Lines and Angles Notes

CONTENTS  Line segment, Ray, Collinear points, Angle  Measure of an angle, Types of angles OA, OB are rays & O is end point.  Measure of an angle: The amount of turning from OA to OB is called the measure of AOB, written as mAOB. An angle is measured in degrees denoted by ‘o’.  Bisector of an angle  Complementary, Supplementary, Adjacent angles   Linear pair of angles   Vertically opposite angles  Line segment: A part (or portion) of a line with two end points  Ray: A part of a line with one end point.  Collinear points: If three or more points lie on the same line other wise these are called non collinear points. B O A  An angle of 360°: If a ray OA starting from its original position OA, rotates about O, in the anticlockwise direction and after making a complete revolution it comes back to its original position, we say that it has rotated through 360 degrees, written as 360°. This complete rotation is divided into 360 equal parts. Each part measures 1°. 1° = 60 minutes, written as 60'. 1' = 60 seconds, written as 60''. We use a protractor to measure an angle.  Types of angles: B  Angle: Right angle Two rays with a common end point form an angle. O (i) A B In this case, AOC = BOC = Acute angle O (ii) 1 AOB. 2  Complementary angles: Two angles are said to be complementary, if the sum of their measures is 90°. A Two complementary angles are called the complement of each other. B Obtuse angle O A Ex. : Angles measuring 20° and 70° are complementary angles.  Supplementary angles: Two angles are said to be supplementary, if the sum of their measures is 180°. (iii) Two supplementary angles are called the supplement of each other. Ex. : Angles measuring 60° and 120° are supplementary angles.  EXAMPLES  Ex.1 Find the measure of an angle which is 20° more than its complement. Sol. Let the measure of the required angle be x°. Reflex angle O Then, measure of its complement = (90 – x)°. A x – (90 – x) = 20 2x = 110 x = 55 Hence, the measure of the required angle is 55°. B (v) Ex.2 Find the measure of an angle which is 40° less than its supplement. Sol. Let the measure of the required angle be x°. Then, measure of its supplement = (180 – x)°. Complete angle (180 – x) – x = 40 2x = 140 x = 70 O A Hence, the measure of the required angle is 70°. (vi) Ex.3 Find the measure of an angle, if six times its complement is 12° less than twice its supplement. A ray OC is called the bisector of AOB, if mAOC = mBOC. Sol. Let the measure of the required angle be x°.  Bisector of an angle: Then, measure of its complement = (90 – x)°. Measure of its supplement = (180 – x)°. 6(90 – x) = 2(180 – x) – 12  540 – 6x = 360 – 2x – 12  4x = 192 x = 48. Hence the measure of the required angle is 48°. Ex.4 Convert 180° in degree, minute & second. Sol. 180° = 179° 59' 60''. Ex.5 Find the measure of the supplement of an angle of 87°28'43''. Sol. We may write, 180° = 179°59'60''. Let the measure of the angle be xº. Then, Sol. xº + xº = 180º Ex.10 An angle is equal to five times its complement. Determine its measure. Let the measure of the given angle be x degrees. Then, its complement is (90 – x)º. Sol. supplement of an angle of (87°28'43'') = an angle of [180° – (87°28'43'')] It is given that : = an angle of [179°59'60'' – 87°28'43''] Angle = 5 × Its complement = an angle of (92°31'17''). Hence, the measure of the required angle   x = 5(90 – x) x = 450 – 5x   6x = 450 x = 75 = (92°31'17''). If A = 36°27'46'' find A + B. Sol. A + B = 36°27'46'' + 28°43'39'' and B = 28°43'39'', Ex.11 An angle is equal to one-third of its supplement. Find its measure. Let the measure of the required angle be x degrees. Then, Sol. = 64°70'85'' = 64°71'25'' = 65°11'25'' (iii) Sol. Its supplement = 180º – x. It is given that : Find the complement of each of the following angles : (i) 58° Angle = (ii) 16° 2 of a right angle 3 (iv) 46° 30' (i) 90° – 58° = 32° 2 (90°) = 90° – 60° = 30° 3 (iv) 90° – 46° 30' Ex.8 Find the measure of an angle which is complement of itself. Sol. Let the measure of the angle be xº, Then, Then, the measure of its complement is given to be xº. Since, the sum of the measures of an angle and its complement is 90º Ex.9 1 (180º – x) 3  x =   4x = 180º  3x = 180º – x x = 45º Thus, the measure of the given angle is 45º. Ex.12 Two supplementary angles are in the ratio 2 : 3. Find the angles. Let the two angles be 2x and 3x in degrees. Then, Sol. = 89° 60' – 46° 30' = 43° 30' 1 (Its supplement) 3  (ii) 90° – 16° = 74° (iii) 90° –  Thus, the measure of the given angles is 75º. Ex.6 Ex.7 2xº = 180º xº = 90º 2x + 3x = 180º    5x = 180º  x = 36º Thus, the two angles are 2x = 2 × 36º = 72º and 3x = 3 × 36º = 108º Ex.13 Write the complement of the following angles: 30º 20´ Sol. Complement of 30º20´ = 90º – 30º 20´  xº + xº = 90º  2xº = 90º xº = 45º = 90º – (30º + 20´) Find the measure of an angle which forms a pair of supplementary angles with itself. = (89º – 30º) + (1º – 20´) = 59º + (60´ – 20´) = 59º + 40´ = 59º 40´ [ 1º = 60´] Ex.14 Find the supplement of the following angles : 134º 30´ 26´´ Sol. Supplement of an angle of 134º 30´ 26´´ = 180º – (134º 30´ 26´´) Given : AOB is a straight line and rays OC, OD and OE stand on it, forming AOC, COD, DOE and EOB. D C E = (179º – 134º) + (1º – 30´ 26´´) = 45º + (60´ – (30´ + 26´´)) [ 1º = 60´] = 45º + (59´ – 30´) + (1´ – 26´´) = 45º + 29´ + 34´´ = 45º29´34´´  Adjacent angles: Two angles are called adjacent angles, if (i) they have the same vertex, (ii) they have a common arm and (iii) their non-common arms are on either side of the common arm. A O B To prove : AOC + COD + DOE + EOB = 180°. Proof : Ray OC stands on line AB.  AOC + COB = 180° AOC + (COD + DOE + EOB) = 180° [COB = COD + DOE + EOB] AOC + COD + DOE + EOB = 180°. Hence, the sum of all the angles formed on the same side of line AB at a point O on it is 180°. Theorem 2 : Prove that the sum of all the angles around a point is 360°. In the given figure, AOC and BOC are adjacent angles having the same vertex O, a common arm OC and their non-common arms OA and OB on either side of OC.  Linear pair of angles: Two adjacent angles are said to form a linear pair of angles, if their non-common arms are two opposite rays. Given : A point O and the rays OA, OB, OC, OD and OE make angles around O. To prove : AOB + BOC + COD + DOE + EOA = 360° Construction : Draw a ray OF opposite to ray OA. Proof : Since ray OB stands on line FA, we have : AOB + BOF = 180° [linear pair] AOB + BOC + COF = 180° ….(i) [BOF = BOC + COF] C In the adjoining figure, AOC and BOC are two adjacent angles whose non-common arms OA and OB are two opposite rays, i.e., BOA is a line AOC and BOC form a linear pair of angles. B F D O A E Theorem 1 : Again, ray OD stands on line FA. Prove that the sum of all the angles formed on the same side of a line at a given point on the line is 180°. FOD + DOA = 180° [linear pair] or FOD + DOE + EOA = 180° …(ii) [DOA = DOE + EOA]  IMPORTANT POINTS Adding (i) and (ii), we get : AOB + BOC + COF + FOD + DOE + EOA = 360° AOB + BOC + COD + DOE + EOA = 360° [COF + FOD = COD] Hence, the sum of all the angles around a point O is 360°.  Vertically opposite angles: Two angles are called a pair of vertically opposite angles, if their arms form two pairs of opposite rays. Let two lines AB and CD intersect at a point O. Then, two pairs of vertically opposite angles are formed : (i) AOC and BOD (ii) AOD and BOC Theorem 3 :  Two angles are called adjacent angles if – (i) they have the same vertex, (ii) they have a common arm, and (iii) uncommon arms are on either side of the common arm.  Two adjacent angles are said to form a linear pair of angles, if the non-common arms are two opposite rays.  If a ray stands on a line, then the sum of the adjacent angles so formed is 180º.  If the sum of two adjacent angles is 180º, then their non-common arms are two opposite rays.  The sum of all the angles round a point is equal to 360º.  Two angle are called a pair of vertically opposite angles, if their arms form two pairs of opposite rays.  If two lines intersect, then the vertically opposite angles are equal. If two lines intersect then the vertically opposite angles are equal.  EXAMPLES  Given : Two lines AB and CD intersect at a point O. To prove : (i) AOC = BOD, (ii) AOD = BOC Proof : Since ray OA stands on line CD, we have: A Ex.15 Two lines AB and CD intersect at O. If AOC = 50°, find AOD, BOD and BOC. D O C B AOC + AOD = 180° [linear pair]. Sol. AOD + AOC = 180° (linear pair) Again, ray OD stands on line AB. AOD + 50° = 180° AOD + BOD = 180° [linear pair] AOD = 130° AOC + AOD = AOD + BOD Also BOD = AOC [each equal to 180°] AOC = BOD Similarly, AOD = BOC (vertically opposite angles) & BOC = AOD = 130° (vertically opposite angles) 130°, 50°, 130°. Ex.16 In the adjoining figure, AOB is a straight line. Find the value of x. Hence, find AOC, COD and BOD. Sol. Since OA and OB are opposite rays. Therefore, AB is a line. Since ray OC stands on line AB. AOC + COB = 180° AOC + COD + BOD = 180° [COB = COD + BOD] (AOC + BOD) + COD = 180° Sol. (3x + 7)° + (2x – 19)° + x° = 180' (linear pair) 90° + COD = 180° 6x – 12) = 180° [ AOC + BOD = 90° (Given)] 6x = 192° COD = 180° – 90° = 90° x = 32° AOC = 3x + 7 = 3(32) + 7 = 96 + 7 = 103° Ex.19 In figure, OP bisects BOC and OQ, AOC. Show that POQ = 90°. COD = 2x – 19 = 2(32) – 19 = 64 – 19 = 45° BOD = x° = 32°. Ex.17 Two lines AB and CD intersect at a point O such that BOC + AOD = 280°, as shown in the figure. Find all the four angles. C B According to question, OP is bisector of BOC 1 = 2 O A Sol. Sol. D or 1 =  AOC = BOD = x(Let) (vertically opposite angles) AOC + (AOD + BOC) + BOD = 360° and OQ is bisector of AOC 3 = 4 or 4 = x + 280° + x = 360° BOC 2 AOC 2 2x = 80° 1 + 4 = x = 40° AOC = BOD = x° = 40°. and BOC = AOD = 280 = 140°. 2 = BOC AOC  2 2 BOC  AOC 180 =  90 2 2 Ex.20 In figure OA and OB are opposite rays : Ex.18 In figure, OA, OB are opposite rays and AOC + BOD = 90°. Find COD. (i) If x = 75, what is the value of y ? (ii) If y = 110, what is the value of x ? Since AOC and BOC form a linear pair. Sol.   90º + COD = 180º Therefore, AOC + BOC = 180º   x + y = 180º [AOC + BOD = 90º (Given)] ...(1) (i) If x = 75, then from (i) 75 + y = 180º y = 105º. (ii) If y = 110 then from (i)    COD = 180º – 90º  COD = 90º Ex.23 In figure ray OE bisects angle AOB and OF is a ray opposite to OE. Show that FOB = FOA. x + 110 = 180   x = 180 – 110 = 70. Ex.21 In figure AOC and BOC form a linear pair. Determine the value of x. Since ray OE bisects angle AOB. Therefore, Sol. EOB = EOA ....(i) Now, ray OB stands on the line EF. Since AOC and BOC form a linear pair. Sol.   AOC + BOC = 180º   4x + 2x = 180º  6x = 180º  180º  x = = 30º 6   EOB + FOB = 180º ...(ii) {linear pair] Again, ray OA stands on the line EF.   EOA + FOA = 180º ....(iii) Form (ii) and (iii), we get   Thus, x = 30º EOB + FOB = EOA + FOA  EOA + FOB = EOA + FOA [ EOB = EOA Ex.22 In figure OA, OB are opposite rays and AOC + BOD = 90º. Find COD.  (from (i)]  FOB = FOA. Ex.24 In figure OE bisects AOC, OF bisects COB and OE OF. Show that A, O, B are collinear. C E Since OA and OB are opposite rays. Therefore, AB is a line. Since ray OC stands on line AB. Therefore, Sol. AOC + COB = 180º [Linear Pairs]   AOC + COD + BOD = 180º F A Sol.  (AOC + BOD) + COD = 180º B Since OE and OF bisect angles AOC and COB respectively. Therefore,  [COB = COD + BOD]  O and AOC = 2EOC ....(i) COB = 2COF ....(ii) Adding (i) and (ii), we get a = c [Vertically opposite angles] AOC + COB = 2EOC + 2COF  c = 35º [a = 35º]   AOC + COB = 2(EOC + COF) Clearly, a + b = 180º   AOC + COB = 2(EOF) [Since a and b are angles of a linear pair]   AOC + COB = 2 × 90º  35º + b = 180º [ OE OF EOF = 90º]   b = 180º – 35º   AOC + COB = 180º b = 145º Since b and d are vertically opposite angles. Therefore, But AOC and COB are adjacent angles. Thus, AOC and COB are adjacent supplementary angles. So, AOC and COB form a linear pair. Consequently OA and OB are two opposite rays. Hence, A, O, B are collinear.  d = b  d = 145º [b = 145º] Ex.27 In Fig., determine the the value of y. Ex.25 If ray OC stands on line AB such that AOC = COB, then show that AOC = 90º. Since ray OC stands on line AB. Therefore, Sol. AOC + COB = 180º [Linear pair] ...(i) Sol. C Since COD and EOF are vertically opposite angles. Therefore, COD = EOF  COD = 5yº [ EOF = 5yº (Given)] Now, OA and OB are opposite rays. A O B But AOC = COB  AOD + DOC + COB = 180º (Given)   AOC + OC = 180º   2AOC = 180º AOC = 90º  2yº + 5yº + 5yº = 180º  12yº = 180º  yº = Ex.26 In Fig., lines 1 and 2 intesect at O, forming angles as shown in the figure. If a = 35º, find the values of b, c, and d. Sol. Since lines 1 and 2 intersect at O. Therefore, Thus, 180º = 15. 12 yº = 15. Ex.28 In Fig., AB and CD are straight lines and OP and OQ are respectively the bisectors of angles BOD and AOC. Show that the rays OP and OQ are in the same line. [Vertically oppsostie angles] In order to prove that OP and OQ are in the same line, it is sufficient to prove that POQ = 180º. Sol.  a = 4b  a = 4 × 21 = 84 Now, OP is the bisector of AOC  1 = 6 Now, OR and OS are in the same line. Therefore. ...(i) ROQ + QOS = 180º and, OQ is the bisector of AOC  3 = 4 [ b = 21] [Linear pair]  2c + a = 180 ....(ii)  2c + 84 = 180 [ b = 84] Clearly, 2 and 5 are vertically opposite angles.  2c = 96 c = 48  2 = 5 Hence, a = 84, b = 21 and c = 48 ....(iii) We know that the sum of the angles formed at a point is 360º. Ex.30 In fig if AOC + BOD = 70º, find COD. C Therefore, D 1 +2 + 3 + 4 + 5 + 6 = 360º   (1 + 6) + (3 + 4) + (2 + 5) = 360º   21 + 23 + 22 = 360º [Using (i), (ii) and (iii)] Sol.    2(1 + 3 + 2) = 360º   1 + 2 + 3 = 180º POQ = 180º B AOC + COD + BOD = 180º or (AOC + BOD) + COD = 180º or 70º + COD = 180º Hence, OP and OQ are in the same straight line. Ex.29 In Fig., two staright lines PQ and RS intersect each other at O. If POT = 75º, find the values of a, b and c. O A or COD = 180º – 70º or COD = 110º Ex.31 In fig. find the value of y. R 4b 2c O 75° Q a 3y b 5y S P T Sol. Since OR and OS are in the same line. Therefore,  ROP + POT + TOS = 180º  4bº + 75º + bº = 180º 5bº + 75º = 180º  5bº = 105º bº = 21 Since PQ and RS intersect at O. Therefore, QOS = POR Sol. 2y 2y + 3y + 5y = 180º 10y = 180º y = 180º = 18º 10º  Parallel lines: The lines which are in same plane and do not intersect each other an y where, i.e. distance between parallel lines is same anywhere r t a b  Transversal line : A line which intersect two or more given lines at distinct points, is called a transversal of the given lines. Here || m and p is transversal line. * Part of transversal which is between the two lines is called intercept (AB).  IMPORTANT POINTS  Two angles on the same side of a transversal are known as the corresponding angles if both lie either above the two lines or below the two lines.  The pairs of interior angles on the same side of the transversal are called pairs of consecutive interior angles.  If a transversal intersect two parallel lines, then each pair of corresponding angles are equal. (n  2) 180º n  If a transversal intersects two parallel lines, then each pair of alternate interior angles are equal .  If a transversal intersects two lines in such a way that a pair of alternate interior angles are equal, then the two lines are parallel.  If a transversal intersects two parallel lines, then each pair of consecutive interior angles are supplementary. 360º 180º each interiorangle But 1 and 2 form a linear pair.  1 + 2 = 180º  1 4 2 3 2 1   3xº + 2xº = 180º 5xº = 180º   x =   1 = 3xº = (3 × 36)º = 108º 180º = 36º 5 and,2 = 2xº = (2 × 36)º = 72º Now, 1 = 4 and 2 = 3 Interior alternate angles Cointerior angles [ Vertically opposite angles]   EXAMPLES  Now, 6 = 2º and 3 = 7 Ex.32 In figure m || n and 1 = 65º. Find 5 and8. 2 1 3 4 m n Sol. [Corresponding angles]    6 = 72º and 7 = 108º  5 = 108º and 8 = 72º Hence, 1 = 108º, 2 = 72º, 3 = 108º, 4 = 72º, 5 = 108º 6 = 72º, 7 = 108º and 8 = 72º. We have, 1 = 3 [Vertically opposite angles] [Corresponding angles] Ex.34 In figure AB || DC and AD || BC. Prove that DAB = DCB. A B  1 = 8  8 = 65º [2 = 72º] Again, 5 = 7 and 8 = 6 5 6 8 7 and,3 = 8  4 = 72º and 3 = 108º [ 1 = 65º (Given)] Now, 5 + 8 = 180º D  5 + 65º = 180º  5 = 180º – 65º = 115º Since AD || BC and AB is a transversal intersecting them at A and B respectively. Therefore Sol. Thus, 5 = 115º and 8 = 65º. DAB + ABC = 180º Ex.33 In figure m || n and angles 1 and 2 are in the raito 3 : 2. Determine all the angles from 1 to 8. m 4 n 5 6 8 7 1 2 3 C [Consecutive interior angles] Again, AB || CD and BC is a transversal interacting them at B and C respectively. Therefore,  ABC + DCB = 180º [Consecutive interior angles] From (i) and (ii), we get Sol. It is given that 1 : 2 = 3 : 2. So, let 1 = 3xº and 2 = 2xº ...(i) DAB + ABC = ABC + DCB   DA B = DCB ....(ii) 108º + 1 = 180º1 = 180º – 108º = 72º Ex.35 In figure AB || CD. Determine a 55° 0 a1 2 C D Again, EF || CD and transversal CE cuts them at E and F respectively.  38°  FEC + ECD = 180º B B A 108° Sol. Clearly, a = 1 + 2 F 112° D ....(ii) Now, 1 = 55º [Alternate s]   2 + 112º = 180º and 2 = 38º [Alternate s]   2 = 180º – 112º  a = 55º + 38º [Using (i)]  2 = 68º   a = 93º. Now, x = 1 + 2 Thus, a = 93º x = 72º + 68º = 140º Ex.36 In figure AB || CD. Determine X. Sol. E 1 2 Through O draw a line  parallel to both AB and CD. Through O, draw a line  parallel to both AB and CD. Then, A  45° B x° 1 0 2 D and Through C, draw CF parallel to both AB and DE. Since AB || CF and the transversal BC cuts them at B and C respectively. Therefore, Sol. 30°  Ex.38 In Figure AB || DE. Prove that ABC + BCD = 180º + CDE. C 1 = 45º [Alternate s] 2 = 30º [Alternate s] ABC + 1 = 180º ....(i) [conseu. interior angles are supplementary]  BOC = 1 + 2 = 45º + 30º = 75º So, x = 360 – BOC Hence, Similarly, DE || CF and tranversal CD intersects them at C and D respectively. Therefore, = 360 – 75 = 285º x = 285º Ex.37 In figure AB || CD. Find the value of x. CDE = 2 ....(ii) Adding (i) and (ii), we get 108° E x° 112° Sol. [Alternate angles] ABC + 1 + 2 = 180º + CDE Draw EF parallel to both AB and CD. Now, AB || EF and transversal AE cuts them at A and E respectively. BAE + FEA = 180º   ABC + BCD = 180º + CDE [ 1 + 2 = BCD] Ex.39 In Figure PQ || RS, PAB = 70º and ACS = 100º. Determine ABC, BAC and CAQ. We have Sol. 2 = 120º and 5 = 60º But 2 = 4 [Vertically opposite angles]   4 = 120º, 5 = 60º   4 + 5 = 120º + 60º = 180º   4 and 5 are supplementary angles.   Consecutive interior angles are supplementary. Since PQ || RS and transversal AB cuts them at A and B respectively. Sol.  ABC = PAB  [Alternate angles] ABC = 70º [PAB = 70º (Given)]  m || n. Ex.41 In figure show that AB || EF. A B Now, PQ || RS and transversal AC cuts them at A and C respectively.  PAC = ACS [Alternate angles]  PAC = 100º [ACS = 100º]  PAB + BAC = 100º 66° 36° 30° E F 150° C D We have, Sol. [PAC = PAB + BAC]   = 36º + 30º = 66º BAC = 30º Now, ray AB stands at A on PQ.  BCD = BCE + ECD 70º + BAC = 100º  PAC + CAQ = 180º  100º + CAQ = 180º  CAQ = 80º   ABC = BCD Thus, lines AB and CD are intersected by the line BC such that ABC = BCD i.e. the alternate angles are equal. Therefore, AB || CD ....(i) Hence, ABC = 70º,   Now, ECD + CEF = 30º + 150º = 180º BAC = 30º and CAQ = 80º. This shows that the sum of the interior angles on the same side of the transversal CE is 180º i.e. they are supplementary. Ex.40 In Figure if 2 = 120º and 5 = 60º, show that m || n.   EF || CD ....(ii) From (i) and (ii), we have AB || CD and CD || EF AB || EF. Hence, AB || EF Ex.42 In figure given that AOC = ACO and BOD = BDO. Prove that AC || DB. Sol.   We have,  AOC = ACO and BOD = BDO But AOC = BOD [Vertically opp. s]  ACO = BOD and BOD = BDO  ACO = BDO Thus, AC and BD are two lines intersected by transversal CD such that ACO = BDO i.e. alternate angles are equal. Therefore, AC || DB. Ex.43 In figure AB || DC if x = Ex.44 In figure lines AB and CD are parallel and P is any point between the two lines. Prove that ABP + CDP = DPB. Through point P draw a line PM parallel to AB or CD. Sol. Now, PM || AB [By construction] ABP = MPB [Alternate angles] ....(i) It is given that CD || AB and PM || AB by construction. Therefore, PM || CD 4 3 y and y = z, 3 8 find the values of x, y and z. [ Lines parallel to the same line are parallel to each other]  CDP = MPD [Alternate angles] ....(ii) Adding (i) and (ii), we get Sol.   Since AB || DC and transversal BD intersects them at B and D respectively. Therefore, ABD = CDB CDB = xº In BCD, we have yº + zº + xº = 180º 3 4 3  zº + zº + × zº = 180 8 3 8 Ex.45 Prove that two lines perpendicular to the same line are parallel to each other. Let lines , m, n be such that  n and Sol. m  n as shown in figure. We have to prove that || m 3 1  zº + zº + zº = 180 8 2  15 zº = 180º 8 4 3 [ x = y and y = z 3 8 x =  ABP + CDP = MPB + MPD = DPB 4 3 z × z= ] 2 3 8  zº = 180º × 8 = 96º 15 3 3 Now, y = z y = × 96º = 36º 8 8 and x = 4 4 y x = × 36º = 48 3 3 Now, n and m  n   1 = 90º and 2 = 90º  1 = 2 Thus, the corresponding angles made by the transversal n with lines  and m are equal. Hence,  || m. Ex.46 Prove that two angles which have their arms parallel are either equal or supplementary. Case III : When one pair of arms is parallel in the same sense and the other in opposite sense. In this case, Given : Two angles ABC and DEF such that BA || ED and BC || EF. Sol. A D To prove :ABC = DEF or ABC + DEF = 180º G B Proof : We have the following three cases: E F Case I : When both pairs of arms are parallel in the same sense fig. in this case, C AB || DE and transversal BC cuts them AB || DE and transversal BC cuts them at B and G respectively   ABC = BGE [Alternate angles] ....(v) Again, BC || FE and transversal DE cuts them   DEF + BGE = 180º [Consecutive interior ....(vi) angles are supplementary] From (v) and (vi), we get  ABC = DGC ...(i) ABC + DEF = 180º [Corresponding angles] Again, BC || EF and transversal DE cuts them at G and E respectively.   DGC = DEF ...(ii) From (i) and (ii), we get Ex.47 In figure bisectors of the exterior angles B and C formed by producing sides AB and AC of ABC intersect each other at the point O. 1 Prove that BOC = 90º – A. 2 A ABC = DEF Case II : When both pairs of arms are parallel in opposite sense in this case, C B D E O DBC = 180º – B Sol.    AB || DE and transversal BC cuts them at B and G respectively. ABC = EGC ....(iii) [Corresponding angles] Again, BC || EF and transversal DE cuts them at G and E respectively. DEF = EGC [Alternate angles] ....(iv) From (iii) and (iv), we get ABC = DEF.  1 DBC = OBC = 90º – B/2 2 Similarly, OCB = 90º – C/2 In OBC, we have, OBC + OCB + BOC = 180º or (90º – B/2) + (90º – C/2) + BOC = 180º BOC = 180º – (180º – B/2 + C/2) = B/2 + C/2 But A + B + C = 180º ...(i) or B + C = 180º – A   1 1 (B + C) = (180º – A) 2 2   1 1 B + C = 90º – A/2 2 2 (ii) Since 4 and 5 are consecutive interior angles. Therefore, 4 + 5 = 180º [ Consecutive supplementary] BOC = 90º – A/2 Ex.48 In fig, given that AB || CD. A 3 C 7 2xº = 180º – 28º 2x = 152º x = 76º  4 = (x + 20)º = (76 + 20)º = 96º 1 4 and, 5 = (x + 8)º = (76 + 8º) = 84º B (iii) We have, 5 and 4 = 8 D 8 are  2xº + 28º = 180º 2 = 4 6 angles  (x + 20)º + (x + 8)º = 180º Hence, from (i), we have, 2 interior [Vertically opposite angles] [Corresponding angles]  2 = 8  (3x – 10)º = (5x – 30)º 3x – 10 = 5x – 30 (i) If 1 = (120 – x)º and 5 = 5xº, find the measures of 1 and 5.  3x – 5x = – 30 + 10 x = 10 (ii) If 4 = (x + 20)º and 5 = (x + 8)º, find the measure of 4 and 5.  2 = (3x – 10)º = (3 × 10 – 10)º = 20º and 8 = (5x – 30)º = (5 × 10 – 30)º = 20º (iii) If 2 = (3x – 10)º and 8 = (5x – 30)º, determine the measures of 2 and 8. (iv) Since 3 and 6 are consecutive interior angles. Therefore (iv) If 1 = (2x + y)º and 6 = (3x – y)º, determine the measures of 2 in terms of y. 3 + 6 = 180º (v) If 2 = (2x + 30)º, 4 = (x + 2y)º and 6 = (3y + 10)º, find the measure of 5. (vi) If 2 = 2 (1), determine 7. (vii)If the ratio of the measures of 3 and 8 is 4 : 5, find the measure 3 and 8 (viii)If the complement of 5 equals the supplement of 4, find the measures of 4 and 5. Sol. (i) since 1 and 5 are the corresponding angles and corresponding angles are equal.  1 = 5  (120 – x)º = 5xº  120º = 6x  x= 120 = 20º 6  1 = (120 – x)º = (120 – 20)º = 100º and, 5 = 5xº = (5 × 20)º = 100º – 2x = – 20 But 1 = 3 1 + 6 = 180º    (2x + y)º + (3x – y)º = 180º    5x = 180º  x = 36.  1 = (2x + y)º = (72 + y)º [ x = 36] But 1 + 2 = 180º   (72 + y)º + 2 = 180º 2 = (180 – (72 + y))º 2 = (108 – y)º. (v) We have, 2 = 4 [Vertically opposite angles] and 4 = 6 [Alternate angles]  2 = 4 = 6 Now, 2 = 4  2x + 30 = x + 2y 2x – x – 2y + 30 = 0  x – 2y + 30 = 0 And, 4 = 6 (viii) We have, Complement of 5 = Supplement of 4 ....(1) (x + 2y) = (3y + 10)  x – y – 10 = 0  90º – 5 = 180º – 4 ....(2)  90º – 5 = 180º – (180º – 5)  4  5  180º   4  180º5    Subtracting (2) from (1), we get (x – 2y + 30) – (x – y – 10) = 0  90º – 5 = 5  – y + 40 = 0 y = 40.   Putting y = 40 in (2), we get x = 50.  2 5 = 90º  5 = 45º  4 + 5 = 180º  4 = (x + 2y)º = (50 + 2 × 40)º = 130º    4 + 45º = 180º 4 = 135º But 4 + 5 = 180º Ex.49 In fig, OP || RS. Determine PQR.  130º + 5 = 180º 5 = 50º (vi) We have, 1 + 2 = 180º  [Linear pairs]  2 = 2 1 1 + 2 1 = 180º  3 1 = 180º 1 = 60º But 1 = 3 [Vertically opposite angles] Sol. Produce OP to intersect RQ in a point T. Now, OT || RS and transversal RT intersect them at T and R respectively.  3 = 60º But 3 = 5 [Alternate angles] and 5 = 7 [Vertically opposite angles]  3 = 7 7 = 60º (vii) [ 3 = 60º] We have, 3 : 8 = 4 : 5. So, let 3 = 4x and 8 = 5x.  5 = 4x and 8 = 5x [3 = 5 (Alternate angles)]  5 + 8 = 4x + 5x   180º = 9x x = 20º RTP = SRT [Alternate angles]  RTP = 130º  PTQ = 180º – 130º = 50º  RTP  PTQ  180º    Linear Pairs  Since, ray QP stands at P on OT.  OPQ + QPT = 180º  110º + QPT = 180º QPT = 70º  3 = 4x = 4 × 20º = 80º and 8 = 5x = 5 × 20º = 100º  PQR = 180º – (70º + 50º) = 60º [ Sum of the angles of a triangle is 180º] Ex.50 Prove that through a given point we can draw only one perpendicular to a given line. Sol. If possible, let PL and PM be two perpendicular from a point P on a line  as shown in fig. P We know that two lines perpendicular to the same line are parallel to each other. Therefore, PL || PM But there cannot be two parallel lines passing through the same point. Therefore, through a given point we can draw only one line perpendicular to a given line.  M L IMPORTANT POINTS TO BE REMEMBERED 1. An angle is the union of two non-collinear rays with a common initial point. 2. An angle whose measure is 90º is called a right angle. 3. An angle whose measure is less than 90º is called an acute angle. 4. An angle whose measure is more than 90º but less than 180 is called an obtuse angle. 5. An angle whose measure is 180º is called a straight angle. 6. An angle whose measure is more than 180º is called a reflex angle. 7. Two angles are complementary if their sum is 90º. 8. Two angles are supplementary if their sum is 180º. 9. Two angles having a common vertex and a common arm are called adjacent angles if their uncommon arms are on either side of the common arm. 10. Two adjacent angles are said to form a linear pair of angles, if their non-common arms are two opposite rays. 11. Two angles are pair of vertically opposite angles if their arms form two pairs of opposite rays. 12. If two lines intersect, then vertically opposite angles are equal. 13. If a transversal intersects two parallel lines, then each pair of (i) corresponding angles are equal (ii) alternate interior angles are equal (iii) interior angles on the same of the same side of the transversal are supplementary. 14. If a transversal intersects two lines such that, either (i) any one pair of corresponding angles are equal, or (ii) any one pair of alternate interior angles are equal, or (iii) any one pair of interior angles on the same side of the transversal are supplementary then the lines are parallel. 15. Lines which are parallel to a given line are parallel to each other.