with 4 right ngles is rectngle. 3.) prllelogrm with 4 equl sides is rhombus.

73 4.) rhombus with 4 equl ngles is squre.

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