{"id":2931,"date":"2022-11-21T16:00:08","date_gmt":"2022-11-21T10:30:08","guid":{"rendered":"https:\/\/www.aplustopper.com\/?p=2931"},"modified":"2022-11-22T16:54:26","modified_gmt":"2022-11-22T11:24:26","slug":"construction-of-bisector-of-an-angle","status":"publish","type":"post","link":"https:\/\/www.aplustopper.com\/construction-of-bisector-of-an-angle\/","title":{"rendered":"How Do You Construct A Bisector Of An Angle"},"content":{"rendered":"<h2><strong>Construction Of The Bisector Of A Given Angle<\/strong><\/h2>\n<p><strong>Given:<\/strong> An angle CAB<br \/>\n<strong>To construct:<\/strong> Bisector of \u2220CAB.<\/p>\n<ul>\n<li><strong>Step 1:<\/strong> Taking A as the centre and with any suitable radius, draw an arc cutting the arms AB and AC of \u2220CAB at D and E respectively.<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/c1.staticflickr.com\/9\/8275\/30023925991_fb3036b9ca_o.png\" alt=\"Construction-of-Bisector-of-Given-Angle\" width=\"344\" height=\"265\" \/><\/li>\n<li><strong>Step 2:<\/strong> Taking D as the centre and any radius more than half of DE, draw an arc.<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/c2.staticflickr.com\/6\/5230\/29479923023_ef006836d5_o.png\" alt=\"Construction-of-Bisector-of-Given-Angle-2\" width=\"343\" height=\"262\" \/><\/li>\n<li><strong>Step 3:<\/strong> Similarly, taking E as the centre and with the same radius (as in step 2), draw an arc intersecting the previous arc at P. Join AP and produce it to get AQ.<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/c2.staticflickr.com\/6\/5674\/30023925831_fe3bed7001_o.png\" alt=\"Construction-of-Bisector-of-Given-Angle-3\" width=\"338\" height=\"256\" \/>\u00a0Thus, ray AQ is the required bisector of \u2220CAB or \u2220BAC.<\/li>\n<\/ul>\n<p><strong>Read More:<\/strong><\/p>\n<ul>\n<li><a href=\"https:\/\/www.aplustopper.com\/construction-of-triangles-with-compass-and-protractor\/\"><span style=\"font-weight: 400;\">Construction of an Equilateral Triangle<\/span><\/a><\/li>\n<li><a href=\"https:\/\/www.aplustopper.com\/construction-of-similar-triangles\/\"><span style=\"font-weight: 400;\">Construction Of Similar Triangle As Per Given Scale Factor<\/span><\/a><\/li>\n<li><a href=\"https:\/\/www.aplustopper.com\/construction-of-line-segment\/\"><span style=\"font-weight: 400;\">Construction Of A Line Segment<\/span><\/a><\/li>\n<li><a href=\"https:\/\/www.aplustopper.com\/construction-of-perpendicular-bisector-of-a-line-segment\/\"><span style=\"font-weight: 400;\">Construction Of Perpendicular Bisector Of A Line Segment<\/span><\/a><\/li>\n<li><a href=\"https:\/\/www.aplustopper.com\/construction-of-angles-using-compass-ruler\/\"><span style=\"font-weight: 400;\">Construction Of An Angle Using Compass And Ruler<\/span><\/a><\/li>\n<\/ul>\n<p><strong>Example 1:<\/strong> \u00a0 \u00a0Using a protractor, draw an angle of measure 78\u00b0. With this angle as given, draw an angle of measure 39\u00b0.<br \/>\n<strong>Solution:<\/strong> \u00a0 \u00a0We follow the following steps to draw an angle of 39\u00b0 from an angle of 78\u00b0.<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/c1.staticflickr.com\/9\/8700\/29084074244_7438bb37a7_o.png\" alt=\"Construction-Of-Bisector-Of-An-Angle-Example-1\" width=\"212\" height=\"205\" \/> <strong>Steps of Construction:<\/strong><\/p>\n<ul>\n<li><strong>Step I:<\/strong> Draw a ray OA as shown in fig.<\/li>\n<li><strong>Step II:<\/strong> With the help of a protractor construct an angle AOB of measure 78\u00b0.<\/li>\n<li><strong>Step III:<\/strong> With centre O and a convenient radius drawn an arc cutting sides OA and OB at P and Q respectively.<\/li>\n<li><strong>Step IV:<\/strong> With centre P and radius more than 1\/2 (PQ), drawn an arc.<\/li>\n<li><strong>Step V:<\/strong> With centre Q and the same radius, as in the previous step, draw another arc intersecting the arc drawn in the previous step at R.<\/li>\n<li><strong>Step VI:<\/strong> Join OR and produce it to form ray OX.<br \/>\nThe angle \u2220AOX so obtained is the required angle of measure 39\u00b0.<\/li>\n<\/ul>\n<p><strong>Verification:<\/strong> Measure \u2220AOX and \u2220BOX. You will find that<br \/>\n\u2220AOX = \u2220BOX = 39\u00b0.<\/p>\n<p><strong>Example 2:<\/strong> \u00a0 \u00a0Using a protractor, draw an angle of measure 128\u00ba. With this angle as given, draw an angle of measure 96\u00ba.<br \/>\n<strong>Solution: \u00a0 \u00a0<\/strong>In order to construct an angle of measure 96\u00ba from an angle of measure 128\u00ba, we follow the following steps:<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/c2.staticflickr.com\/8\/7503\/29084074164_ec51e287bc_o.png\" alt=\"Construction-Of-Bisector-Of-An-Angle-Example-2\" width=\"258\" height=\"167\" \/> <strong>Steps of Construction:\u00a0<\/strong><\/p>\n<ul>\n<li><strong>Step I:<\/strong> Draw an angle \u2220AOB of measure 128\u00ba by using a protractor.<\/li>\n<li><strong>Step II:<\/strong> With centre O and a convenient radius draw an arc cutting OA and OB at P and Q respectively.<\/li>\n<li><strong>Steps III:<\/strong> With centre P and radius more than 1\/2 (PQ), draw an arc.<\/li>\n<li><strong>Step IV:<\/strong> With centre Q and the same radius, as in step III, draw another arc intersecting the previously drawn arc at R.<\/li>\n<li><strong>Steps V:<\/strong> Join OR and produce it to form ray OX. The \u2220AOX so obtained is of measure (128\u00ba\/2) i.e. 64\u00ba.<\/li>\n<li><strong>Step VI:<\/strong> With centre S (the point where ray OX cuts the arc (PQ) and radius more than 1\/2 (QS), draw an arc.<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/c1.staticflickr.com\/9\/8374\/29084074044_f52e08c75c_o.png\" alt=\"Construction-Of-Bisector-Of-An-Angle-Example-2-1\" width=\"255\" height=\"219\" \/><\/li>\n<li><strong>Step VII:<\/strong> With centre Q and the same radius, as in step VI, draw another arc intersecting the arc drawn in step VI at T.<\/li>\n<li><strong>Step VIII:<\/strong> Join OT and produce it form OY.<br \/>\nClearly, \u2220XOY = 1\/2 \u2220XOB = 1\/2 (64\u00ba) = 32\u00ba.<br \/>\n\u2234 \u2220AOT = \u2220AOX + \u2220XOY = 64\u00ba + 32\u00ba = 96\u00ba<br \/>\nThen, \u2220AOY is the desired angle.<\/li>\n<\/ul>\n<p><strong>Verification:<\/strong> Measure \u2220AOX, \u2220XOY and \u2220AOY. You will find \u2220AOY = 96\u00ba.<\/p>\n","protected":false},"excerpt":{"rendered":"<p>Construction Of The Bisector Of A Given Angle Given: An angle CAB To construct: Bisector of \u2220CAB. Step 1: Taking A as the centre and with any suitable radius, draw an arc cutting the arms AB and AC of \u2220CAB at D and E respectively. Step 2: Taking D as the centre and any radius [&hellip;]<\/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":[1273,1182,1276,1275,1274,292],"yoast_head":"<!-- This site is optimized with the Yoast SEO Premium plugin v17.8 (Yoast SEO v17.8) - https:\/\/yoast.com\/wordpress\/plugins\/seo\/ -->\n<title>How Do You Construct A Bisector Of An Angle - A Plus Topper<\/title>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" href=\"https:\/\/www.aplustopper.com\/construction-of-bisector-of-an-angle\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"How Do You Construct A Bisector Of An Angle\" \/>\n<meta property=\"og:description\" content=\"Construction Of The Bisector Of A Given Angle Given: An angle CAB To construct: Bisector of \u2220CAB. 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