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CHAPTER 6 J A LINES AND ANGLES Points to Remember : 1. If a ray stands on a line, then the sum of the two adjacent angles so formed is 180° and vice-versa. This property is known as the Linear Pair Axiom. 2. If two lines intersect each other, then the vertically opposite angles are equal. 3. If a transversal intersects two parallel lines, then (a) each pair of corresponding angles is equal. (b) each pair of alternate interior angles is equal. (c) each pair of interior angles on the same side of the transversal is supplementary. 4. If a transversal intersects two line such that, either (a) any one pair of corresponding angles is equal, or (b) any one pair of alternate interior angles is equal, or (c) any one pair of interior angles on the same side of of the transversal is supplementary, then the lines are parallel. 5. Two intersecting lines cannot both be parallel to the same line. 6. Lines which are parallel to a given line are parallel to each other. 7. The sum of three angles of a triangle is 180°. 8. If a side of a triangle is produced, the exterior angle so formed is equal to the sum of the two interior opposite angles. T I B J A ILLUSTRATIVE EXAMPLES Example 1. If two lines intersects, prove that vertically opposite angles are equal. Solution. Given : Two lines AB and CD intersects at O. To prove : 1 = 3 and 2 = 4 Proof : OB stands on line COD. 1 + 2 = 180° ( linear pair) ...(1) Also, OD stands on line AOB 2 + 3 = 180° ( linear pair) ...(2) from (1) and (2) 1 + 2 = 2 + 3 1 = 3 Similarly, 2 = 4 Hence proved M A Example 2. The complement of an angle is half of itself. Find the angle and its complement. Solution. Let the given angle be x. Then, its complement is 90° – x. According to given question, 90° – x = 1 x 2 MATHEMATICS–IX x x 90 2 3x 2 90 90 x 2 3 LINES AND ANGLES 55 x = 60° Hence, the given angle is 60° and its complement is 90° – 60° i.e. 30°. Example 3. Two supplementary angles are in the ratio 3 : 2. Find the angles. Solution. Let the angles be 3x and 2x. according to given question, 3x + 2x = 180° ( angles are supplementary) x J A 180º 36º 5 5x = 180° Angles are 3 x 3 36º 108 Ans. and 2 x 2 36º 72 J A Example 4. What value of x would make AOB a line in given figure if AOC = (3x –10)° and BOC = (7x + 30)°? Solution. B Since, OB and OA are opposite rays BOC + AOC = 180º 7x + 30° + 3x – 10° = 180° 10x + 20° = 180° 10x = 180° – 20° 10x = 160º x T I 160 10 x 16 Ans. Example 5. Rays OA, OB, OC, OD and OE have the common initial point O. Show that : AOB + BOC + COD + DOE + EOA = 360° Solution. Let us draw a ray OF oposite to ray OA. M A B C 3 4 F D 2 0 5 1 6 A E so, 1 + 2 + 3 = 180° and 4 + 5 + 6 = 180° ( linear pair axiom) adding (1) and (2), (1 + 2 + 3) + (4 + 5 + 6) = 180° + 180° 1 + 2 + (3+ 4) + 5 + 6 = 360° AOB + BOC + COD + DOE + EOA = 360° Hence shown. 56 LINES AND ANGLES ...(1) ...(2) MATHEMATICS–IX Example 6. Prove that if a transversal intersects two parallel lines, then each pair of interior angle on the same side of the transversal is supplementary. Solution. Given : A transversal l intersects two parallel lines AB and CD at P and Q. J A J A 4 + 5 = 180° and 3 + 6 = 180º Proof : Ray QD stands on line l 1 + 4 = 180° ...(1) ( linear pair axiom) also, 1 = 5 ...(2) ( corresponding angles) from (1) and (2) 4 + 5 = 180° Again, ray QC stands on line l 2 + 3 = 180° ...(3) ( linear pair axiom) and, 2 = 6 ...(4) ( corresponding angles) from (3) and (4) 3 + 6 = 180° Hence proved. Example 7. If two parallel lines are intersected by a transversal, prove that the bisectors of the two pairs of interior angles encloses a rectangle. Solution. Given : Two parallel lines AB and CD are intersected by a transversal at P and R respectively. PQ, RQ, RS and PS are bisectors of APR, PRC, PRD and BPR respectively. To prove : T I M A B To prove : PQRS is a rectangle Proof : Since AB || CD and l is a transversal APR = PRD ( alt. interior angles) 1 1 APR = PRD QPR = PRS 2 2 But, these are alternate interior angles. PQ || RS. Similarly QR || PS. PQRS is a parallelogram. MATHEMATICS–IX LINES AND ANGLES 57 Now, ray PR stands on AB APR + BPR = 180° ( linear pair) J A 1 1 APR + BPR = 90° 2 2 QPR + SPR = 90° QPS = 90° Thus, PQRS is a parallelogram, one of whose angle is 90° PQRS is a rectangle. Hence proved. Example 8. In the given figure lines XY and MN intersect at O. If POY = 90° and a : b = 2 : 3, find c. B J A —NCERT Since XY is a line a + b + 90° = 180° ( linear pair) a + b = 180° – 90° a + b = 90° But a : b = 2 : 3. Let a = 2x and b = 3x. 2x + 3x = 90° 5x = 90° x = 18° a = 2x = 2 × 18° = 36° and b = 3x = 3 × 18° = 54° Now, OM and ON are opposite rays. MON is a line. Since ray OX stands on MN MOX + XON = 180° ( linear pair) c + b = 180° c + 54° = 180° C = 180° – 154° = 126° c = 126° Ans. Example 9. It is given that XYZ = 64° and XY is produced to point P. Draw a figure from the given information. If ray YQ bisects ZYP, find XYQ and reflex QYP. —NCERT Solution. Since XY is produced to point P, XP is a straight line. Since, YZ stands on XP, XYZ + ZYP = 180° ( linear pair) 64° + ZYP =1 80° ZYP = 180° – 64° = 116° Since ray YQ bisects ZYP Solution. T I M A 116 58 2 Now, XYQ = XYZ + ZYQ XYQ = 64° + 58° = 122° and reflex QYP = 360° – QYP = 360° – 58° = 302° Ans. 58 QYP ZYQ LINES AND ANGLES MATHEMATICS–IX Example 10. In the given figure, if AB || CD, CD || EF and y : z = 3 : 7, find x. —NCERT J A Since, AB || CD and CD || EF AB || CD || EF. Now, CD || EF and PR is a transversal. PQD = PRF 180° – y = z y + z = 180° also, y : z = 3 : 7. Let y = 3a and z = 7a 3a + 7a = 180° 10a = 180° a = 18° y = 3 × 18° = 54° and z = 7 × 18° = 126° Now, AB || CD and PQ is a transversal. x + y = 180° ( consecutive interior angles are supplementary) x + 54° = 180° x = 180° – 54° = 126° x = 126° Ans. Example 11. In the given figure, X = 62°, XYZ = 54°. If YO and ZO are the bisectors of XYZ and XZY respectively of XYZ, find OZY and YOZ. —NCERT Solution. T I M A Solution. B J A In XYZ, YXZ + XYZ + XZY = 180° ( angle sum property of a triangle) 62° + 54° + XZY = 180° XZY = 180° – 62° – 54° = 64° Since, YO and ZO are bisectors of XYZ and XZY, OYZ 1 1 XYZ 54 27 2 2 1 1 XZY 64 32 2 2 In OYZ, we have YOZ + OYZ + OZY = 180° ( angle sum property of a triangle) YOZ + 27° + 32° = 180° YOZ = 180° – 27° – 32° = 121° Hence, OZY = 32° and YOZ = 121° Ans. and, OZY MATHEMATICS–IX LINES AND ANGLES 59 Example 12. In the given figure, if PQ PS, PQ || SR, SQR = 28° and QRT = 65°, then find the values of x and y. —NCERT J A J A QRT = RQS + QSR ( Exterior angle property in SRQ) 65° = 28° + QSR QSR = 65° – 28° = 37° Since, PQ || SR and the transversal PS intersects then at P and S respectively. PSR + SPQ = 180° ( sum of consecutive interior angles is 180°) (PSQ + QSR) + 90° = 180° y + 37° + 90° = 180° y = 180° – 37° – 90° = 53° Now, in the right SPQ, we have PQS + PSQ = 90° x + 53°= 90° x = 90° – 53° = 37° Hence, x = 37° and y = 53° Ans. Example 13. In each of the following figures, AB || CD, find x. Solution. B T I A B 50° x° O M A C (i) Solution. 60 25° D (i) Through O, Draw EOF || AB || CD. then, 1 + 2 = x Now, EO || AB and BO is the transversal. 1 + 50º = 180º ( Interior angles on the same side of the transversal) A B 50° E 1 O 2 C F 25° D LINES AND ANGLES MATHEMATICS–IX 1 = 180° – 50° 1 = 130° Again, EO || CD and OD is a transversal 2 + 25° = 180° 2 = 180° – 25° 2 = 155° adding (1) and (2) 1 + 2 = 130° + 155° = 285° ...(1) J A ...(2) i.e. x 285º Ans. (ii) Produce AB to intersect CE and F. A 105° B 125° G F x° E Now, ABE + EBF = 180° J A D C ( Linear pair) 125° + EBF = 180° EBF = 180° – 125° = 55° Again, CD || FG and CF is a transversal DCF + CFG = 180° ( Interior angles on same side of transversal) 105° + CFG = 180º CFG = 180° – 105° = 75° T I Also, BFE = CFG = 75° B ( vertically opposite angles) Now, In BEF, BEF + EBF + BFE = 180° ( angle sum property) x° + 55° + 75° = 180° xº + 130° = 180° x = 180º – 130° M A x 50 Ans. Example 14. The side QR of PQR is produced to a point S. If the bisectors of PQR and PRS meet at point 1 T, then prove that QTR = QPR. —NCERT 2 Solution. Side QR of PQR is produced to S, Ext. PRS =P + Q ( Exterior angle sum property of a triangle) 1 1 1 Ext PRS = P + Q 2 2 2 1 P + 1 ...(1) 2 Again, In QRT., Ext. TRS = T + 1 ( same as above) 2 = T + 1 ...(2) T P 2 = MATHEMATICS–IX 2 1 1 Q LINES AND ANGLES 2 R S 61 Equating (1) and (2), we get 1 P + 1 = T + 1 2 1 1 T = P or QTR = QPR 2 2 Hence Proved. Example 15. Bisectors of exterior angles of ABC (obtained by producing sides AB and AC) meet at O. Prove 1 that BOC = 90° – A. 2 Solution. Given : ABC in which bisectors of exterior angles meet at O (as shown) 1 To prove : BOC = 90 – A 2 Proof : In OBC 1 + 2 + 0 = 180º ( angle sum property of a triangle) 1 1 CBD + BCE + O = 180° 2 2 CBD + BCE + 2O = 360° (180º – B) + (180º – C) + 2O = 360° B + C = 2O 180 – A = 2O 1 O = (180 – A) 2 1 BOC = 90° – A. Hence proved. 2 T I B PRACTICE EXERCISE 1. 2. 3. 4. 5. 6. J A J A If an angle is 24° less than its complement. find its measure. An angle is 40° less than one-third of its supplement. Find the angle and its supplement. Two supplementary angles are in the ratio 11 : 7. Find them. Find the angle whose supplement is four times its complement. Find the measure of an angle if, three times its supplement is 60° more than six times its complement. In the following figure, it is given that 2a – 5b = 10º find a and b. M A a A b B 7. In the following figures, what value of x will make AOB a straight line? 62 LINES AND ANGLES MATHEMATICS–IX 8. In thefollowing figure, COA = 90º and AOB is a straight line. Find x and y. J A J A 9. Prove that the bisectors of the angles of a linear pair are at right angles. 10. If the bisectors of two adjacent angles form a right angle then prove that their non-common arms are in the same straight line. 11. In given figure, AOB is a line. Ray OD is perpendicular to AB . OC is another ray lying between OA and OD. Prove that DOC = 1 (BOC – AOC) 2 D B 90° T I B A 12. In the following figure, AB and CD intersects at O and BOE = 70°. Find value of a, b and c. M A 13. In the given figure 1 : 2 = 5 : 4 and AB || CD. Find all the labelled angles. MATHEMATICS–IX 7 8 LINES AND ANGLES 63 14. Find missing x in the following diagrams. Given AB || CD. B A B 140° T I 85° J A J A E x° D C (v) 15. In the following figure AB || CD. Find the values of a, b and c. M A 16. In the following figure, show that AB || CD. E 64 F 50° A 30° 20° C 160° LINES AND ANGLES D B MATHEMATICS–IX 17. In the following figure, AB || CD and CD || EF. Also EA AB. If BEF = 65°, find the value of a, b and c. J A J A x x 18. The angles of a triangle are 50 , 60 and (2x – 15)°, find the angles. 2 3 19. If a transversal cuts two parallel lines and is perpendicular to one of them, show that it will be perpendicular to the other also. 20. If two parallel lines are intersected by a transversal, show that the bisectors of any corresponding angles are parallel. 21. If two lines are intersected by a transversal in such a way that the bisectors of a pair of corresponding angles are parallel, show that the lines are parallel to each other. —NCERT 22. Prove that the bisectors of a pair of alternate angles of two parallel lines are themselves parallel. 23. Prove that if the arms of an angle are, respectively, parallel to the arms of another angle, then the angles either have equal measure or they are supplementary. 24. PQ and RS are two mirrors placed parallel to each other. An incident ray AB strikes the mirror PQ at B, the reflected ray moves along the path BC and strikes the mirror RS at C and again reflects back along CD. Prove that AB || CD. —NCERT T I B P A M A R B Q D C S 25. If two straight lines are perpendicular to the same line, prove that they are parallel to each other. 26. One of the angles of a triangle is 55°. Find the remaining two angles, if their diffrence is 35°. 27. In the given figures find x : MATHEMATICS–IX LINES AND ANGLES 65 28. Sides QP and RQ of PQR are produced to points S and T respectively. If SPR = 140° and POT = 105°, find PRQ. J A J A 29. A square ABCD is surmounted by an equilateral triangle EDC. Find x. E x° D C B A 30. In the following figure, AB || DC. If x y M A General Instructions : y 4 and y z . Find the values of x, y and z. 2 9 T I B MM : 30 B x z C D A PRACTICE TEST Time : 1 hour Q. 1-4 carry 2 marks, Q. 5-8 carry 3 marks and Q. 9-10 carry 5 marks each. 1. What value of x would make AOB a straight line? 66 C (2x–15)° x° (3x + 3)° A D O LINES AND ANGLES B MATHEMATICS–IX 2. If two parallel lines are intersected by a transversal, prove that the bisectors of two interior alternate angles are parallel. 3. Find the value of x°. 4. Find the value of x°, it is given that AB || CD. B J A J A 5. If two straight lines are perpendicular to the same line, prove that they are parallel to each other. 6. In the given figure, show that AB || CD. T I A 80° M A 45° 35° E B C 145° D F 7. In the following figure, AB || CD, find a and b. A E B 55° b C a F 135° G D 8. Angles A, B and C of a triangle satisfy B – A = 30° and C – B = 45°. Find all the angles. 9. Prove that the sum of three angles of a triangle is 180°. MATHEMATICS–IX LINES AND ANGLES 67 10. In the given figure, the sides AB and AC of ABC are produced to points E and D respectively. If bisectors BO and CO of CBE and BCD respectively meet at O, then prove that : BOC = 90° – J A 1 BAC 2 A B C E J A D O B ANSWERS OF PRACTICE EXERCISE 1. 33° 2. 15°, 165° 3. 110°, 70° 6. 130°, 50° 7. (i) x = 20° (ii) x = 33° (iii) x = 15° T I 5. 20° 8. x = 33°, y = 11° 13. 1 = 3 = 5 = 7 = 80°, 2 = 4 = 6 = 8 = 100° 12. a = 22°, b = 44°, c = 92° 14. (i) x = 115° (ii) x = 40° (iii) x = 55° (iv) x = 80° 15. a = 100°, b = 35°, c = 45° (v) x = 135° 17. a = 25°, b = 115°, c = 115° 18. 65°, 70°, 45° 26. a = 80°, b = 45° 27. (i) 65° (ii) 95° 29. x = 45° 4. 60° 28. 65° M A 30. 24°, 48°, 108° ANSWERS OF PRACTICE TEST 1. x = 32° 3. x = 110° 7. a = 55°, b = 80° 8. A = 25°, B = 55°, C = 100° 68 4. 290° LINES AND ANGLES MATHEMATICS–IX