Page 1 Q u e s t i o n : 3 0 In ?ABC, if 3 ?A = 4 ?B = 6 ?C then A : B : C = ? a 3 : 4 : 6 b 4 : 3 : 2 c Page 2 Q u e s t i o n : 3 0 In ?ABC, if 3 ?A = 4 ?B = 6 ?C then A : B : C = ? a 3 : 4 : 6 b 4 : 3 : 2 c 2 : 3 : 4 d 6 : 4 : 3 S o l u t i o n : LCM of 3, 4 and 6 = 12 3 ?A = 4 ?B = 6 ?C Given Dividing throughout by 12, we get 3 ?A 12 = 4 ?B 12 = 6 ?C 12 ? ?A 4 = ?B 3 = ?C 2 Let ?A 4 = ?B 3 = ?C 2 = k , where k is some constant Then, ?A = 4k, ?B = 3k, ?C = 2k ? ?A : ?B : ?C = 4k : 3k : 2k = 4 : 3 : 2 Hence, the correct answer is option b . Q u e s t i o n : 3 1 In a ?ABC, if ?A - ?B = 42° and ?B - ?C = 21° then ?B = ? a 32° b 63° c 53° d 95° Figure S o l u t i o n : c 53° Let ?A - ?B = 42° . . . (i) and ?B - ?C = 21° . . . (ii)Adding (i) and (ii), we get: ?A - ?C = 63° ?B = ?A -42° [Using (i)] ?C = ?A -63° [Using (iii)] ? ?A + ?B + ?C = 180° [Sum of the angles of a triangle] ? ?A + ?A -42°+ ?A -63° = 180° ? 3 ?A -105° = 180° ? 3 ?A = 285° ? ?A = 95° ? ?B = (95 -42)° ? ?B = 53° Q u e s t i o n : 3 2 In ?ABC, side BC is produced to D. If ?ABC = 50° and ?ACD = 110° then ?A = ? a 160° b 60° c 80° d 30° S o l u t i o n : ? ?A + ?B = ?ACD ? ?A +50° = 110° ? ?A = 60° Hence, the correct answer is option b . Q u e s t i o n : 3 3 Side BC of ?ABC has been produced to D on left and to E on right hand side of BC such that ?ABD = 125° and ?ACE = 130°. Then ?A = ? a 50° b 55° c 65° d 75° S o l u t i o n : d 75° We have : ? ?ABD + ?ABC = 180° [ ? DE is a straight line] ? 125°+ ?ABC = 180° ? ?ABC = 55° Also, Page 3 Q u e s t i o n : 3 0 In ?ABC, if 3 ?A = 4 ?B = 6 ?C then A : B : C = ? a 3 : 4 : 6 b 4 : 3 : 2 c 2 : 3 : 4 d 6 : 4 : 3 S o l u t i o n : LCM of 3, 4 and 6 = 12 3 ?A = 4 ?B = 6 ?C Given Dividing throughout by 12, we get 3 ?A 12 = 4 ?B 12 = 6 ?C 12 ? ?A 4 = ?B 3 = ?C 2 Let ?A 4 = ?B 3 = ?C 2 = k , where k is some constant Then, ?A = 4k, ?B = 3k, ?C = 2k ? ?A : ?B : ?C = 4k : 3k : 2k = 4 : 3 : 2 Hence, the correct answer is option b . Q u e s t i o n : 3 1 In a ?ABC, if ?A - ?B = 42° and ?B - ?C = 21° then ?B = ? a 32° b 63° c 53° d 95° Figure S o l u t i o n : c 53° Let ?A - ?B = 42° . . . (i) and ?B - ?C = 21° . . . (ii)Adding (i) and (ii), we get: ?A - ?C = 63° ?B = ?A -42° [Using (i)] ?C = ?A -63° [Using (iii)] ? ?A + ?B + ?C = 180° [Sum of the angles of a triangle] ? ?A + ?A -42°+ ?A -63° = 180° ? 3 ?A -105° = 180° ? 3 ?A = 285° ? ?A = 95° ? ?B = (95 -42)° ? ?B = 53° Q u e s t i o n : 3 2 In ?ABC, side BC is produced to D. If ?ABC = 50° and ?ACD = 110° then ?A = ? a 160° b 60° c 80° d 30° S o l u t i o n : ? ?A + ?B = ?ACD ? ?A +50° = 110° ? ?A = 60° Hence, the correct answer is option b . Q u e s t i o n : 3 3 Side BC of ?ABC has been produced to D on left and to E on right hand side of BC such that ?ABD = 125° and ?ACE = 130°. Then ?A = ? a 50° b 55° c 65° d 75° S o l u t i o n : d 75° We have : ? ?ABD + ?ABC = 180° [ ? DE is a straight line] ? 125°+ ?ABC = 180° ? ?ABC = 55° Also, ?ACE + ?ACB = 180° ? 130°+ ?ACB = 180° ? ?ACB = 50° ? ?BAC + ?ABC + ?ACB = 180° [Sum of the angles of a triangle] ? ?BAC +55°+50° = 180° ? ?BAC = 75° ? ?A = 75° Q u e s t i o n : 3 4 In the given figure, the sides CB and BA of ?ABC have been produced to D and E, respectively, such that ?ABD = 110° and ?CAE = 135°. Then ?ACB = ? a 65° b 45° c 55° d 35° S o l u t i o n : a 65° We have : ?ABD + ?ABC = 180° [ ? CBD is a straight line] ? 100°+ ?ABC = 180° ? ?ABC = 70° Side AB of triangle ABC is produced to E. ? ?CAE = ?ABC + ?ACB ? 135° = 70°+ ?ACB ? ?ACB = 65° Q u e s t i o n : 3 5 The sides BC, CA and AB of ?ABC have been produced to D, E and F, respectively. ?BAE + ?CBF + ?ACD = ? a 240° b 300° c 320° d 360° S o l u t i o n : d 360° We have : ?1 + ?BAE = 180° . . . (i) ?2 + ?CBF = 180° . . . (ii) and ?3 + ?ACD = 180° . . . (iii) Adding (i), (ii) and (iii), we get:( ?1 + ?2 + ?3)+( ?BAE + ?CBF+ ?ACD) = 540° ? 180°+ ?BAE + ?CBF+ ?ACD = 540° [ ? ?1 + ?2 + ?3 = 180°] ? ?BAE + ?CBF+ ?ACD = 360° Q u e s t i o n : 3 6 The the given figure, EAD ? BCD. Ray FAC cuts ray EAD at a point A such that ?EAF = 30°. Also, in ?BAC, ?BAC = x° and ?ABC = (x + 10)°. Then, the value of x is a 20 b 25 c 30 d 35 S o l u t i o n : In the given figure, ?CAD = ?EAF Verticallyoppositeangles ? ?CAD = 30° In ?ABD, ?ABD + ?BAD + ?ADB = 180° Anglesumproperty ? (x + 10)° + (x° + 30°) + 90° = 180° ? 2x° + 130° = 180° ? 2x° = 180° - 130° = 50° ? x = 25 Thus, the value of x is 25. Page 4 Q u e s t i o n : 3 0 In ?ABC, if 3 ?A = 4 ?B = 6 ?C then A : B : C = ? a 3 : 4 : 6 b 4 : 3 : 2 c 2 : 3 : 4 d 6 : 4 : 3 S o l u t i o n : LCM of 3, 4 and 6 = 12 3 ?A = 4 ?B = 6 ?C Given Dividing throughout by 12, we get 3 ?A 12 = 4 ?B 12 = 6 ?C 12 ? ?A 4 = ?B 3 = ?C 2 Let ?A 4 = ?B 3 = ?C 2 = k , where k is some constant Then, ?A = 4k, ?B = 3k, ?C = 2k ? ?A : ?B : ?C = 4k : 3k : 2k = 4 : 3 : 2 Hence, the correct answer is option b . Q u e s t i o n : 3 1 In a ?ABC, if ?A - ?B = 42° and ?B - ?C = 21° then ?B = ? a 32° b 63° c 53° d 95° Figure S o l u t i o n : c 53° Let ?A - ?B = 42° . . . (i) and ?B - ?C = 21° . . . (ii)Adding (i) and (ii), we get: ?A - ?C = 63° ?B = ?A -42° [Using (i)] ?C = ?A -63° [Using (iii)] ? ?A + ?B + ?C = 180° [Sum of the angles of a triangle] ? ?A + ?A -42°+ ?A -63° = 180° ? 3 ?A -105° = 180° ? 3 ?A = 285° ? ?A = 95° ? ?B = (95 -42)° ? ?B = 53° Q u e s t i o n : 3 2 In ?ABC, side BC is produced to D. If ?ABC = 50° and ?ACD = 110° then ?A = ? a 160° b 60° c 80° d 30° S o l u t i o n : ? ?A + ?B = ?ACD ? ?A +50° = 110° ? ?A = 60° Hence, the correct answer is option b . Q u e s t i o n : 3 3 Side BC of ?ABC has been produced to D on left and to E on right hand side of BC such that ?ABD = 125° and ?ACE = 130°. Then ?A = ? a 50° b 55° c 65° d 75° S o l u t i o n : d 75° We have : ? ?ABD + ?ABC = 180° [ ? DE is a straight line] ? 125°+ ?ABC = 180° ? ?ABC = 55° Also, ?ACE + ?ACB = 180° ? 130°+ ?ACB = 180° ? ?ACB = 50° ? ?BAC + ?ABC + ?ACB = 180° [Sum of the angles of a triangle] ? ?BAC +55°+50° = 180° ? ?BAC = 75° ? ?A = 75° Q u e s t i o n : 3 4 In the given figure, the sides CB and BA of ?ABC have been produced to D and E, respectively, such that ?ABD = 110° and ?CAE = 135°. Then ?ACB = ? a 65° b 45° c 55° d 35° S o l u t i o n : a 65° We have : ?ABD + ?ABC = 180° [ ? CBD is a straight line] ? 100°+ ?ABC = 180° ? ?ABC = 70° Side AB of triangle ABC is produced to E. ? ?CAE = ?ABC + ?ACB ? 135° = 70°+ ?ACB ? ?ACB = 65° Q u e s t i o n : 3 5 The sides BC, CA and AB of ?ABC have been produced to D, E and F, respectively. ?BAE + ?CBF + ?ACD = ? a 240° b 300° c 320° d 360° S o l u t i o n : d 360° We have : ?1 + ?BAE = 180° . . . (i) ?2 + ?CBF = 180° . . . (ii) and ?3 + ?ACD = 180° . . . (iii) Adding (i), (ii) and (iii), we get:( ?1 + ?2 + ?3)+( ?BAE + ?CBF+ ?ACD) = 540° ? 180°+ ?BAE + ?CBF+ ?ACD = 540° [ ? ?1 + ?2 + ?3 = 180°] ? ?BAE + ?CBF+ ?ACD = 360° Q u e s t i o n : 3 6 The the given figure, EAD ? BCD. Ray FAC cuts ray EAD at a point A such that ?EAF = 30°. Also, in ?BAC, ?BAC = x° and ?ABC = (x + 10)°. Then, the value of x is a 20 b 25 c 30 d 35 S o l u t i o n : In the given figure, ?CAD = ?EAF Verticallyoppositeangles ? ?CAD = 30° In ?ABD, ?ABD + ?BAD + ?ADB = 180° Anglesumproperty ? (x + 10)° + (x° + 30°) + 90° = 180° ? 2x° + 130° = 180° ? 2x° = 180° - 130° = 50° ? x = 25 Thus, the value of x is 25. Hence, the correct answer is option b . Q u e s t i o n : 3 7 In the given figure, two rays BD and CE intersect at a point A. The side BC of ?ABC have been produced on both sides to points F and G respectively. If ?ABF = x°, ?ACG = y° and ?DAE = z° then z = ? a x + y – 180 b x + y + 180 c 180 – (x + y) d x + y + 360° S o l u t i o n : In the given figure, ?ABF + ?ABC = 180° Linearpairofangles ? x° + ?ABC = 180° ? ?ABC = 180° - x° .....1 Also, ?ACG + ?ACB = 180° Linearpairofangles ? y° + ?ACB = 180° ? ?ACB = 180° - y° .....2 Also, ?BAC = ?DAE = z° .....3 Verticallyoppositeangles In ?ABC, ?BAC + ?ABC + ?ACB = 180° Anglesumproperty ? z° + 180° - x° + 180° - y° = 180° Using(1), (2)and(3) ? z = x + y - 180 Hence, the correct answer is option a . Q u e s t i o n : 3 8 In the given figure, lines AB and CD intersect at a point O. The sides CA and OB have been produced to E and F respectively such that ?OAE = x° and ?DBF = y°. If ?OCA = 80°, ?COA = 40° and ?BDO = 70° then x° + y° = ? a 190° b 230° c 210° d 270° S o l u t i o n : In the given figure, ?BOD = ?COA Verticallyoppositeangles ? ?BOD = 40° .....1 In ?ACO, ?OAE = ?OCA + ?COA Exteriorangleofatriangleisequaltothesumoftwooppositeinteriorangles ? x° = 80° + 40° = 120° .....2 In ?BDO, Page 5 Q u e s t i o n : 3 0 In ?ABC, if 3 ?A = 4 ?B = 6 ?C then A : B : C = ? a 3 : 4 : 6 b 4 : 3 : 2 c 2 : 3 : 4 d 6 : 4 : 3 S o l u t i o n : LCM of 3, 4 and 6 = 12 3 ?A = 4 ?B = 6 ?C Given Dividing throughout by 12, we get 3 ?A 12 = 4 ?B 12 = 6 ?C 12 ? ?A 4 = ?B 3 = ?C 2 Let ?A 4 = ?B 3 = ?C 2 = k , where k is some constant Then, ?A = 4k, ?B = 3k, ?C = 2k ? ?A : ?B : ?C = 4k : 3k : 2k = 4 : 3 : 2 Hence, the correct answer is option b . Q u e s t i o n : 3 1 In a ?ABC, if ?A - ?B = 42° and ?B - ?C = 21° then ?B = ? a 32° b 63° c 53° d 95° Figure S o l u t i o n : c 53° Let ?A - ?B = 42° . . . (i) and ?B - ?C = 21° . . . (ii)Adding (i) and (ii), we get: ?A - ?C = 63° ?B = ?A -42° [Using (i)] ?C = ?A -63° [Using (iii)] ? ?A + ?B + ?C = 180° [Sum of the angles of a triangle] ? ?A + ?A -42°+ ?A -63° = 180° ? 3 ?A -105° = 180° ? 3 ?A = 285° ? ?A = 95° ? ?B = (95 -42)° ? ?B = 53° Q u e s t i o n : 3 2 In ?ABC, side BC is produced to D. If ?ABC = 50° and ?ACD = 110° then ?A = ? a 160° b 60° c 80° d 30° S o l u t i o n : ? ?A + ?B = ?ACD ? ?A +50° = 110° ? ?A = 60° Hence, the correct answer is option b . Q u e s t i o n : 3 3 Side BC of ?ABC has been produced to D on left and to E on right hand side of BC such that ?ABD = 125° and ?ACE = 130°. Then ?A = ? a 50° b 55° c 65° d 75° S o l u t i o n : d 75° We have : ? ?ABD + ?ABC = 180° [ ? DE is a straight line] ? 125°+ ?ABC = 180° ? ?ABC = 55° Also, ?ACE + ?ACB = 180° ? 130°+ ?ACB = 180° ? ?ACB = 50° ? ?BAC + ?ABC + ?ACB = 180° [Sum of the angles of a triangle] ? ?BAC +55°+50° = 180° ? ?BAC = 75° ? ?A = 75° Q u e s t i o n : 3 4 In the given figure, the sides CB and BA of ?ABC have been produced to D and E, respectively, such that ?ABD = 110° and ?CAE = 135°. Then ?ACB = ? a 65° b 45° c 55° d 35° S o l u t i o n : a 65° We have : ?ABD + ?ABC = 180° [ ? CBD is a straight line] ? 100°+ ?ABC = 180° ? ?ABC = 70° Side AB of triangle ABC is produced to E. ? ?CAE = ?ABC + ?ACB ? 135° = 70°+ ?ACB ? ?ACB = 65° Q u e s t i o n : 3 5 The sides BC, CA and AB of ?ABC have been produced to D, E and F, respectively. ?BAE + ?CBF + ?ACD = ? a 240° b 300° c 320° d 360° S o l u t i o n : d 360° We have : ?1 + ?BAE = 180° . . . (i) ?2 + ?CBF = 180° . . . (ii) and ?3 + ?ACD = 180° . . . (iii) Adding (i), (ii) and (iii), we get:( ?1 + ?2 + ?3)+( ?BAE + ?CBF+ ?ACD) = 540° ? 180°+ ?BAE + ?CBF+ ?ACD = 540° [ ? ?1 + ?2 + ?3 = 180°] ? ?BAE + ?CBF+ ?ACD = 360° Q u e s t i o n : 3 6 The the given figure, EAD ? BCD. Ray FAC cuts ray EAD at a point A such that ?EAF = 30°. Also, in ?BAC, ?BAC = x° and ?ABC = (x + 10)°. Then, the value of x is a 20 b 25 c 30 d 35 S o l u t i o n : In the given figure, ?CAD = ?EAF Verticallyoppositeangles ? ?CAD = 30° In ?ABD, ?ABD + ?BAD + ?ADB = 180° Anglesumproperty ? (x + 10)° + (x° + 30°) + 90° = 180° ? 2x° + 130° = 180° ? 2x° = 180° - 130° = 50° ? x = 25 Thus, the value of x is 25. Hence, the correct answer is option b . Q u e s t i o n : 3 7 In the given figure, two rays BD and CE intersect at a point A. The side BC of ?ABC have been produced on both sides to points F and G respectively. If ?ABF = x°, ?ACG = y° and ?DAE = z° then z = ? a x + y – 180 b x + y + 180 c 180 – (x + y) d x + y + 360° S o l u t i o n : In the given figure, ?ABF + ?ABC = 180° Linearpairofangles ? x° + ?ABC = 180° ? ?ABC = 180° - x° .....1 Also, ?ACG + ?ACB = 180° Linearpairofangles ? y° + ?ACB = 180° ? ?ACB = 180° - y° .....2 Also, ?BAC = ?DAE = z° .....3 Verticallyoppositeangles In ?ABC, ?BAC + ?ABC + ?ACB = 180° Anglesumproperty ? z° + 180° - x° + 180° - y° = 180° Using(1), (2)and(3) ? z = x + y - 180 Hence, the correct answer is option a . Q u e s t i o n : 3 8 In the given figure, lines AB and CD intersect at a point O. The sides CA and OB have been produced to E and F respectively such that ?OAE = x° and ?DBF = y°. If ?OCA = 80°, ?COA = 40° and ?BDO = 70° then x° + y° = ? a 190° b 230° c 210° d 270° S o l u t i o n : In the given figure, ?BOD = ?COA Verticallyoppositeangles ? ?BOD = 40° .....1 In ?ACO, ?OAE = ?OCA + ?COA Exteriorangleofatriangleisequaltothesumoftwooppositeinteriorangles ? x° = 80° + 40° = 120° .....2 In ?BDO, ?DBF = ?BDO + ?BOD Exteriorangleofatriangleisequaltothesumoftwooppositeinteriorangles ? y° = 70° + 40° = 110° Using(1) .....3 Adding 2 and 3 , we get x° + y° = 120° + 110° = 230° Hence, the correct answer is option b . Q u e s t i o n : 3 9 In a ?ABC, it is given that ?A : ?B : ?C = 3 : 2 : 1 and ?ACD = 90°. If BC is produced to E then ?ECD = ? a 60° b 50° c 40° d 25° S o l u t i o n : Let ?A = (3x)°, ?B = (2x)° and ?C = x° Then, 3x +2x +x = 180° [Sum of the angles of a triangle] ? 6x = 180° ? x = 30° Hence, the angles are ?A = 3 ×30° = 90°, ?B = 2 ×30° = 60° and ?C = 30° Side BC of triangle ABC is produced to E. ? ?ACE = ?A + ?B ? ?ACD + ?ECD = 90°+60° ? 90°+ ?ECD = 150° ? ?ECD = 60° Hence, the correct answer is option a . Q u e s t i o n : 4 0 In the given figure, BO and CO are the bisectors of ?B and ?C respectively. If ?A = 50° then ?BOC = ? a 130° b 100° c 115° d 120° S o l u t i o n : c 115° In ? ABC , we have: ?A + ?B + ?C = 180° [Sum of the angles of a triangle] ? 50°+ ?B + ?C = 180° ? ?B + ?C = 130° ? 1 2 ?B + 1 2 ?C = 65° . . . (i) In ? OBC , we have: ?OBC + ?OCB + ?BOC = 180° ? 1 2 ?B + 1 2 ?C + ?BOC = 180° [Using (i)] ? 65°+ ?BOC = 180° ? ?BOC = 115° Q u e s t i o n : 4 1 In the given figure, side BC of ?ABC has been produced to a point D. If ?A = 3y°, ?B = x°, ?C = 5y° and ?CBD = 7y°. Then, the value of x is a 60 b 50 c 45 d 35 S o l u t i o n : D i s c l a i m e r: In the question ?ACD should be 7y°. In the given figure, ?ACB + ?ACD = 180° Linearpairofangles ? 5y° + 7y° = 180°Read More

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