{"id":512,"date":"2020-08-17T04:02:48","date_gmt":"2020-08-17T04:02:48","guid":{"rendered":"http:\/\/themindpalace.in\/?p=512"},"modified":"2021-08-28T09:41:55","modified_gmt":"2021-08-28T09:41:55","slug":"lines-and-angles","status":"publish","type":"post","link":"https:\/\/themindpalace.in\/index.php\/2020\/08\/17\/lines-and-angles\/","title":{"rendered":"Lines and Angles"},"content":{"rendered":"\n<p><a href=\"#summary\">Summary of lines and angles<\/a><\/p>\n\n\n\n<p><a href=\"#solved exercise\">Solved exercise of lines and angles<\/a><\/p>\n\n\n\n<figure class=\"wp-block-embed is-type-video is-provider-youtube wp-block-embed-youtube wp-embed-aspect-16-9 wp-has-aspect-ratio\"><div class=\"wp-block-embed__wrapper\">\n<div class=\"epyt-video-wrapper\"><iframe loading=\"lazy\"  id=\"_ytid_46104\"  width=\"480\" height=\"270\"  data-origwidth=\"480\" data-origheight=\"270\" src=\"https:\/\/www.youtube.com\/embed\/2XRviYYOooA?enablejsapi=1&#038;autoplay=0&#038;cc_load_policy=0&#038;cc_lang_pref=&#038;iv_load_policy=1&#038;loop=0&#038;rel=1&#038;fs=1&#038;playsinline=0&#038;autohide=2&#038;theme=dark&#038;color=red&#038;controls=1&#038;disablekb=0&#038;\" class=\"__youtube_prefs__  epyt-is-override  no-lazyload\" title=\"YouTube player\"  allow=\"fullscreen; accelerometer; autoplay; clipboard-write; encrypted-media; gyroscope; picture-in-picture; web-share\" referrerpolicy=\"strict-origin-when-cross-origin\" allowfullscreen data-no-lazy=\"1\" data-skipgform_ajax_framebjll=\"\"><\/iframe><\/div>\n<\/div><\/figure>\n\n\n\n<h1 class=\"wp-block-heading\">Summary<\/h1>\n\n\n\n<p>This chapter tells you to find out the properties of angel formed when two lines intersect each other and also angles formed when two or more line intersect at distinct point.<\/p>\n\n\n\n<h4 class=\"wp-block-heading\"><strong><span class=\"has-inline-color has-luminous-vivid-orange-color\">Uses of Lines and angels<\/span><\/strong><\/h4>\n\n\n\n<p>Suppose you want to build a hut to keep it in your school exhibition using a bamboo sticks. We should keep some sticks&nbsp; parallel to each other and some slanted&nbsp; so with the help of these angles and lines we can do this.<\/p>\n\n\n\n<p>To study the refraction of property of light from one medium to another we use the property of intersecting line and parallel line.<\/p>\n\n\n\n<div class=\"wp-block-image\"><figure class=\"aligncenter size-large is-resized\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/themindpalace.in\/wp-content\/uploads\/2020\/08\/bamboo-house.jpg\" alt=\"\" class=\"wp-image-513\" width=\"335\" height=\"223\" srcset=\"https:\/\/themindpalace.in\/wp-content\/uploads\/2020\/08\/bamboo-house.jpg 1024w, https:\/\/themindpalace.in\/wp-content\/uploads\/2020\/08\/bamboo-house-300x200.jpg 300w, https:\/\/themindpalace.in\/wp-content\/uploads\/2020\/08\/bamboo-house-768x512.jpg 768w\" sizes=\"(max-width: 335px) 100vw, 335px\" \/><figcaption>bamboo house<\/figcaption><\/figure><\/div>\n\n\n\n<h4 class=\"wp-block-heading\"><strong><span class=\"has-inline-color has-vivid-purple-color\">Some of the basic terms and definition<\/span><\/strong><\/h4>\n\n\n\n<p><strong>Point: <\/strong>A point is shown by dot , an exact location in space.<\/p>\n\n\n\n<div class=\"wp-block-image\"><figure class=\"aligncenter size-large is-resized\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/themindpalace.in\/wp-content\/uploads\/2020\/08\/point.jpg\" alt=\"\" class=\"wp-image-514\" width=\"279\" height=\"254\" srcset=\"https:\/\/themindpalace.in\/wp-content\/uploads\/2020\/08\/point.jpg 407w, https:\/\/themindpalace.in\/wp-content\/uploads\/2020\/08\/point-300x273.jpg 300w\" sizes=\"(max-width: 279px) 100vw, 279px\" \/><figcaption>points<\/figcaption><\/figure><\/div>\n\n\n\n<p><strong><span style=\"text-decoration: underline;\"><span class=\"has-inline-color has-vivid-purple-color\">Line<\/span><\/span><\/strong>: A line is made of points, a line&nbsp; is denoted by l, m, n.<\/p>\n\n\n\n<div class=\"wp-block-image\"><figure class=\"aligncenter size-large is-resized\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/themindpalace.in\/wp-content\/uploads\/2020\/08\/line.jpg\" alt=\"\" class=\"wp-image-515\" width=\"526\" height=\"24\" srcset=\"https:\/\/themindpalace.in\/wp-content\/uploads\/2020\/08\/line.jpg 526w, https:\/\/themindpalace.in\/wp-content\/uploads\/2020\/08\/line-300x14.jpg 300w, https:\/\/themindpalace.in\/wp-content\/uploads\/2020\/08\/line-520x24.jpg 520w\" sizes=\"(max-width: 526px) 100vw, 526px\" \/><figcaption>line<\/figcaption><\/figure><\/div>\n\n\n\n<p><strong><span class=\"has-inline-color has-vivid-purple-color\">Collinear&nbsp; points<\/span><\/strong> If two or more points lies on the same line .<\/p>\n\n\n\n<div class=\"wp-block-image\"><figure class=\"aligncenter size-large\"><img loading=\"lazy\" decoding=\"async\" width=\"534\" height=\"24\" src=\"https:\/\/themindpalace.in\/wp-content\/uploads\/2020\/08\/COLLINEAR-POINT.jpg\" alt=\"\" class=\"wp-image-516\" srcset=\"https:\/\/themindpalace.in\/wp-content\/uploads\/2020\/08\/COLLINEAR-POINT.jpg 534w, https:\/\/themindpalace.in\/wp-content\/uploads\/2020\/08\/COLLINEAR-POINT-300x13.jpg 300w, https:\/\/themindpalace.in\/wp-content\/uploads\/2020\/08\/COLLINEAR-POINT-520x24.jpg 520w\" sizes=\"(max-width: 534px) 100vw, 534px\" \/><figcaption>collinear point<\/figcaption><\/figure><\/div>\n\n\n\n<p><strong><span class=\"has-inline-color has-vivid-purple-color\">Non Collinear points<\/span><\/strong> If two or more points &nbsp;which do not lies on the same line.<\/p>\n\n\n\n<div class=\"wp-block-image\"><figure class=\"aligncenter size-large is-resized\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/themindpalace.in\/wp-content\/uploads\/2020\/08\/non-collinear-point.jpg\" alt=\"\" class=\"wp-image-517\" width=\"225\" height=\"148\"\/><figcaption>non collinear<\/figcaption><\/figure><\/div>\n\n\n\n<p><strong><span class=\"has-inline-color has-vivid-purple-color\">line segment<\/span><\/strong>: A line with two end points is called line segment.<\/p>\n\n\n\n<div class=\"wp-block-image\"><figure class=\"aligncenter size-large is-resized\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/themindpalace.in\/wp-content\/uploads\/2020\/08\/line-seg2.jpg\" alt=\"\" class=\"wp-image-520\" width=\"405\" height=\"53\" srcset=\"https:\/\/themindpalace.in\/wp-content\/uploads\/2020\/08\/line-seg2.jpg 631w, https:\/\/themindpalace.in\/wp-content\/uploads\/2020\/08\/line-seg2-300x39.jpg 300w\" sizes=\"(max-width: 405px) 100vw, 405px\" \/><figcaption>line segment<\/figcaption><\/figure><\/div>\n\n\n\n<p><strong><span class=\"has-inline-color has-vivid-purple-color\">&nbsp;Ray<\/span><\/strong>: A part of line with one end is called as ray.<\/p>\n\n\n\n<div class=\"wp-block-image\"><figure class=\"aligncenter size-large\"><img loading=\"lazy\" decoding=\"async\" width=\"621\" height=\"38\" src=\"https:\/\/themindpalace.in\/wp-content\/uploads\/2020\/08\/RAY.jpg\" alt=\"\" class=\"wp-image-521\" srcset=\"https:\/\/themindpalace.in\/wp-content\/uploads\/2020\/08\/RAY.jpg 621w, https:\/\/themindpalace.in\/wp-content\/uploads\/2020\/08\/RAY-300x18.jpg 300w\" sizes=\"(max-width: 621px) 100vw, 621px\" \/><figcaption>ray<\/figcaption><\/figure><\/div>\n\n\n\n<h3 class=\"wp-block-heading\"><strong><u><span class=\"has-inline-color has-vivid-red-color\">Notation<\/span><\/u><\/strong><\/h3>\n\n\n\n<p>Point<\/p>\n\n\n\n<div class=\"wp-block-image\"><figure class=\"aligncenter size-large is-resized\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/themindpalace.in\/wp-content\/uploads\/2020\/08\/point-1.jpg\" alt=\"\" class=\"wp-image-522\" width=\"220\" height=\"201\" srcset=\"https:\/\/themindpalace.in\/wp-content\/uploads\/2020\/08\/point-1.jpg 407w, https:\/\/themindpalace.in\/wp-content\/uploads\/2020\/08\/point-1-300x273.jpg 300w\" sizes=\"(max-width: 220px) 100vw, 220px\" \/><figcaption>point<\/figcaption><\/figure><\/div>\n\n\n\n<div class=\"wp-block-image\"><figure class=\"aligncenter size-large is-resized\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/themindpalace.in\/wp-content\/uploads\/2020\/08\/LINE-SEGMENT-NOTITION.jpg\" alt=\"\" class=\"wp-image-524\" width=\"241\" height=\"224\" srcset=\"https:\/\/themindpalace.in\/wp-content\/uploads\/2020\/08\/LINE-SEGMENT-NOTITION.jpg 408w, https:\/\/themindpalace.in\/wp-content\/uploads\/2020\/08\/LINE-SEGMENT-NOTITION-300x279.jpg 300w\" sizes=\"(max-width: 241px) 100vw, 241px\" \/><\/figure><\/div>\n\n\n\n<div class=\"wp-block-image\"><figure class=\"aligncenter size-large is-resized\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/themindpalace.in\/wp-content\/uploads\/2020\/08\/ray-notation.jpg\" alt=\"\" class=\"wp-image-525\" width=\"618\" height=\"269\" srcset=\"https:\/\/themindpalace.in\/wp-content\/uploads\/2020\/08\/ray-notation.jpg 869w, https:\/\/themindpalace.in\/wp-content\/uploads\/2020\/08\/ray-notation-300x131.jpg 300w, https:\/\/themindpalace.in\/wp-content\/uploads\/2020\/08\/ray-notation-768x335.jpg 768w\" sizes=\"(max-width: 618px) 100vw, 618px\" \/><\/figure><\/div>\n\n\n\n<div class=\"wp-block-image\"><figure class=\"aligncenter size-large is-resized\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/themindpalace.in\/wp-content\/uploads\/2020\/08\/LINE-SEGMENT-NOTITION-1.jpg\" alt=\"\" class=\"wp-image-526\" width=\"265\" height=\"246\" srcset=\"https:\/\/themindpalace.in\/wp-content\/uploads\/2020\/08\/LINE-SEGMENT-NOTITION-1.jpg 408w, https:\/\/themindpalace.in\/wp-content\/uploads\/2020\/08\/LINE-SEGMENT-NOTITION-1-300x279.jpg 300w\" sizes=\"(max-width: 265px) 100vw, 265px\" \/><\/figure><\/div>\n\n\n\n<p><strong>Angle and there types<\/strong><\/p>\n\n\n\n<p><strong>Angle:<\/strong> An angle is formed when two rays&nbsp; originate from the same end point.<\/p>\n\n\n\n<div class=\"wp-block-image\"><figure class=\"aligncenter size-large is-resized\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/themindpalace.in\/wp-content\/uploads\/2020\/08\/angles-org.jpg\" alt=\"\" class=\"wp-image-529\" width=\"356\" height=\"220\" srcset=\"https:\/\/themindpalace.in\/wp-content\/uploads\/2020\/08\/angles-org.jpg 792w, https:\/\/themindpalace.in\/wp-content\/uploads\/2020\/08\/angles-org-300x186.jpg 300w, https:\/\/themindpalace.in\/wp-content\/uploads\/2020\/08\/angles-org-768x476.jpg 768w\" sizes=\"(max-width: 356px) 100vw, 356px\" \/><figcaption>angles<br><\/figcaption><\/figure><\/div>\n\n\n\n<p><strong><span style=\"color:#b61197\" class=\"has-inline-color\">The different types of angles are<\/span><\/strong><\/p>\n\n\n\n<p><strong>RIGHT ANGLE:<\/strong> An angle that is exactly 90\u02da<\/p>\n\n\n\n<div class=\"wp-block-image\"><figure class=\"aligncenter size-large is-resized\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/themindpalace.in\/wp-content\/uploads\/2020\/08\/right-angle.jpg\" alt=\"\" class=\"wp-image-530\" width=\"399\" height=\"380\" srcset=\"https:\/\/themindpalace.in\/wp-content\/uploads\/2020\/08\/right-angle.jpg 906w, https:\/\/themindpalace.in\/wp-content\/uploads\/2020\/08\/right-angle-300x286.jpg 300w, https:\/\/themindpalace.in\/wp-content\/uploads\/2020\/08\/right-angle-768x732.jpg 768w\" sizes=\"(max-width: 399px) 100vw, 399px\" \/><figcaption>right angle<\/figcaption><\/figure><\/div>\n\n\n\n<p><strong>ACUTE ANGLE:<\/strong> An angle that is less than 90\u02da<\/p>\n\n\n\n<div class=\"wp-block-image\"><figure class=\"aligncenter size-large is-resized\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/themindpalace.in\/wp-content\/uploads\/2020\/08\/acute-angle.jpg\" alt=\"\" class=\"wp-image-531\" width=\"300\" height=\"286\" srcset=\"https:\/\/themindpalace.in\/wp-content\/uploads\/2020\/08\/acute-angle.jpg 906w, https:\/\/themindpalace.in\/wp-content\/uploads\/2020\/08\/acute-angle-300x286.jpg 300w, https:\/\/themindpalace.in\/wp-content\/uploads\/2020\/08\/acute-angle-768x732.jpg 768w\" sizes=\"(max-width: 300px) 100vw, 300px\" \/><figcaption>acute angle<\/figcaption><\/figure><\/div>\n\n\n\n<p><strong>OBTUSE ANGLE:<\/strong> An angle that is greater than90\u02daand less than 180.<\/p>\n\n\n\n<div class=\"wp-block-image\"><figure class=\"aligncenter size-large is-resized\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/themindpalace.in\/wp-content\/uploads\/2020\/08\/obtuse-angle.jpg\" alt=\"\" class=\"wp-image-532\" width=\"391\" height=\"373\" srcset=\"https:\/\/themindpalace.in\/wp-content\/uploads\/2020\/08\/obtuse-angle.jpg 906w, https:\/\/themindpalace.in\/wp-content\/uploads\/2020\/08\/obtuse-angle-300x286.jpg 300w, https:\/\/themindpalace.in\/wp-content\/uploads\/2020\/08\/obtuse-angle-768x732.jpg 768w\" sizes=\"(max-width: 391px) 100vw, 391px\" \/><figcaption>obtuse angle<br><\/figcaption><\/figure><\/div>\n\n\n\n<p><strong>STRAIGHT ANGLE \/ STRAIGHT LINE:<\/strong> An angle that is exactly 180 \u02da<\/p>\n\n\n\n<div class=\"wp-block-image\"><figure class=\"aligncenter size-large is-resized\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/themindpalace.in\/wp-content\/uploads\/2020\/08\/straight-angle.jpg\" alt=\"\" class=\"wp-image-533\" width=\"471\" height=\"449\" srcset=\"https:\/\/themindpalace.in\/wp-content\/uploads\/2020\/08\/straight-angle.jpg 906w, https:\/\/themindpalace.in\/wp-content\/uploads\/2020\/08\/straight-angle-300x286.jpg 300w, https:\/\/themindpalace.in\/wp-content\/uploads\/2020\/08\/straight-angle-768x732.jpg 768w\" sizes=\"(max-width: 471px) 100vw, 471px\" \/><figcaption>Straight line angle<\/figcaption><\/figure><\/div>\n\n\n\n<p><strong>REFLEX ANGLE:<\/strong> An angle that is greater than 180\u02da<br>and less than 360 \u02da<\/p>\n\n\n\n<p>The sum of two adjacent angels form a pair of 180<sup>0 <\/sup>is called linear pair of angels.<\/p>\n\n\n\n<div class=\"wp-block-image\"><figure class=\"aligncenter size-large is-resized\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/themindpalace.in\/wp-content\/uploads\/2020\/08\/reflex-angle.jpg\" alt=\"\" class=\"wp-image-534\" width=\"332\" height=\"317\" srcset=\"https:\/\/themindpalace.in\/wp-content\/uploads\/2020\/08\/reflex-angle.jpg 906w, https:\/\/themindpalace.in\/wp-content\/uploads\/2020\/08\/reflex-angle-300x286.jpg 300w, https:\/\/themindpalace.in\/wp-content\/uploads\/2020\/08\/reflex-angle-768x732.jpg 768w\" sizes=\"(max-width: 332px) 100vw, 332px\" \/><figcaption>reflex angle<\/figcaption><\/figure><\/div>\n\n\n\n<p><strong>FULL ANGLE:<\/strong> An angle that is exactly 360 \u02da<\/p>\n\n\n\n<div class=\"wp-block-image\"><figure class=\"aligncenter size-large is-resized\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/themindpalace.in\/wp-content\/uploads\/2020\/08\/full-angle.jpg\" alt=\"\" class=\"wp-image-535\" width=\"473\" height=\"451\" srcset=\"https:\/\/themindpalace.in\/wp-content\/uploads\/2020\/08\/full-angle.jpg 906w, https:\/\/themindpalace.in\/wp-content\/uploads\/2020\/08\/full-angle-300x286.jpg 300w, https:\/\/themindpalace.in\/wp-content\/uploads\/2020\/08\/full-angle-768x732.jpg 768w\" sizes=\"(max-width: 473px) 100vw, 473px\" \/><figcaption>full angle<\/figcaption><\/figure><\/div>\n\n\n\n<h4 class=\"wp-block-heading\"><span class=\"has-inline-color has-luminous-vivid-orange-color\">The different types of lines are<\/span><\/h4>\n\n\n\n<ol type=\"1\"><li>PARALLEL LINES: They do not intersect when produced indefinitely in either direction<\/li><\/ol>\n\n\n\n<div class=\"wp-block-image\"><figure class=\"aligncenter size-large is-resized\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/themindpalace.in\/wp-content\/uploads\/2020\/08\/PARALLEL-LINE.jpg\" alt=\"\" class=\"wp-image-537\" width=\"530\" height=\"130\" srcset=\"https:\/\/themindpalace.in\/wp-content\/uploads\/2020\/08\/PARALLEL-LINE.jpg 760w, https:\/\/themindpalace.in\/wp-content\/uploads\/2020\/08\/PARALLEL-LINE-300x74.jpg 300w\" sizes=\"(max-width: 530px) 100vw, 530px\" \/><figcaption>parallel lines<\/figcaption><\/figure><\/div>\n\n\n\n<ol type=\"1\"><li>INTERSECTING LINES: Are lines that meet a point<\/li><\/ol>\n\n\n\n<div class=\"wp-block-image\"><figure class=\"aligncenter size-large is-resized\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/themindpalace.in\/wp-content\/uploads\/2020\/08\/INTERSECTING-LINE.jpg\" alt=\"\" class=\"wp-image-538\" width=\"531\" height=\"193\" srcset=\"https:\/\/themindpalace.in\/wp-content\/uploads\/2020\/08\/INTERSECTING-LINE.jpg 760w, https:\/\/themindpalace.in\/wp-content\/uploads\/2020\/08\/INTERSECTING-LINE-300x109.jpg 300w\" sizes=\"(max-width: 531px) 100vw, 531px\" \/><figcaption>intersecting line<\/figcaption><\/figure><\/div>\n\n\n\n<p>PERPENDICULAR LINES: They intersect each other at right angles<\/p>\n\n\n\n<div class=\"wp-block-image\"><figure class=\"aligncenter size-large is-resized\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/themindpalace.in\/wp-content\/uploads\/2020\/08\/perpendicular-org.jpg\" alt=\"\" class=\"wp-image-540\" width=\"520\" height=\"155\" srcset=\"https:\/\/themindpalace.in\/wp-content\/uploads\/2020\/08\/perpendicular-org.jpg 760w, https:\/\/themindpalace.in\/wp-content\/uploads\/2020\/08\/perpendicular-org-300x89.jpg 300w\" sizes=\"(max-width: 520px) 100vw, 520px\" \/><figcaption>perpendicular line<\/figcaption><\/figure><\/div>\n\n\n\n<p>ADJACENT ANGLES:Share a common ray and a common vertex.<\/p>\n\n\n\n<div class=\"wp-block-image\"><figure class=\"aligncenter size-large is-resized\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/themindpalace.in\/wp-content\/uploads\/2020\/08\/ADJACENT-ANGLES.jpg\" alt=\"\" class=\"wp-image-541\" width=\"386\" height=\"211\" srcset=\"https:\/\/themindpalace.in\/wp-content\/uploads\/2020\/08\/ADJACENT-ANGLES.jpg 931w, https:\/\/themindpalace.in\/wp-content\/uploads\/2020\/08\/ADJACENT-ANGLES-300x165.jpg 300w, https:\/\/themindpalace.in\/wp-content\/uploads\/2020\/08\/ADJACENT-ANGLES-768x422.jpg 768w\" sizes=\"(max-width: 386px) 100vw, 386px\" \/><figcaption>adjacent angle<\/figcaption><\/figure><\/div>\n\n\n\n<p>VERTICAL ANGLES: Are pairs of opposite angles made by intersecting lines.<\/p>\n\n\n\n<div class=\"wp-block-image\"><figure class=\"aligncenter size-large is-resized\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/themindpalace.in\/wp-content\/uploads\/2020\/08\/VERTICAL-ANGLES.jpg\" alt=\"\" class=\"wp-image-542\" width=\"469\" height=\"263\" srcset=\"https:\/\/themindpalace.in\/wp-content\/uploads\/2020\/08\/VERTICAL-ANGLES.jpg 910w, https:\/\/themindpalace.in\/wp-content\/uploads\/2020\/08\/VERTICAL-ANGLES-300x168.jpg 300w, https:\/\/themindpalace.in\/wp-content\/uploads\/2020\/08\/VERTICAL-ANGLES-768x431.jpg 768w\" sizes=\"(max-width: 469px) 100vw, 469px\" \/><figcaption>vertical angles<\/figcaption><\/figure><\/div>\n\n\n\n<h2 class=\"wp-block-heading\" id=\"Solved-exercise\"><strong><span class=\"has-inline-color has-vivid-purple-color\">Solved Exercise 3.1<\/span><\/strong><\/h2>\n\n\n\n<p>1. In figure, lines AB and CD intersect at 0. If \u2220AOC + \u2220BOE = 70\u00b0 and \u2220BOD = 40\u00b0, find \u2220BOE and reflex \u2220COE.<\/p>\n\n\n\n<p><strong><span class=\"has-inline-color has-vivid-red-color\">Given<\/span>:<\/strong> \u2220AOC + \u2220BOE = 70\u00b0<\/p>\n\n\n\n<div class=\"wp-block-image\"><figure class=\"aligncenter size-large\"><img loading=\"lazy\" decoding=\"async\" width=\"376\" height=\"257\" src=\"https:\/\/themindpalace.in\/wp-content\/uploads\/2020\/09\/1-pic.jpg\" alt=\"\" class=\"wp-image-925\" srcset=\"https:\/\/themindpalace.in\/wp-content\/uploads\/2020\/09\/1-pic.jpg 376w, https:\/\/themindpalace.in\/wp-content\/uploads\/2020\/09\/1-pic-300x205.jpg 300w\" sizes=\"(max-width: 376px) 100vw, 376px\" \/><\/figure><\/div>\n\n\n\n<p>&nbsp;&nbsp; \u2220BOD = 40\u00b0<\/p>\n\n\n\n<p><strong>To find: <\/strong>\u2220BOE and Reflex \u2220COE &nbsp;<\/p>\n\n\n\n<p><span class=\"has-inline-color has-vivid-purple-color\">Solution:<\/span><br>Since AB is a straight line,<\/p>\n\n\n\n<p>\u2234 \u2220AOC + \u2220COE + \u2220EOB = 180\u00b0<\/p>\n\n\n\n<p>(\u2220AOC + \u2220BOE) + \u2220COE = 180<\/p>\n\n\n\n<p>70\u00b0 + \u2220COE = 180\u00b0&nbsp;<\/p>\n\n\n\n<p>[(Given)]<\/p>\n\n\n\n<p>\u2220COE = 180\u00b0 \u2013 70\u00b0<\/p>\n\n\n\n<p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; = 110\u00b0<\/p>\n\n\n\n<p>\u2234 Reflex \u2220COE = 360\u00b0 \u2013 110\u00b0<\/p>\n\n\n\n<p>&nbsp; &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; = 250\u00b0<\/p>\n\n\n\n<p>AB and CD intersect at O<\/p>\n\n\n\n<div class=\"wp-block-image\"><figure class=\"aligncenter size-large\"><img loading=\"lazy\" decoding=\"async\" width=\"349\" height=\"249\" src=\"https:\/\/themindpalace.in\/wp-content\/uploads\/2020\/09\/2-pic.jpg\" alt=\"\" class=\"wp-image-926\" srcset=\"https:\/\/themindpalace.in\/wp-content\/uploads\/2020\/09\/2-pic.jpg 349w, https:\/\/themindpalace.in\/wp-content\/uploads\/2020\/09\/2-pic-300x214.jpg 300w\" sizes=\"(max-width: 349px) 100vw, 349px\" \/><\/figure><\/div>\n\n\n\n<p>\u2234\u2220COA = \u2220BOD [Vertically opposite angles<\/p>\n\n\n\n<p>But \u2220BOD = 40\u00b0 [Given]<\/p>\n\n\n\n<p>\u2234 \u2220COA = 40\u00b0<\/p>\n\n\n\n<p>Also, \u2220AOC + \u2220BOE = 70\u00b0<\/p>\n\n\n\n<p>\u2234 40\u00b0 + \u2220 BOE = 70\u00b0<\/p>\n\n\n\n<p>&nbsp;&nbsp;&nbsp;&nbsp; \u2220BOE = 70\u00b0 -40\u00b0 = 30\u00b0<\/p>\n\n\n\n<p><strong>Thus, \u2220BOE = 30\u00b0 and reflex \u2220COE = 250\u00b0.<\/strong><\/p>\n\n\n\n<p><strong><span class=\"has-inline-color has-very-dark-gray-color\">2. In figure, lines XY and MN intersect at 0. If \u2220POY = 90\u00b0 , and a : b = 2 : 3. find c.<\/span><\/strong><\/p>\n\n\n\n<div class=\"wp-block-image\"><figure class=\"aligncenter size-large\"><img loading=\"lazy\" decoding=\"async\" width=\"242\" height=\"217\" src=\"https:\/\/themindpalace.in\/wp-content\/uploads\/2020\/09\/2-prb-1.jpg\" alt=\"\" class=\"wp-image-928\"\/><\/figure><\/div>\n\n\n\n<p><strong>Given:&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; <\/strong>a : b = 2 : 3 , \u2220POY = 90\u00b0<\/p>\n\n\n\n<p><strong>To find:&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; <\/strong>a, b, and c<\/p>\n\n\n\n<p><strong><span class=\"has-inline-color has-vivid-purple-color\">Solution<\/span><\/strong><\/p>\n\n\n\n<p>Since XOY is a straight line<\/p>\n\n\n\n<p>\u2234 b+a+\u2220POY= 180\u00b0 [ Linear pair]<\/p>\n\n\n\n<p>But \u2220POY = 90\u00b0 [Given]<\/p>\n\n\n\n<p>\u2234 b + a = 180\u00b0 \u2013 90\u00b0<\/p>\n\n\n\n<p>&nbsp;&nbsp; b + a = 90\u00b0<\/p>\n\n\n\n<p>Let a=2x and b= 3x<\/p>\n\n\n\n<p>Then we have<\/p>\n\n\n\n<p>\u21d2 2x + 3x = 90<sup>0<\/sup><\/p>\n\n\n\n<p>\u21d2 5x= 90<sup>0<\/sup><\/p>\n\n\n\n<p>X=90\/5<\/p>\n\n\n\n<p><strong>X= 18<sup>0<\/sup><\/strong><\/p>\n\n\n\n<p>Then a= 2x<\/p>\n\n\n\n<p>b= 3x Now c= a + 90<sup>0<\/sup> [V O A]<\/p>\n\n\n\n<p>=2 X 18<sup>0<\/sup>= 36<sup>0<\/sup><\/p>\n\n\n\n<p>= 3 x 18<sup>0 <\/sup>=54<sup>0<\/sup><\/p>\n\n\n\n<p>C= 36<sup>0<\/sup> + 90<sup>0<\/sup> =126<sup>0<\/sup><\/p>\n\n\n\n<p><strong>\u2234 a = 36<sup>0<\/sup> , b =&nbsp; 54<sup>0<\/sup> and C=126<sup>0<\/sup><\/strong><\/p>\n\n\n\n<p>3. In figure, \u2220PQR = \u2220PRQ, then prove that \u2220PQS = \u2220PRT.<\/p>\n\n\n\n<div class=\"wp-block-image\"><figure class=\"aligncenter size-large\"><img loading=\"lazy\" decoding=\"async\" width=\"268\" height=\"213\" src=\"https:\/\/themindpalace.in\/wp-content\/uploads\/2020\/09\/3-prb.jpg\" alt=\"\" class=\"wp-image-930\"\/><\/figure><\/div>\n\n\n\n<p><strong>Given:<\/strong>\u2220PQR = \u2220PRQ<\/p>\n\n\n\n<p><strong>To prove:<\/strong>\u2220PQS = \u2220PRT<\/p>\n\n\n\n<p><strong>Proof<\/strong>:ST is a straight line.<\/p>\n\n\n\n<p>\u2234 \u2220PQR + \u2220PQS = 180\u00b0 \u2026(1) [Linear pair]<\/p>\n\n\n\n<p>Similarly, \u2220PRT + \u2220PRQ = 180\u00b0 \u2026(2) [Linear Pair]<\/p>\n\n\n\n<p>From (1) and (2), we have<\/p>\n\n\n\n<p>\u2220PQS + \u2220PQR = \u2220PRT + \u2220PRQ<\/p>\n\n\n\n<p>But \u2220PQR = \u2220PRQ [Given<\/p>\n\n\n\n<p>\u2234 \u2220PQS = \u2220PRT<\/p>\n\n\n\n<p>4. In figure, if x + y = w + \u21d2, then prove that AOB is a line<\/p>\n\n\n\n<div class=\"wp-block-image\"><figure class=\"aligncenter size-large\"><img loading=\"lazy\" decoding=\"async\" width=\"205\" height=\"233\" src=\"https:\/\/themindpalace.in\/wp-content\/uploads\/2020\/09\/prb-4.jpg\" alt=\"\" class=\"wp-image-932\"\/><\/figure><\/div>\n\n\n\n<p><strong>Given: <\/strong>x + y = z + w<\/p>\n\n\n\n<p>T<strong>o prove:<\/strong>\u2234 AOB is a straight line<\/p>\n\n\n\n<p><strong>Proof:<\/strong>Sum of all the angles at a point = 360\u00b0<\/p>\n\n\n\n<p>x+ y+ z + w = 360<sup>0<\/sup> [complete angle]<\/p>\n\n\n\n<p>x + y + x + y = 360<sup>0<\/sup> [x + y = z + w]<\/p>\n\n\n\n<p>2x + 2y = 360<sup>0<\/sup><\/p>\n\n\n\n<p>2(x + y) = 360<sup>0<\/sup><\/p>\n\n\n\n<p>(x + y) =360<sup>0<\/sup>\/2<\/p>\n\n\n\n<p>(x + y) = 180<sup>0<\/sup><\/p>\n\n\n\n<p>By linear pair axioms AOB&nbsp; is a straight line<\/p>\n\n\n\n<p>5. In figure, POQ is a line. Ray OR is perpendicular to line PQ. OS is another ray lying between rays OP and OR. Prove that<\/p>\n\n\n\n<div class=\"wp-block-image\"><figure class=\"aligncenter size-large\"><img loading=\"lazy\" decoding=\"async\" width=\"235\" height=\"152\" src=\"https:\/\/themindpalace.in\/wp-content\/uploads\/2020\/09\/5-prb.jpg\" alt=\"\" class=\"wp-image-933\"\/><\/figure><\/div>\n\n\n\n<figure class=\"wp-block-image size-large\"><img loading=\"lazy\" decoding=\"async\" width=\"257\" height=\"58\" src=\"https:\/\/themindpalace.in\/wp-content\/uploads\/2020\/09\/image-4.png\" alt=\"\" class=\"wp-image-934\"\/><\/figure>\n\n\n\n<p><strong>Given<\/strong><\/p>\n\n\n\n<p><strong>To prove<\/strong><\/p>\n\n\n\n<p><strong>Proof:<\/strong><\/p>\n\n\n\n<p>POQ is a straight line. [Given]<\/p>\n\n\n\n<p>POQ is a straight line. [Given]<\/p>\n\n\n\n<p>But OR \u22a5 PQ   \u2234 \u2220ROQ = 90\u00b0<\/p>\n\n\n\n<p>\u21d2 \u2220POS + \u2220ROS + 90\u00b0 = 180\u00b0<\/p>\n\n\n\n<p>\u21d2 \u2220POS + \u2220ROS = 90\u00b0<\/p>\n\n\n\n<p>\u21d2 \u2220ROS = 90\u00b0 \u2013 \u2220POS \u2026 (1)<\/p>\n\n\n\n<p>Now, we have \u2220ROS + \u2220ROQ = \u2220QOS<\/p>\n\n\n\n<p>\u21d2 \u2220ROS + 90\u00b0 = \u2220QOS<\/p>\n\n\n\n<p>\u21d2 \u2220ROS = \u2220QOS \u2013 90\u00b0 \u2026\u2026(2)<\/p>\n\n\n\n<p>Adding (1) and (2), we have<\/p>\n\n\n\n<p>2 \u2220ROS = (\u2220QOS \u2013 \u2220POS)<\/p>\n\n\n\n<p>2 \u2220ROS = (\u2220QOS \u2013 \u2220POS)<\/p>\n\n\n\n<p><strong>\u2234 \u2220ROS =1\/2 (\u2220QOS &#8211; \u2220POS)<\/strong><\/p>\n\n\n\n<p>&nbsp;6. It is given that \u2220XYZ = 64\u00b0 and XY is produced to point P. Draw a figure from the given information. If ray YQ bisects \u2220ZYP, find \u2220XYQ and reflex \u2220QYP.<\/p>\n\n\n\n<p><strong>Given: <\/strong>\u2220XYZ = 64\u00b0 ,YQ is a bisector of \u2220ZYP<\/p>\n\n\n\n<p><strong>To find: <\/strong>\u2220XYQ and reflex \u2220QYP<\/p>\n\n\n\n<p><strong>Solution:<\/strong><\/p>\n\n\n\n<p>XYP is a straight line.<\/p>\n\n\n\n<p>\u2234 \u2220XYZ + \u2220ZYQ + \u2220QYP = 180\u00b0<\/p>\n\n\n\n<p>\u21d2 64\u00b0 + \u2220ZYQ + \u2220QYP = 180\u00b0  [YQ bisects \u2220ZYP so, \u2220QYP = \u2220ZYQ]<\/p>\n\n\n\n<p>\u21d2 2\u2220QYP = 180\u00b0 \u2013 64\u00b0<\/p>\n\n\n\n<p>\u21d2 2\u2220QYP = 116\u00b0<\/p>\n\n\n\n<p>\u21d2 \u2220QYP = &#8220;116\u00b0 &#8221; <strong>\/<\/strong><strong>2<\/strong><\/p>\n\n\n\n<p>\u21d2 \u2220QYP = 58\u00b0<\/p>\n\n\n\n<p>\u2234 Reflex \u2220QYP = 360\u00b0 \u2013 58\u00b0 = 302\u00b0<\/p>\n\n\n\n<p>Since \u2220XYQ = \u2220XYZ + \u2220ZYQ<\/p>\n\n\n\n<p>\u21d2 \u2220XYQ = 64\u00b0 + \u2220QYP [\u2235\u2220XYZ = 64\u00b0(Given) and \u2220ZYQ = \u2220QYP]<\/p>\n\n\n\n<p>\u21d2 \u2220XYQ = 64\u00b0 + 58\u00b0<\/p>\n\n\n\n<p>&nbsp; &nbsp;= 122\u00b0 [\u2220QYP = 58\u00b0]<\/p>\n\n\n\n<p><strong>Thus, \u2220XYQ = 122\u00b0 and reflex \u2220QYP = 302\u00b0.<\/strong><\/p>\n\n\n\n<p>Exercise 3.2<\/p>\n\n\n\n<div class=\"wp-block-image\"><figure class=\"aligncenter size-large\"><img loading=\"lazy\" decoding=\"async\" width=\"243\" height=\"379\" src=\"https:\/\/themindpalace.in\/wp-content\/uploads\/2020\/09\/1-3.jpg\" alt=\"\" class=\"wp-image-979\" srcset=\"https:\/\/themindpalace.in\/wp-content\/uploads\/2020\/09\/1-3.jpg 243w, https:\/\/themindpalace.in\/wp-content\/uploads\/2020\/09\/1-3-192x300.jpg 192w\" sizes=\"(max-width: 243px) 100vw, 243px\" \/><\/figure><\/div>\n\n\n\n<p>1. In figure, find the values of x and y and then show that AB || CD.<\/p>\n\n\n\n<p>To find: values of x and y and AB || CD.<\/p>\n\n\n\n<p>Solution:<\/p>\n\n\n\n<p>In the figure, we have CD and PQ intersect at F<\/p>\n\n\n\n<p>\u2234 y = 130\u00b0 \u2026(1)[Vertically opposite angles]<\/p>\n\n\n\n<p>PQ is a straight line and EA stands on it<\/p>\n\n\n\n<p>\u2220AEP + \u2220AEQ = 180\u00b0 [Linear pair]<\/p>\n\n\n\n<p>50\u00b0 + x = 180\u00b0<\/p>\n\n\n\n<p>\u21d2 x = 180\u00b0 \u2013 50\u00b0<\/p>\n\n\n\n<p>\u21d2 x = 130\u00b0 \u2026(2)<\/p>\n\n\n\n<p>From (1) and (2), x = y = 130\u00b0<\/p>\n\n\n\n<p>As they are pair of alternate interior angles.<\/p>\n\n\n\n<p><strong>\u2234 AB || CD<\/strong><\/p>\n\n\n\n<p>2. In figure, if AB || CD, CD || EF and y : z = 3 : 7, find x.<\/p>\n\n\n\n<div class=\"wp-block-image\"><figure class=\"aligncenter size-large\"><img loading=\"lazy\" decoding=\"async\" width=\"233\" height=\"203\" src=\"https:\/\/themindpalace.in\/wp-content\/uploads\/2020\/09\/2-2.jpg\" alt=\"\" class=\"wp-image-980\"\/><\/figure><\/div>\n\n\n\n<p>Given: AB || CD, CD || EF and y : z = 3 : 7<\/p>\n\n\n\n<p>To find: x<\/p>\n\n\n\n<p><strong>Solution<\/strong>:AB || CD, and CD || EF [Given]<\/p>\n\n\n\n<p>\u2234 AB || EF<\/p>\n\n\n\n<p>\u2234 x = z [Alternate interior angles] \u2026.(1)\u21d2 K= &#8220;180<sup>0<\/sup> &#8221; \/10<\/p>\n\n\n\n<p>Again, AB || CD \u21d2 K= 18<sup>0<\/sup><\/p>\n\n\n\n<p>\u21d2 x + y = 180\u00b0 [Co-interior angles Then, Y = 3k,  Z = 7k,<\/p>\n\n\n\n<p>\u21d2 z + y = 180\u00b0 \u2026 (2) [By (1)] <\/p>\n\n\n\n<p>\u21d2 Y =3 x 18<sup>0<\/sup><\/p>\n\n\n\n<p>\u21d2 Y =54<sup>0<\/sup> <\/p>\n\n\n\n<p>Z = 7k,<\/p>\n\n\n\n<p>\u21d2 z = 7 x 18<sup>0<\/sup><\/p>\n\n\n\n<p>\u21d2 Z = 126\u00b0<\/p>\n\n\n\n<p>But y : z = 3 : 7<\/p>\n\n\n\n<p>Let k be any variable<\/p>\n\n\n\n<p>3k + 7k =180<\/p>\n\n\n\n<p>10k=180<\/p>\n\n\n\n<p>3. In figure, if AB || CD, EF \u22a5 CD and \u2220GED = 126\u00b0, find \u2220AGE, \u2220GEF and \u2220FGE.<\/p>\n\n\n\n<div class=\"wp-block-image\"><figure class=\"aligncenter size-large\"><img loading=\"lazy\" decoding=\"async\" width=\"274\" height=\"207\" src=\"https:\/\/themindpalace.in\/wp-content\/uploads\/2020\/09\/3-1.jpg\" alt=\"\" class=\"wp-image-981\"\/><\/figure><\/div>\n\n\n\n<p>Given: AB || CD, EF \u22a5 CD and \u2220GED = 126\u00b0<\/p>\n\n\n\n<p>To find: \u2220AGE, \u2220GEF and \u2220FGE.<\/p>\n\n\n\n<p><strong>Solution<\/strong>:AB || CD and GE is a transversal.<\/p>\n\n\n\n<p>\u2234 \u2220AGE = \u2220GED [Alternate interior angles<\/p>\n\n\n\n<p>But \u2220GED = 126\u00b0 [Given<\/p>\n\n\n\n<p>\u2234\u2220AGE = 126\u00b0<\/p>\n\n\n\n<p>Also, \u2220GEF + \u2220FED = \u2220GED<\/p>\n\n\n\n<p>\u2220GEF + 90\u00b0 = 126\u00b0 [\u2235 EF \u22a5 CD (given)]<\/p>\n\n\n\n<p>x = z [Alternate interior angles]\u2026 (1) Again, AB || CD<\/p>\n\n\n\n<p>\u21d2 x + y = 180\u00b0 [Co-interior angles]<\/p>\n\n\n\n<p>\u2220GEF = 126\u00b0 -90\u00b0 = 36\u00b0<\/p>\n\n\n\n<p>Now, AB || CD and GE is a transversal<\/p>\n\n\n\n<p>\u2234 \u2220FGE + \u2220GED = 180\u00b0 [Co-interior angles<\/p>\n\n\n\n<p>\u2220FGE + 126\u00b0 = 180\u00b0<\/p>\n\n\n\n<p>\u2220FGE = 180\u00b0 \u2013 126\u00b0 = 54\u00b0<\/p>\n\n\n\n<p>Thus, \u2220AGE = 126\u00b0,<\/p>\n\n\n\n<p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; <strong>\u2220GEF= 36\u00b0 and \u2220FGE = 54\u00b0.<\/strong><\/p>\n\n\n\n<p>4. In figure, if PQ || ST, \u2220 PQR = 110\u00b0 and \u2220 RST = 130\u00b0, find \u2220QRS.<\/p>\n\n\n\n<div class=\"wp-block-image\"><figure class=\"aligncenter size-large\"><img loading=\"lazy\" decoding=\"async\" width=\"268\" height=\"282\" src=\"https:\/\/themindpalace.in\/wp-content\/uploads\/2020\/09\/4-2.jpg\" alt=\"\" class=\"wp-image-982\"\/><\/figure><\/div>\n\n\n\n<p>Given: PQ || ST, \u2220 PQR = 110\u00b0 and \u2220 RST = 130\u00b0<\/p>\n\n\n\n<p>To find: \u2220QRS.<\/p>\n\n\n\n<p><strong>Solution<\/strong>:Draw a line EF parallel to ST through R.<\/p>\n\n\n\n<p>Since PQ || ST [Given]<br>&nbsp;&nbsp;&nbsp; and EF || ST [Construction]<br>\u2234 PQ || EF and QR is a transversal<\/p>\n\n\n\n<p>\u21d2 \u2220PQR = \u2220QRF [Alternate interior angles]<\/p>\n\n\n\n<p>&nbsp;But \u2220PQR = 110\u00b0 [Given]<br>\u2234\u2220QRF = \u2220QRS + \u2220SRF = 110\u00b0 \u2026(1)<br>Again ST || EF and RS is a transversal<\/p>\n\n\n\n<p>\u2234 \u2220RST + \u2220SRF = 180\u00b0 [Co-interior angles] &nbsp;<\/p>\n\n\n\n<p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; 130\u00b0 + \u2220SRF = 180\u00b0\u21d2 \u2220SRF = 180\u00b0 \u2013 130\u00b0 = 50\u00b0                  <\/p>\n\n\n\n<p>  \u21d2 \u2220QRS = 110\u00b0 \u2013 50\u00b0 = 60\u00b0=Thus, <strong>\u2220QRS = 60\u00b0<\/strong>.<br>   Now, from (1), we have \u2220QRS + 50\u00b0 = 110\u00b0<br><br><\/p>\n\n\n\n<p>5. In figure, if AB || CD, \u2220APQ = 50\u00b0 and \u2220PRD = 127\u00b0, find x and y<\/p>\n\n\n\n<div class=\"wp-block-image\"><figure class=\"aligncenter size-large\"><img loading=\"lazy\" decoding=\"async\" width=\"291\" height=\"257\" src=\"https:\/\/themindpalace.in\/wp-content\/uploads\/2020\/09\/5-1.jpg\" alt=\"\" class=\"wp-image-983\"\/><\/figure><\/div>\n\n\n\n<p>Given: AB || CD, \u2220APQ = 50\u00b0 and \u2220PRD = 127\u00b0,<\/p>\n\n\n\n<p>To find: x and y.<\/p>\n\n\n\n<p><strong>Solution:<\/strong> We have AB || CD and PQ is a transversal<\/p>\n\n\n\n<p>\u2234 \u2220APQ = \u2220PQR&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; [Alternate interior angles]<\/p>\n\n\n\n<p>\u21d2 50\u00b0 = x [ \u2235 \u2220APQ = 50\u00b0 (given)]<\/p>\n\n\n\n<p>AB || CD and PR is a transversal.<br>\u2234 \u2220APR = \u2220PRD [Alternate interior angles]<\/p>\n\n\n\n<p>\u21d2 \u2220APR = 127\u00b0 [ \u2235 \u2220PRD = 127\u00b0 (given)]<\/p>\n\n\n\n<p>\u21d2 \u2220APQ + \u2220QPR = 127\u00b0<br>\u21d2 50\u00b0 + y = 127\u00b0 [ \u2235 \u2220APQ = 50\u00b0 (given)]<br>\u21d2 y = 127\u00b0- 50\u00b0 = 77\u00b0<br><strong>Thus, x = 50\u00b0 and y = 77\u00b0.<\/strong><\/p>\n\n\n\n<p>6. In figure, 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.<br><br><\/p>\n\n\n\n<div class=\"wp-block-image\"><figure class=\"aligncenter size-large\"><img loading=\"lazy\" decoding=\"async\" width=\"267\" height=\"332\" src=\"https:\/\/themindpalace.in\/wp-content\/uploads\/2020\/09\/6.jpg\" alt=\"\" class=\"wp-image-985\" srcset=\"https:\/\/themindpalace.in\/wp-content\/uploads\/2020\/09\/6.jpg 267w, https:\/\/themindpalace.in\/wp-content\/uploads\/2020\/09\/6-241x300.jpg 241w\" sizes=\"(max-width: 267px) 100vw, 267px\" \/><\/figure><\/div>\n\n\n\n<p>Given:<\/p>\n\n\n\n<p>To find:<\/p>\n\n\n\n<p>CONSTRUCTION: Draw ray BL \u22a5PQ and CM \u22a5 RS<\/p>\n\n\n\n<p><strong>Solution<\/strong>:\u2235 PQ || RS \u21d2 BL || CM&nbsp; [\u2235 BL || PQ and CM || RS]<br>Now, BL || CM and BC is a transversal.<\/p>\n\n\n\n<p>\u2234 \u2220LBC = \u2220MCB \u2026(1) [Alternate interior angles]<br>Since, angle of incidence = Angle of reflection<br>\u2220ABL = \u2220LBC and \u2220MCB = \u2220MCD<br>\u21d2 \u2220ABL = \u2220MCD \u2026(2) [By (1)]<br>Adding (1) and (2), we get<br>\u2220LBC + \u2220ABL = \u2220MCB + \u2220MCD<br>\u21d2 \u2220ABC = \u2220BCD<\/p>\n\n\n\n<p>i.e., a pair of alternate interior angles are equal.<br>\u2234 AB || CD.<br><\/p>\n\n\n\n<p><strong>Angle Sum Property of Triangle:<\/strong><\/p>\n\n\n\n<div class=\"wp-block-image\"><figure class=\"aligncenter size-large\"><img loading=\"lazy\" decoding=\"async\" width=\"261\" height=\"243\" src=\"https:\/\/themindpalace.in\/wp-content\/uploads\/2020\/09\/11.jpg\" alt=\"\" class=\"wp-image-986\"\/><\/figure><\/div>\n\n\n\n<p>STATEMENT<strong>:<\/strong>&nbsp;The sum of the angles of a triangle is 180\u00ba.<\/p>\n\n\n\n<p><strong>Given:<\/strong>&nbsp;A&nbsp; \u0394 ABC.<\/p>\n\n\n\n<p><strong>To Prove:&nbsp; <\/strong>\u22201 + \u22202 + \u22203 = 180\u00b0<\/p>\n\n\n\n<p><strong>Construction:&nbsp; <\/strong>Let us draw a line DAE&nbsp; parallel to BC.<\/p>\n\n\n\n<p><strong>Proof:<\/strong> Statement                                                                 Reason<\/p>\n\n\n\n<p>DAE ||BC and PQ and AB Is the transversal                   <\/p>\n\n\n\n<p>Hence, \u22204 = \u22202                                                                         (alternate interior angles)<\/p>\n\n\n\n<p>&nbsp; DAE ||BC and AC Is the transversal                                 (alternate interior angles<\/p>\n\n\n\n<p>&nbsp;&nbsp;\u22205 = \u22203<\/p>\n\n\n\n<p>Now, DAE is a straight line<\/p>\n\n\n\n<p>\u22204 + \u22201 +\u22205 = 180\u00b0<\/p>\n\n\n\n<p>\u21d2 \u22201 + \u22204 +\u22205 = 180\u00b0                                                          <\/p>\n\n\n\n<p>Angles on the same side<\/p>\n\n\n\n<p>of DAE at the point A<\/p>\n\n\n\n<p>\u22204 = \u22202 &nbsp;and&nbsp; \u22205 = \u22203<\/p>\n\n\n\n<p><strong>Hence, the sum of the angles of a triangles 180\u00b0<\/strong><\/p>\n\n\n\n<p><strong>Exterior angle property of a triangle theorem<\/strong><\/p>\n\n\n\n<p>STATEMENT<strong>:<\/strong>&nbsp;If a side of a triangle is produced, then the exterior angle so formed is equal to the sum of the two interior opposite angles.<\/p>\n\n\n\n<p><strong>Given<\/strong><strong>: <\/strong>A&nbsp; \u0394 &nbsp;ABC. whose side BC has been produced to D.<\/p>\n\n\n\n<p>&nbsp; Forming exterior angle \u2220ACD<\/p>\n\n\n\n<p><strong>To Prove:&nbsp; <\/strong>\u2220ACD = \u2220BAC + \u2220CBA&nbsp; [\u22204 = \u22201 + \u22202<\/p>\n\n\n\n<p><strong>Proof:<\/strong>. Statement                                                               Reason<\/p>\n\n\n\n<p>We know that \u0394 &nbsp;ABC                                              from the angle sum property<\/p>\n\n\n\n<p>&nbsp;\u22201 + \u22202 + \u22203 = 180\u00b0\u2026..(i)<\/p>\n\n\n\n<p>Also&nbsp;&nbsp; \u22203 +\u22204 = 180\u00b0\u2026..(ii)                                       form a linear pair<\/p>\n\n\n\n<p>From equation (i) and (ii) it follows that:<\/p>\n\n\n\n<p>\u22203 +\u22204 = \u22201 + \u22202 + \u22203<\/p>\n\n\n\n<p>\u2234 \u22204 = \u22201 + \u22202<\/p>\n\n\n\n<p><strong>Hence, proved<\/strong><\/p>\n\n\n\n<p>1. In figure, sides QP and RQ of \u2206PQR are produced to points S and T, &nbsp;&nbsp;respectively. If \u2220SPR = 135\u00b0 and \u2220PQT = 110\u00b0, find \u2220PRQ.<\/p>\n\n\n\n<div class=\"wp-block-image\"><figure class=\"aligncenter size-large\"><img loading=\"lazy\" decoding=\"async\" width=\"182\" height=\"174\" src=\"https:\/\/themindpalace.in\/wp-content\/uploads\/2020\/09\/111.jpg\" alt=\"\" class=\"wp-image-987\"\/><\/figure><\/div>\n\n\n\n<p><strong>Given:<\/strong>&nbsp; \u2220SPR = 135\u00b0 and \u2220PQT = 110\u00b0,<\/p>\n\n\n\n<p><strong>To find: <\/strong>\u2220PRQ.<\/p>\n\n\n\n<p><strong>Solution<\/strong>:We have, \u2220TQP + \u2220PQR = 180\u00b0&nbsp; [Linear pair]<\/p>\n\n\n\n<p>\u21d2 110\u00b0 + \u2220PQR = 180\u00b0<br>\u21d2 \u2220PQR = 180\u00b0 \u2013 110\u00b0 = 70\u00b0<\/p>\n\n\n\n<p>Since, the side QP of \u2206PQR is produced to S.<\/p>\n\n\n\n<p>\u21d2 \u2220PQR + \u2220PRQ = 135\u00b0<br>[Exterior angle property of a triangle]<\/p>\n\n\n\n<p>\u21d2 70\u00b0 + \u2220PRQ = 135\u00b0 [\u2220PQR = 70\u00b0]<br>\u21d2 \u2220PRQ = 135\u00b0 \u2013 70\u00b0 \u21d2 \u2220PRQ = 65\u00b0<\/p>\n\n\n\n<p>2. In figure, \u2220X = 62\u00b0, \u2220XYZ = 54\u00b0, if YO and ZO are the bisectors of \u2220XYZ and \u2220XZY respectively of \u2206XYZ, find \u2220OZY and \u2220YOZ.<\/p>\n\n\n\n<div class=\"wp-block-image\"><figure class=\"aligncenter size-large\"><img loading=\"lazy\" decoding=\"async\" width=\"262\" height=\"206\" src=\"https:\/\/themindpalace.in\/wp-content\/uploads\/2020\/09\/222.jpg\" alt=\"\" class=\"wp-image-988\"\/><\/figure><\/div>\n\n\n\n<p><strong>Given:<\/strong> \u2220X = 62\u00b0, \u2220XYZ = 54\u00b0, YO and ZO are the bisectors of \u2220XYZ and \u2220XZY<\/p>\n\n\n\n<p><strong>To find:<\/strong>\u2220OZY and \u2220YOZ.<\/p>\n\n\n\n<p><strong>Solution:<\/strong>In \u2206XYZ, we have \u2220XYZ + \u2220YZX + \u2220ZXY = 180\u00b0<br>[Angle sum property of a triangle]<\/p>\n\n\n\n<p>But \u2220XYZ = 54\u00b0 and \u2220ZXY = 62\u00b0<\/p>\n\n\n\n<p>\u2234 54\u00b0 + \u2220YZX + 62\u00b0 = 180\u00b0<\/p>\n\n\n\n<p>\u21d2 \u2220YZX = 180\u00b0 \u2013 54\u00b0 \u2013 62\u00b0 = 64\u00b0<\/p>\n\n\n\n<p>YO and ZO are the bisectors of \u2220XYZ and \u2220XZY respectively.<\/p>\n\n\n\n<p>\u2234 \u2220OYZ = \u2220XYZ\/ 2 = 54\u00b0\/ 2 = 27\u00b0<br>and \u2220OZY = \u2220YZX \/ 2 = 64\u00b0\/ 2 = 32\u00b0<\/p>\n\n\n\n<p>in \u2206OYZ, we have<br>\u2220YOZ + \u2220OYZ + \u2220OZY = 180\u00b0 [Angle sum property of a triangle<\/p>\n\n\n\n<p>\u21d2 \u2220YOZ + 27\u00b0 + 32\u00b0 = 180\u00b0<br>\u21d2 \u2220YOZ = 180\u00b0 -27\u00b0 \u2013 32\u00b0 = 121\u00b0<\/p>\n\n\n\n<p><strong>Thus, \u2220OZY = 32\u00b0 and \u2220YOZ = 121\u00b0<\/strong><\/p>\n\n\n\n<p>3. In figure, if AB || DE, \u2220BAC = 35\u00b0 and \u2220CDE = 53\u00b0&nbsp; , find \u2220DCE.<\/p>\n\n\n\n<div class=\"wp-block-image\"><figure class=\"aligncenter size-large\"><img loading=\"lazy\" decoding=\"async\" width=\"211\" height=\"204\" src=\"https:\/\/themindpalace.in\/wp-content\/uploads\/2020\/09\/333.jpg\" alt=\"\" class=\"wp-image-989\"\/><\/figure><\/div>\n\n\n\n<p><strong>Given:<\/strong> AB || DE, \u2220BAC = 35\u00b0 and \u2220CDE = 53\u00b0<\/p>\n\n\n\n<p><strong>To find: <\/strong>\u2220DCE.<\/p>\n\n\n\n<p><strong>Solution:<\/strong>AB || DE and AE is a transversal.<br>So, \u2220BAC = \u2220AED&nbsp; [Alternate interior angles]<\/p>\n\n\n\n<p>and \u2220BAC = 35\u00b0 [Given]<br>&nbsp;&nbsp;&nbsp;&nbsp; \u2234 \u2220AED = 35\u00b0<\/p>\n\n\n\n<p>&nbsp;in \u2206CDE, we have<\/p>\n\n\n\n<p>&nbsp;\u2220CDE + \u2220DEC + \u2220DCE = 180\u00b0&nbsp;&nbsp;&nbsp;&nbsp; [Angle sum property of a triangle]<\/p>\n\n\n\n<p>\u2234 53\u00b0 + 35\u00b0 + \u2220DCE =180\u00b0[\u2235 \u2220DEC = \u2220AED = 35\u00b0 and\u2220CDE = 53\u00b0 (Given)]<\/p>\n\n\n\n<p>\u21d2 \u2220DCE = 180\u00b0 \u2013 53\u00b0 \u2013 35\u00b0 = 92\u00b0<br>Thus, <strong>\u2220DCE = 92\u00b0<\/strong><\/p>\n\n\n\n<p>4. In figure, if lines PQ and RS intersect at point T, such that \u2220 PRT = 40\u00b0, \u2220 RPT = 95\u00b0 and<\/p>\n\n\n\n<p>\u2220TSQ = 75\u00b0, find \u2220 SQT.<\/p>\n\n\n\n<div class=\"wp-block-image\"><figure class=\"aligncenter size-large\"><img loading=\"lazy\" decoding=\"async\" width=\"256\" height=\"207\" src=\"https:\/\/themindpalace.in\/wp-content\/uploads\/2020\/09\/444.jpg\" alt=\"\" class=\"wp-image-990\"\/><\/figure><\/div>\n\n\n\n<p><strong>Given:<\/strong> \u2220PRT = 40\u00b0, \u2220 RPT = 95\u00b0 and \u2220TSQ = 75\u00b0 <strong>&nbsp;<\/strong><\/p>\n\n\n\n<p><strong>To find: <\/strong>\u2220SQT.<\/p>\n\n\n\n<p><strong>Solution<\/strong>:<\/p>\n\n\n\n<p>In \u2206PRT, we have \u2220P + \u2220R + \u2220PTR = 180\u00b0[Angle sum property of a triangle]<\/p>\n\n\n\n<p>\u21d2 95\u00b0 + 40\u00b0 + \u2220PTR =180\u00b0 [ \u2235 \u2220P = 95\u00b0, \u2220R = 40\u00b0 (given)]<\/p>\n\n\n\n<p>\u21d2 \u2220PTR = 180\u00b0 \u2013 95\u00b0 \u2013 40\u00b0 = 45\u00b0<br>But PQ and RS intersect at T.<\/p>\n\n\n\n<p>\u2234 \u2220PTR = \u2220QTS [Vertically opposite angles<\/p>\n\n\n\n<p>\u2234 \u2220QTS = 45\u00b0 [ \u2235 \u2220PTR = 45\u00b0]<\/p>\n\n\n\n<p>in \u2206 TQS, we have \u2220TSQ + \u2220STQ + \u2220SQT = 180\u00b0 [Angle sum property of a triangle]<\/p>\n\n\n\n<p>\u2234 75\u00b0 + 45\u00b0 + \u2220SQT = 180\u00b0 [ \u2235 \u2220TSQ = 75\u00b0 and \u2220STQ = 45\u00b0]<\/p>\n\n\n\n<p>\u21d2 \u2220SQT = 180\u00b0 \u2013 75\u00b0 \u2013 45\u00b0 = 60\u00b0<br>Thus, <strong>\u2220SQT = 60\u00b0<\/strong><\/p>\n\n\n\n<p>5. In figure, if PQ \u22a5 PS, PQ||SR, \u2220SQR = 2S\u00b0 and \u2220QRT = 65\u00b0, then find the values of x and y.<\/p>\n\n\n\n<div class=\"wp-block-image\"><figure class=\"aligncenter size-large\"><img loading=\"lazy\" decoding=\"async\" width=\"270\" height=\"231\" src=\"https:\/\/themindpalace.in\/wp-content\/uploads\/2020\/09\/555.jpg\" alt=\"\" class=\"wp-image-991\"\/><\/figure><\/div>\n\n\n\n<p><strong>Given<\/strong><\/p>\n\n\n\n<p><strong>To find:<\/strong><\/p>\n\n\n\n<p><strong>Solution<\/strong>: In \u2206 QRS, the side SR is produced to T.<\/p>\n\n\n\n<p>\u2234 \u2220QRT = \u2220RQS + \u2220RSQ [Exterior angle property of a triangle]<\/p>\n\n\n\n<p>But \u2220RQS = 28\u00b0 and \u2220QRT = 65\u00b0<\/p>\n\n\n\n<p>So, 28\u00b0 + \u2220RSQ = 65\u00b0<br>\u21d2 \u2220RSQ = 65\u00b0 \u2013 28\u00b0 = 37\u00b0<\/p>\n\n\n\n<p>Since, PQ || SR and QS is a transversal.<\/p>\n\n\n\n<p>\u2234 \u2220PQS = \u2220RSQ = 37\u00b0 [Alternate interior angles]<\/p>\n\n\n\n<p>\u21d2 x = 37\u00b0<\/p>\n\n\n\n<p>PQ \u22a5 PS \u21d2 AP = 90\u00b0<br>&nbsp;in \u2206PQS, \u2220P + \u2220PQS + \u2220PSQ = 180\u00b0 [Angle sum property of a triangle]<\/p>\n\n\n\n<p>\u21d2 90\u00b0 + 37\u00b0 + y = 180\u00b0<\/p>\n\n\n\n<p>\u21d2 y = 180\u00b0 \u2013 90\u00b0 \u2013 37\u00b0 = 53\u00b0<\/p>\n\n\n\n<p><strong>Thus, x = 37\u00b0 and y = 53\u00b0<\/strong><\/p>\n\n\n\n<p>6. In figure, the side QR of \u2206PQR is produced to a point S. If the bisectors of \u2220PQR and \u2220PRS meet at point T, then prove that \u2220QTR =&nbsp;1\/2\u2220QPR<\/p>\n\n\n\n<div class=\"wp-block-image\"><figure class=\"aligncenter size-large\"><img loading=\"lazy\" decoding=\"async\" width=\"278\" height=\"222\" src=\"https:\/\/themindpalace.in\/wp-content\/uploads\/2020\/09\/666.jpg\" alt=\"\" class=\"wp-image-992\"\/><\/figure><\/div>\n\n\n\n<p>Solution:In \u2206PQR, side QR is produced to S, so by exterior angle property, \u2220PRS = \u2220P + \u2220PQR<\/p>\n\n\n\n<p>\u21d2&nbsp;1\/2\u2220PRS =&nbsp;1\/2\u2220P +&nbsp;12\u2220PQR<\/p>\n\n\n\n<p>Now\u21d2 \u2220TRS =&nbsp;1\/2\u2220P + \u2220TQR \u2026(1)[\u2235 QT and RT are bisectors of \u2220PQR and &nbsp; \u2220PRS respectively.]<\/p>\n\n\n\n<p>\u2220TRS = \u2220TQR + \u2220T \u2026(2)[Exterior angle property of a triangle]<\/p>\n\n\n\n<p>From (1) and (2),<\/p>\n\n\n\n<p>we have \u2220TQR +&nbsp;1\/2\u2220P = \u2220TQR + \u2220T<\/p>\n\n\n\n<p>\u21d2&nbsp;1\/2\u2220P = \u2220T<\/p>\n\n\n\n<p>\u21d2&nbsp;1\/2\u2220QPR = \u2220QTR or <strong>\u2220QTR =&nbsp;<\/strong>1\/2<strong>\u2220QPR<\/strong><\/p>\n","protected":false},"excerpt":{"rendered":"<p>Summary of lines and angles Solved exercise of lines and angles Summary This chapter tells you to find out the properties of angel formed when two lines intersect each other and also angles formed&#46;&#46;&#46;<\/p>\n","protected":false},"author":3,"featured_media":1129,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"_jetpack_memberships_contains_paid_content":false,"footnotes":""},"categories":[111,475,14],"tags":[67,66,68],"cp_meta_data":{"_edit_lock":["1630143715:2"],"_thumbnail_id":["1129"],"_edit_last":["2"],"_layout":["inherit"],"_oembed_95287caaddeb112cd4edfcbd8e525566":["<iframe title=\"Introduction of Computers  Part1\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/SzIGR3gp_F4?feature=oembed\" frameborder=\"0\" allow=\"accelerometer; autoplay; encrypted-media; gyroscope; picture-in-picture\" allowfullscreen><\/iframe>"],"_oembed_time_95287caaddeb112cd4edfcbd8e525566":["1597637035"],"_oembed_5e596f2e1724d644f729a20d7e668c1b":["<iframe title=\"Lines and Angles part 2,  NCERT 9th class Mathematics\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/2XRviYYOooA?feature=oembed\" frameborder=\"0\" allow=\"accelerometer; autoplay; clipboard-write; encrypted-media; gyroscope; picture-in-picture\" allowfullscreen><\/iframe>"],"_oembed_time_5e596f2e1724d644f729a20d7e668c1b":["1626444081"],"_jetpack_related_posts_cache":["a:1:{s:32:\"8f6677c9d6b0f903e98ad32ec61f8deb\";a:2:{s:7:\"expires\";i:1775856469;s:7:\"payload\";a:3:{i:0;a:1:{s:2:\"id\";i:1931;}i:1;a:1:{s:2:\"id\";i:1303;}i:2;a:1:{s:2:\"id\";i:1145;}}}}"],"_oembed_89e9e4b8c0f858dbccdf7baab7902e0c":["<iframe title=\"Lines and Angles part 2,  NCERT 9th class Mathematics\" width=\"640\" height=\"360\" src=\"https:\/\/www.youtube.com\/embed\/2XRviYYOooA?feature=oembed\" frameborder=\"0\" allow=\"accelerometer; autoplay; clipboard-write; encrypted-media; gyroscope; picture-in-picture\" allowfullscreen><\/iframe>"],"_oembed_time_89e9e4b8c0f858dbccdf7baab7902e0c":["1609568322"],"_last_editor_used_jetpack":["block-editor"]},"jetpack_sharing_enabled":true,"jetpack_featured_media_url":"https:\/\/themindpalace.in\/wp-content\/uploads\/2020\/08\/vertical-angles.png","jetpack-related-posts":[],"_links":{"self":[{"href":"https:\/\/themindpalace.in\/index.php\/wp-json\/wp\/v2\/posts\/512"}],"collection":[{"href":"https:\/\/themindpalace.in\/index.php\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/themindpalace.in\/index.php\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/themindpalace.in\/index.php\/wp-json\/wp\/v2\/users\/3"}],"replies":[{"embeddable":true,"href":"https:\/\/themindpalace.in\/index.php\/wp-json\/wp\/v2\/comments?post=512"}],"version-history":[{"count":15,"href":"https:\/\/themindpalace.in\/index.php\/wp-json\/wp\/v2\/posts\/512\/revisions"}],"predecessor-version":[{"id":2595,"href":"https:\/\/themindpalace.in\/index.php\/wp-json\/wp\/v2\/posts\/512\/revisions\/2595"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/themindpalace.in\/index.php\/wp-json\/wp\/v2\/media\/1129"}],"wp:attachment":[{"href":"https:\/\/themindpalace.in\/index.php\/wp-json\/wp\/v2\/media?parent=512"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/themindpalace.in\/index.php\/wp-json\/wp\/v2\/categories?post=512"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/themindpalace.in\/index.php\/wp-json\/wp\/v2\/tags?post=512"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}