{"id":2778,"date":"2022-11-21T16:00:40","date_gmt":"2022-11-21T10:30:40","guid":{"rendered":"https:\/\/www.aplustopper.com\/?p=2778"},"modified":"2022-11-22T16:55:13","modified_gmt":"2022-11-22T11:25:13","slug":"linear-pair-of-angles","status":"publish","type":"post","link":"https:\/\/www.aplustopper.com\/linear-pair-of-angles\/","title":{"rendered":"Linear Pair Of Angles"},"content":{"rendered":"
Two adjacent angles are said to form a linear pair of angles, if their non-common arms are two opposite rays. Theorem 1:<\/strong> Theorem 2:<\/strong> Example 1: \u00a0 \u00a0<\/strong>In the adjoining figure, AOB is a straight line. Find the value of x. Hence, find \u2220AOC, \u2220COD and \u2220BOD. Example 2: \u00a0 \u00a0<\/strong>In figure, OA, OB are opposite rays and \u2220AOC + \u2220BOD = 90\u00b0. Find \u2220COD. Example 3: \u00a0 \u00a0<\/strong>In figure, OP bisects \u2220BOC and OQ, \u2220AOC. Show that \u2220POQ = 90\u00b0. Example 4: \u00a0 \u00a0<\/strong>In figure OA and OB are opposite rays : Example 5: \u00a0 \u00a0<\/strong>In figure \u2220AOC and \u2220BOC form a linear pair. Determine the value of x. Example 6: \u00a0 \u00a0<\/strong>In figure OA, OB are opposite rays and Example 7: \u00a0 \u00a0<\/strong>In figure ray OE bisects angle \u2220AOB and OF is a ray opposite to OE. Show that Example 8: \u00a0 \u00a0<\/strong>In figure OE bisects \u2220AOC, OF bisects \u2220COB and OE \u22a5OF. Show that A, O, B are collinear. Example 9: \u00a0 \u00a0<\/strong>If ray OC stands on line AB such that Example 10: \u00a0 \u00a0<\/strong>In fig if \u2220AOC + \u2220BOD = 70\u00ba, find \u2220COD. Example 11: \u00a0 \u00a0<\/strong>In fig. find the value of y. <\/p>\n","protected":false},"excerpt":{"rendered":" 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. In the adjoining figure, \u2220AOC and \u2220BOC are two adjacent angles whose non-common arms OA and OB are two opposite rays, i.e., BOA is a line \u2234 \u2220AOC and \u2220BOC form […]<\/p>\n","protected":false},"author":3,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"_genesis_hide_title":false,"_genesis_hide_breadcrumbs":false,"_genesis_hide_singular_image":false,"_genesis_hide_footer_widgets":false,"_genesis_custom_body_class":"","_genesis_custom_post_class":"","_genesis_layout":"","footnotes":""},"categories":[5],"tags":[1193,1197,1196,1195,1177,1194],"yoast_head":"\n
\n
\nIn the adjoining figure, \u2220AOC and \u2220BOC are two adjacent angles whose non-common arms OA and OB are two opposite rays, i.e., BOA is a line
\n\u2234 \u2220AOC and \u2220BOC form a linear pair of angles<\/strong>.<\/p>\n
\nProve that the sum of all the angles formed on the same side of a line at a given point on the line is 180\u00b0.
\nGiven:<\/strong> AOB is a straight line and rays OC, OD and OE stand on it, forming \u2220AOC, \u2220COD, \u2220DOE and \u2220EOB.
\n
\nTo prove:<\/strong> \u2220AOC + \u2220COD + \u2220DOE + \u2220EOB\u00a0= 180\u00b0.
\nProof:<\/strong> Ray OC stands on line AB.
\n\u2234 \u2220AOC + \u2220COB = 180\u00b0
\n\u21d2 \u2220AOC + (\u2220COD + \u2220DOE + \u2220EOB) = 180\u00b0
\n[\u2235 \u2220COB = \u2220COD + \u2220DOE + \u2220EOB]
\n\u21d2 \u2220AOC + \u2220COD + \u2220DOE + \u2220EOB = 180\u00b0.
\nHence, the sum of all the angles formed on the same side of line AB at a point O on it is 180\u00b0.<\/p>\n
\nProve that the sum of all the angles around a point is 360\u00b0.
\nGiven:<\/strong> A point O and the rays OA, OB, OC, OD and OE make angles around O.
\nTo prove:<\/strong> \u2220AOB + \u2220BOC + \u2220COD + \u2220DOE + \u2220EOA = 360\u00b0
\nConstruction:<\/strong> Draw a ray OF opposite to ray OA.
\nProof:<\/strong> Since ray OB stands on line FA,
\n
\nwe have, \u2220AOB + \u2220BOF = 180\u00b0 \u00a0 [linear pair]
\n\u2234 \u2220AOB + \u2220BOC + \u2220COF = 180\u00b0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u2026.(i)
\n[\u2235 \u2220BOF = \u2220BOC + \u2220COF]
\nAgain, ray OD stands on line FA.
\n\u2234 \u2220FOD + \u2220DOA = 180\u00b0 [linear pair]
\nor \u2220FOD + \u2220DOE + \u2220EOA = 180\u00b0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u2026(ii)
\n[\u2235 \u2220DOA = \u2220DOE + \u2220EOA]
\nAdding (i) and (ii), we get,
\n\u2220AOB + \u2220BOC + \u2220COF + \u2220FOD + \u2220DOE + \u2220EOA = 360\u00b0
\n\u2234 \u2220AOB + \u2220BOC + \u2220COD + \u2220DOE + \u2220EOA = 360\u00b0
\n[\u2235 \u2220COF + \u2220FOD = \u2220COD]
\nHence, the sum of all the angles around a point O is 360\u00b0.<\/p>\nLinear Pair Of Angles Example Problems With Solutions<\/strong><\/h2>\n
\n
\nSolution: \u00a0 \u00a0<\/strong>(3x + 7)\u00b0 + (2x \u2013 19)\u00b0 + x\u00b0 = 180′ (linear pair)
\n\u21d2 6x \u2013 12) = 180\u00b0
\n\u21d2 6x = 192\u00b0
\n\u21d2 x = 32\u00b0
\n\u2234 \u2220AOC = 3x + 7 = 3(32) + 7 = 96 + 7 = 103\u00b0
\n\u2220COD = 2x \u2013 19 = 2(32) \u2013 19 = 64 \u2013 19 = 45\u00b0
\n\u2220BOD = x\u00b0 = 32\u00b0.<\/p>\n
\n
\nSolution: \u00a0 \u00a0<\/strong>Since OA and OB are opposite rays. Therefore, AB is a line. Since ray OC stands on line AB.
\n\u2234 \u2220AOC + \u2220COB = 180\u00b0
\n\u21d2 \u2220AOC + \u2220COD + \u2220BOD = 180\u00b0
\n[\u2235 \u2220COB = \u2220COD + \u2220BOD]
\n\u21d2 (\u2220AOC + \u2220BOD) + \u2220COD = 180\u00b0
\n\u21d2 90\u00b0 + \u2220COD = 180\u00b0
\n[\u2235 \u2220AOC + \u2220BOD = 90\u00b0 (Given)]
\n\u21d2 \u2220COD = 180\u00b0 \u2013 90\u00b0 = 90\u00b0<\/p>\n
\n
\nSolution: \u00a0 \u00a0<\/strong>According to question, OP is bisector of \u2220BOC
\n<\/p>\n
\n
\n(i) If x = 75, what is the value of y ?
\n(ii) If y = 110, what is the value of x ?
\nSolution: \u00a0 \u00a0<\/strong>Since \u2220AOC and \u2220BOC form a linear pair.
\nTherefore, \u2220AOC + \u2220BOC = 180\u00ba
\n\u21d2 x + y = 180\u00ba \u00a0 \u00a0 \u00a0 \u00a0…(1)
\n(i) If x = 75, then from (i)
\n75 + y = 180\u00ba
\ny = 105\u00ba.
\n(ii) If y = 110 then from (i)
\nx + 110 = 180
\n\u21d2 x = 180 \u2013 110 = 70.<\/p>\n
\n
\nSolution: \u00a0 \u00a0<\/strong>Since \u2220AOC and \u2220BOC form a linear pair.
\n\u2234 \u2220AOC + \u2220BOC = 180\u00ba
\n\u21d2 4x + 2x = 180\u00ba
\n\u21d2 6x = 180\u00ba
\n\u21d2 x = 180\/6 = 30\u00ba
\nThus, x = 30\u00ba<\/p>\n
\n\u2220AOC + \u2220BOD = 90\u00ba. Find \u2220COD.
\n
\nSolution: \u00a0 \u00a0<\/strong>Since OA and OB are opposite rays. Therefore, AB is a line. Since ray OC stands on line AB. Therefore,
\n\u2220AOC + \u2220COB = 180\u00ba \u00a0 \u00a0 \u00a0[Linear Pairs]
\n\u21d2 \u2220AOC + \u2220COD + \u2220BOD = 180\u00ba
\n[\u2235 \u2220COB = \u2220COD + \u2220BOD]
\n\u21d2 (\u2220AOC + \u2220BOD) + \u2220COD = 180\u00ba
\n\u21d2 90\u00ba + \u2220COD = 180\u00ba
\n[\u2235 \u2220AOC + \u2220BOD = 90\u00ba (Given)]
\n\u21d2 \u2220COD = 180\u00ba \u2013 90\u00ba
\n\u21d2 \u2220COD = 90\u00ba<\/p>\n
\n\u2220FOB = \u2220FOA.
\n
\nSolution: \u00a0 \u00a0<\/strong>Since ray OE bisects angle AOB. Therefore,
\n\u2220EOB = \u2220EOA ….(i)
\nNow, ray OB stands on the line EF.
\n\u2234 \u2220EOB + \u2220FOB = 180\u00ba \u00a0 \u00a0 …(ii) \u00a0 \u00a0 \u00a0 [linear pair]
\nAgain, ray OA stands on the line EF.
\n\u2234 \u2220EOA + \u2220FOA = 180\u00ba \u00a0 \u00a0….(iii)
\nForm (ii) and (iii), we get
\n\u2220EOB + \u2220FOB = \u2220EOA + \u2220FOA
\n\u21d2 \u2220EOA + \u2220FOB = \u2220EOA + \u2220FOA
\n[\u2235 \u2220EOB = \u2220EOA (from (i)]
\n\u21d2 \u2220FOB = \u2220FOA.<\/p>\n
\n
\nSolution: \u00a0 \u00a0<\/strong>Since OE and OF bisect angles AOC and COB respectively. Therefore,
\n\u2220AOC = 2\u2220EOC ….(i)
\nand \u2220COB = 2\u2220COF ….(ii)
\nAdding (i) and (ii), we get
\n\u2220AOC + \u2220COB = 2\u2220EOC + 2\u2220COF
\n\u21d2 \u2220AOC + \u2220COB = 2(\u2220EOC + \u2220COF)
\n\u21d2 \u2220AOC + \u2220COB = 2(\u2220EOF)
\n\u21d2 \u2220AOC + \u2220COB = 2 \u00d7 90\u00ba
\n[\u2235 OE \u22a5 OF \u2234 \u2220EOF = 90\u00ba]
\n\u21d2 \u2220AOC + \u2220COB = 180\u00ba
\nBut \u2220AOC and \u2220COB are adjacent angles.
\nThus, \u2220AOC and \u2220COB are adjacent supplementary angles. So, \u2220AOC and \u2220COB form a linear pair. Consequently OA and OB are two opposite rays. Hence, A, O, B are collinear.<\/p>\n
\n\u2220AOC = \u2220COB, then show that
\n\u2220AOC = 90\u00ba.
\nSolution: \u00a0 \u00a0<\/strong>Since ray OC stands on line AB. Therefore,
\n\u2220AOC + \u2220COB = 180\u00ba [Linear pair] …(i)
\n
\nBut \u2220AOC = \u2220COB \u00a0 \u00a0 (Given)
\n\u2234 \u2220AOC + \u2220 OC = 180\u00ba
\n\u21d2 2\u2220AOC = 180\u00ba
\n\u21d2 \u2220AOC = 90\u00ba<\/p>\n
\n
\nSolution: \u00a0 \u00a0<\/strong>\u2220AOC + \u2220COD + \u2220BOD = 180\u00ba
\nor \u00a0 (\u2220AOC + \u2220BOD) + \u2220COD = 180\u00ba
\nor \u00a0 70\u00ba + \u2220COD = 180\u00ba
\nor \u00a0 \u2220COD = 180\u00ba \u2013 70\u00ba
\nor \u00a0 \u2220COD = 110\u00ba<\/p>\n
\n
\nSolution: \u00a0 \u00a0<\/strong>2y + 3y + 5y = 180\u00ba
\n\u21d2 10y = 180\u00ba
\n\u21d2 y = 180\u00b0\/10\u00ba = 18\u00ba<\/p>\n