{"id":4320,"date":"2023-02-08T05:41:24","date_gmt":"2023-02-08T05:41:24","guid":{"rendered":"https:\/\/theexampillar.com\/ncert\/?p=4320"},"modified":"2023-02-08T05:42:35","modified_gmt":"2023-02-08T05:42:35","slug":"ncert-solutions-class-9-maths-chapter-6-lines-and-angles-ex-6-3","status":"publish","type":"post","link":"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-9-maths-chapter-6-lines-and-angles-ex-6-3\/","title":{"rendered":"NCERT Solutions Class 9 Maths Chapter 6 Lines And Angles Ex 6.3"},"content":{"rendered":"<h2 style=\"text-align: center;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\">NCERT Solutions Class 9 Maths\u00a0<\/span><br \/>\n<span style=\"color: #000000; font-family: georgia, palatino, serif;\">Chapter &#8211; 6 (Lines And Angles)\u00a0<\/span><\/h2>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\">The NCERT Solutions in English Language for Class 9 Mathematics\u00a0<strong>Chapter &#8211; 6 Lines And Angles <\/strong>Exercise 6.3 has been provided here to help the students in solving the questions from this exercise.\u00a0<\/span><\/p>\n<p><span style=\"color: #000000;\"><strong>Chapter 6: Lines and Angles<\/strong><\/span><\/p>\n<ul>\n<li><a href=\"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-9-maths-chapter-6-lines-and-angles-ex-6-1\/\" target=\"_blank\" rel=\"noopener\">NCERT Solution Class 9 Maths Ex &#8211; 6.1<\/a><\/li>\n<li><a href=\"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-9-maths-chapter-6-lines-and-angles-ex-6-2\/\" target=\"_blank\" rel=\"noopener\">NCERT Solution Class 9 Maths Ex &#8211; 6.2<\/a><\/li>\n<\/ul>\n<h2 style=\"text-align: center;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\">Exercise &#8211; 6.3\u00a0<\/span><\/h2>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>1. In Fig. 6.39, sides QP and RQ of \u0394PQR are produced to points S and T, respectively. If \u2220SPR = 135\u00b0 and \u2220PQT = 110\u00b0, find \u2220PRQ.<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-4345\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-6.3-Q1.png\" alt=\"NCERT Class 9 Solutions Maths\" width=\"194\" height=\"191\" \/><br \/>\n<\/strong><strong>Answer &#8211;\u00a0 <\/strong>It is given in the question that:<\/span><br \/>\n<span style=\"color: #000000;\"><span lang=\"EN-US\">\u2220<\/span><span lang=\"EN-US\">SPR = 135\u00b0 <\/span><span lang=\"EN-US\">and, <\/span><span lang=\"EN-US\">\u2220<\/span><span lang=\"EN-US\">PQT = 110<sup>o<br \/>\n<\/sup><\/span><span lang=\"EN-US\">Now, according to the question,<br \/>\n<\/span><span lang=\"EN-US\">\u2220<\/span><span lang=\"EN-US\">SPR +<\/span>\u00a0<span lang=\"EN-US\">\u2220<\/span><span lang=\"EN-US\">QPR = 180\u00b0\u00a0(SQ is a straight line)<br \/>\n<\/span><span lang=\"EN-US\">135\u00b0\u00a0+<\/span>\u00a0<span lang=\"EN-US\">\u2220<\/span><span lang=\"EN-US\">QPR = 180\u00b0<br \/>\n<\/span><span lang=\"EN-US\">\u2220<\/span><span lang=\"EN-US\">QPR = 45\u00b0<br \/>\n<\/span><span lang=\"EN-US\">And,<br \/>\n<\/span><span lang=\"EN-US\">\u2220<\/span><span lang=\"EN-US\">PQT +<\/span>\u00a0<span lang=\"EN-US\">\u2220<\/span><span lang=\"EN-US\">PQR = 180\u00b0\u00a0(TR is a straight line)<br \/>\n<\/span><span lang=\"EN-US\">110\u00b0\u00a0+<\/span>\u00a0<span lang=\"EN-US\">\u2220<\/span><span lang=\"EN-US\">PQR = 180\u00b0<br \/>\n<\/span><span lang=\"EN-US\">\u2220<\/span><span lang=\"EN-US\">PQR = 70\u00b0<br \/>\n<\/span><span lang=\"EN-US\">Now,<br \/>\n<\/span><span lang=\"EN-US\">\u2220<\/span><span lang=\"EN-US\">PQR +<\/span>\u00a0<span lang=\"EN-US\">\u2220<\/span><span lang=\"EN-US\">QPR +<\/span>\u00a0<span lang=\"EN-US\">\u2220<\/span><span lang=\"EN-US\">PRQ = 180\u00b0\u00a0(Sum of the interior angles of the triangle)<br \/>\n<\/span><span lang=\"EN-US\">70\u00b0 + 45\u00b0\u00a0+<\/span>\u00a0<span lang=\"EN-US\">\u2220<\/span><span lang=\"EN-US\">PRQ = 180\u00b0<br \/>\n<\/span><span lang=\"EN-US\">115\u00b0\u00a0+<\/span>\u00a0<span lang=\"EN-US\">\u2220<\/span><span lang=\"EN-US\">PRQ = 180\u00b0<br \/>\n<\/span><span lang=\"EN-US\">\u2220<\/span><span lang=\"EN-US\">PRQ = 65\u00b0<\/span><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>2. In Fig. 6.40, \u2220X = 62\u00b0, \u2220XYZ = 54\u00b0. If YO and ZO are the bisectors of \u2220XYZ and \u2220XZY, respectively of \u0394 XYZ, find \u2220OZY and \u2220YOZ.<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-4346\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-6.3-Q2.png\" alt=\"NCERT Class 9 Solutions Maths\" width=\"169\" height=\"165\" \/><br \/>\n<\/strong><strong>Answer &#8211; <\/strong>We know the sum of the interior angles of the triangle.<\/span><br \/>\n<span style=\"color: #000000;\">So, \u2220YXZ +\u2220XYZ +\u2220XZY = 180\u00b0<\/span><br \/>\n<span style=\"color: #000000;\">62\u00b0+ 54\u00b0 +\u2220XZY = 180\u00b0<\/span><br \/>\n<span style=\"color: #000000;\">116\u00b0 + \u2220XZY = 180\u00b0<\/span><br \/>\n<span style=\"color: #000000;\">\u2220XZY = 180\u00b0 &#8211; 116\u00b0<\/span><br \/>\n<span style=\"color: #000000;\">\u2220XZY = 64\u00b0<\/span><br \/>\n<span style=\"color: #000000;\">Now, we know that ZO is the bisector, so, <\/span><br \/>\n<span style=\"color: #000000;\">\u2220OZY = \u00bd \u2220XZY <\/span><br \/>\n<span style=\"color: #000000;\">\u2234 \u2220OZY = <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{1}{2}\" alt=\"\\frac{1}{2}\" align=\"absmiddle\" \/> \u00d7 32\u00b0<\/span><br \/>\n<span style=\"color: #000000;\">Similarly, YO is a bisector, and so, <\/span><br \/>\n<span style=\"color: #000000;\">\u2220OYZ = \u00bd \u2220XYZ<\/span><br \/>\n<span style=\"color: #000000;\">\u2220OYZ = <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{1}{2}\" alt=\"\\frac{1}{2}\" align=\"absmiddle\" \/> \u00d7 54\u00b0 <\/span><br \/>\n<span style=\"color: #000000;\">\u2220OYZ = 27\u00b0 <\/span><br \/>\n<span style=\"color: #000000;\">Now, as the sum of the interior angles of the triangle, <\/span><br \/>\n<span style=\"color: #000000;\">\u2220OZY +\u2220OYZ +\u2220O = 180\u00b0<\/span><br \/>\n<span style=\"color: #000000;\">\u2220O = 180\u00b0 &#8211; 32\u00b0 &#8211; 27\u00b0 <\/span><br \/>\n<span style=\"color: #000000;\">Hence, \u2220O = 121\u00b0<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>3. In Fig. 6.41, if AB || DE, \u2220BAC = 35\u00b0 and \u2220CDE = 53\u00b0, find \u2220DCE.<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-4347\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-6.3-Q3.png\" alt=\"NCERT Class 9 Solutions Maths\" width=\"182\" height=\"162\" \/><br \/>\n<\/strong><strong>Answer &#8211; <\/strong>We know that AE is a transversal since AB || DE<\/span><br \/>\n<span style=\"color: #000000;\">Here \u2220BAC and \u2220AED are alternate interior angles.<\/span><br \/>\n<span style=\"color: #000000;\">Hence, \u2220BAC = \u2220AED<\/span><br \/>\n<span style=\"color: #000000;\">It is given that \u2220BAC = 35\u00b0<\/span><br \/>\n<span style=\"color: #000000;\">\u2220AED = 35\u00b0<\/span><br \/>\n<span style=\"color: #000000;\">Now, consider the triangle CDE. We know that the sum of the interior angles of a triangle is 180\u00b0.<\/span><br \/>\n<span style=\"color: #000000;\">\u2234 \u2220DCE + \u2220CED + \u2220CDE = 180\u00b0<\/span><br \/>\n<span style=\"color: #000000;\">\u2220DCE + 35\u00b0 + 53\u00b0 = 180\u00b0<\/span><br \/>\n<span style=\"color: #000000;\">\u2220DCE + 88\u00b0 = 180\u00b0<\/span><br \/>\n<span style=\"color: #000000;\">\u2220DCE = 180\u00b0 &#8211; 88\u00b0<\/span><br \/>\n<span style=\"color: #000000;\">\u2220DCE = 92\u00b0<\/span><br \/>\n<span style=\"color: #000000;\">Hence, \u2220DCE = 92\u00b0<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>4. In Fig. 6.42, if lines PQ and RS intersect at point T, such that \u2220PRT = 40\u00b0, \u2220RPT = 95\u00b0, and \u2220TSQ = 75\u00b0, find \u2220SQT.<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-4348\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-6.3-Q4.png\" alt=\"NCERT Class 9 Solutions Maths\" width=\"280\" height=\"188\" \/><br \/>\n<\/strong><strong>Answer &#8211; <\/strong>Consider triangle PRT.<\/span><br \/>\n<span style=\"color: #000000;\">\u2220PRT + \u2220RPT + \u2220PTR = 180\u00b0<\/span><br \/>\n<span style=\"color: #000000;\">\u2220PTR + 95\u00b0 + 40\u00b0 = 180\u00b0<\/span><br \/>\n<span style=\"color: #000000;\">\u2220PTR + 135\u00b0 = 180\u00b0<\/span><br \/>\n<span style=\"color: #000000;\">\u2220PTR = 180\u00b0 &#8211; 135\u00b0\u00a0<\/span><br \/>\n<span style=\"color: #000000;\">\u2220PTR = 45\u00b0\u00a0<\/span><br \/>\n<span style=\"color: #000000;\">Now \u2220PTR will be equal to \u2220STQ as they are vertically opposite angles. <\/span><br \/>\n<span style=\"color: #000000;\">So, \u2220PTR = \u2220STQ = 45\u00b0<\/span><br \/>\n<span style=\"color: #000000;\">Again, in triangle STQ,<\/span><br \/>\n<span style=\"color: #000000;\">\u2220TSQ + \u2220PTR + \u2220SQT = 180\u00b0<\/span><br \/>\n<span style=\"color: #000000;\">75\u00b0 + 45\u00b0 + \u2220SQT = 180\u00b0<\/span><br \/>\n<span style=\"color: #000000;\">120\u00b0 + \u2220SQT = 180\u00b0<\/span><br \/>\n<span style=\"color: #000000;\">\u2220SQT = 180\u00b0 &#8211; 120\u00b0\u00a0<\/span><br \/>\n<span style=\"color: #000000;\">\u2220SQT = 60\u00b0\u00a0<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>5. In Fig. 6.43, if PQ \u22a5 PS, PQ || SR, \u2220SQR = 28\u00b0 and \u2220QRT = 65\u00b0, then find the values of x and y.<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-4349\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-6.3-Q5.png\" alt=\"NCERT Class 9 Solutions Maths\" width=\"208\" height=\"175\" \/><br \/>\n<\/strong><strong>Answer &#8211; <\/strong>PQ \u22a5 PS, PQ || SR, \u2220SQR = 28\u00b0 and \u2220QRT = 65\u00b0<\/span><br \/>\n<span style=\"color: #000000;\">We know when two\u00a0parallel lines\u00a0are cut by a transversal,\u00a0alternate interior angles formed are equal. <\/span><br \/>\n<span style=\"color: #000000;\">According to the\u00a0angle sum property\u00a0of a triangle,\u00a0sum of the\u00a0interior angles of a triangle is 360\u00b0. <\/span><br \/>\n<span style=\"color: #000000;\">Since, PQ || SR and QR is the transversal,<\/span><br \/>\n<span style=\"color: #000000;\">\u2220PQR = \u2220QRT [Alternate interior angles]<\/span><br \/>\n<span style=\"color: #000000;\">\u2220PQS + \u2220SQR = \u2220QRT [From figure]<\/span><br \/>\n<span style=\"color: #000000;\">x + 28\u00b0 = 65\u00b0<\/span><br \/>\n<span style=\"color: #000000;\">x = 65\u00b0 &#8211; 28\u00b0<\/span><br \/>\n<span style=\"color: #000000;\">x = 37\u00b0<\/span><br \/>\n<span style=\"color: #000000;\">Now, in \u25b3PQS<\/span><br \/>\n<span style=\"color: #000000;\">\u2220PQS + \u2220PSQ + \u2220QPS = 180\u00b0 [Angle sum property of a triangle]<\/span><br \/>\n<span style=\"color: #000000;\">37\u00b0 + y + 90\u00b0 = 180\u00b0 [Since.\u00a0\u2220QPS =\u00a090\u00b0]<\/span><br \/>\n<span style=\"color: #000000;\">y = 180\u00b0 &#8211; 127\u00b0<\/span><br \/>\n<span style=\"color: #000000;\">y = 53\u00b0<\/span><br \/>\n<span style=\"color: #000000;\">Hence, x = 37\u00b0 and y = 53\u00b0<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>6. In Fig. 6.44, the side QR of \u0394PQR is produced to a point S. If the bisectors of \u2220PQR and \u2220PRS meet at point T, then prove that \u2220QTR = \u00bd \u2220QPR.<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-4350\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/01\/NCERT-Solutions-Class-9-Maths-Ex-6.3-Q6.png\" alt=\"NCERT Class 9 Solutions Maths\" width=\"150\" height=\"164\" \/><br \/>\n<\/strong><strong>Answer &#8211; <\/strong>Given: TQ and TR are the\u00a0bisectors of \u2220PQR and \u2220PRS respectively<\/span><br \/>\n<span style=\"color: #000000;\">To Prove: \u2220QTR = 1\/2 \u2220QPR<\/span><br \/>\n<span style=\"color: #000000;\">According to the\u00a0exterior angle theorem of a triangle, 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.<\/span><br \/>\n<span style=\"color: #000000;\">\u2220PRS = 2\u2220TRS\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0&#8212;&#8212;&#8212;&#8212;&#8211;\u00a0 \u00a0(i) [Since TR is the angle bisector\u00a0of\u00a0\u2220PRS]<\/span><br \/>\n<span style=\"color: #000000;\">\u2220PQR = 2\u2220TQR\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0&#8212;&#8212;&#8212;&#8212;&#8212; (ii) [Since TQ is the angle bisector of \u2220PQR]<\/span><br \/>\n<span style=\"color: #000000;\">Now, in \u25b3TQR<\/span><br \/>\n<span style=\"color: #000000;\">\u2220TRS = \u2220TQR +\u2220QTR [Exterior angle theorem of a triangle] <\/span><br \/>\n<span style=\"color: #000000;\">\u2220QTR = \u2220TRS &#8211; \u2220TQR\u00a0 \u00a0 \u00a0 \u00a0&#8212;&#8212;&#8212;&#8212;&#8212;&#8211; (iii)<\/span><br \/>\n<span style=\"color: #000000;\">Similarly, in \u25b3PQR<\/span><br \/>\n<span style=\"color: #000000;\">\u2220PRS = \u2220QPR + \u2220PQR [Exterior angle theorem of a triangle]<\/span><br \/>\n<span style=\"color: #000000;\">2\u2220TRS = \u2220QPR + 2\u2220TQR [From (i) and (ii)]<\/span><br \/>\n<span style=\"color: #000000;\">\u2220QPR = 2\u2220TRS &#8211; 2\u2220TQR<\/span><br \/>\n<span style=\"color: #000000;\">\u2220QPR = 2(\u2220TRS &#8211; \u2220TQR)<\/span><br \/>\n<span style=\"color: #000000;\">\u2220QPR = 2\u2220QTR [From (iii)]<\/span><br \/>\n<span style=\"color: #000000;\">\u2220QTR = 1\/2 \u2220QPR<\/span><br \/>\n<span style=\"color: #000000;\">Hence proved,\u00a0\u2220QTR = 1\/2 \u2220QPR.<\/span><\/p>\n<p style=\"text-align: justify;\">\n","protected":false},"excerpt":{"rendered":"<p>NCERT Solutions Class 9 Maths\u00a0 Chapter &#8211; 6 (Lines And Angles)\u00a0 The NCERT Solutions in English Language for Class 9 Mathematics\u00a0Chapter &#8211; 6 Lines And Angles Exercise 6.3 has been provided here to help the students in solving the questions from this exercise.\u00a0 Chapter 6: Lines and Angles NCERT Solution Class 9 Maths Ex &#8211; [&hellip;]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[698],"tags":[711,702,186,712,701,5],"class_list":["post-4320","post","type-post","status-publish","format-standard","hentry","category-class-9-maths","tag-class-9th-chapter-6-lines-and-angles-in-english","tag-ncert-solution-class-9","tag-ncert-solutions","tag-ncert-solutions-class-9-maths-chapter-6-in-english","tag-ncert-solutions-class-9-maths-in-english","tag-ncert-solutions-in-english"],"yoast_head":"<!-- This site is optimized with the Yoast SEO Premium plugin v22.9 (Yoast SEO v27.5) - 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