{"id":126664,"date":"2023-09-03T11:42:00","date_gmt":"2023-09-03T06:12:00","guid":{"rendered":"https:\/\/www.kopykitab.com\/blog\/?p=126664"},"modified":"2023-12-07T10:53:54","modified_gmt":"2023-12-07T05:23:54","slug":"rd-sharma-class-9-solutions-chapter-11-exercise-11-2","status":"publish","type":"post","link":"https:\/\/www.kopykitab.com\/blog\/rd-sharma-class-9-solutions-chapter-11-exercise-11-2\/","title":{"rendered":"RD Sharma Class 9 Solutions Chapter 11 Exercise 11.2 (Updated for 2024)"},"content":{"rendered":"\n<p><img class=\"alignnone wp-image-126667 size-full\" src=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/09\/RD-Sharma-Class-9-Solutions-Chapter-11-Exercise-11.2.jpg\" alt=\"RD Sharma Class 9 Solutions Chapter 11 Exercise 11.2\" width=\"1200\" height=\"675\" srcset=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/09\/RD-Sharma-Class-9-Solutions-Chapter-11-Exercise-11.2.jpg 1200w, https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/09\/RD-Sharma-Class-9-Solutions-Chapter-11-Exercise-11.2-768x432.jpg 768w\" sizes=\"(max-width: 1200px) 100vw, 1200px\" \/><\/p>\n<p><strong>RD Sharma Class 9 Solutions Chapter 11 Exercise 11.2:&nbsp;<\/strong>Worried about your upcoming maths test or the piled-up maths assignments? Don&#8217;t worry about them and start studying the <a href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-class-9-solutions-for-maths\/\" target=\"_blank\" rel=\"noopener\">RD Sharma Solutions Class 9 Maths<\/a>. You will find solutions to all your problems at once. The solutions of <a href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-9-maths-chapter-11-coordinate-geometry\/\" target=\"_blank\" rel=\"noopener\">RD Sharma Class 9 Solutions Chapter 11<\/a> Exercise 11.2 are designed by subject matter experts and are very credible. To know more, read the whole blog.&nbsp;<\/p>\n<div id=\"ez-toc-container\" class=\"ez-toc-v2_0_47_1 counter-hierarchy ez-toc-counter ez-toc-grey ez-toc-container-direction\">\n<div class=\"ez-toc-title-container\">\n<p class=\"ez-toc-title\">Table of Contents<\/p>\n<span class=\"ez-toc-title-toggle\"><a href=\"#\" class=\"ez-toc-pull-right ez-toc-btn ez-toc-btn-xs ez-toc-btn-default ez-toc-toggle\" aria-label=\"ez-toc-toggle-icon-1\"><label for=\"item-69e7bb42d3750\" aria-label=\"Table of Content\"><span style=\"display: flex;align-items: center;width: 35px;height: 30px;justify-content: center;direction:ltr;\"><svg style=\"fill: #000000;color:#000000\" xmlns=\"http:\/\/www.w3.org\/2000\/svg\" class=\"list-377408\" width=\"20px\" height=\"20px\" viewBox=\"0 0 24 24\" fill=\"none\"><path d=\"M6 6H4v2h2V6zm14 0H8v2h12V6zM4 11h2v2H4v-2zm16 0H8v2h12v-2zM4 16h2v2H4v-2zm16 0H8v2h12v-2z\" fill=\"currentColor\"><\/path><\/svg><svg style=\"fill: #000000;color:#000000\" class=\"arrow-unsorted-368013\" xmlns=\"http:\/\/www.w3.org\/2000\/svg\" width=\"10px\" height=\"10px\" viewBox=\"0 0 24 24\" version=\"1.2\" baseProfile=\"tiny\"><path d=\"M18.2 9.3l-6.2-6.3-6.2 6.3c-.2.2-.3.4-.3.7s.1.5.3.7c.2.2.4.3.7.3h11c.3 0 .5-.1.7-.3.2-.2.3-.5.3-.7s-.1-.5-.3-.7zM5.8 14.7l6.2 6.3 6.2-6.3c.2-.2.3-.5.3-.7s-.1-.5-.3-.7c-.2-.2-.4-.3-.7-.3h-11c-.3 0-.5.1-.7.3-.2.2-.3.5-.3.7s.1.5.3.7z\"\/><\/svg><\/span><\/label><input  type=\"checkbox\" id=\"item-69e7bb42d3750\"><\/a><\/span><\/div>\n<nav><ul class='ez-toc-list ez-toc-list-level-1 eztoc-visibility-hide-by-default' ><li class='ez-toc-page-1 ez-toc-heading-level-2'><a class=\"ez-toc-link ez-toc-heading-1\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-class-9-solutions-chapter-11-exercise-11-2\/#download-rd-sharma-class-9-solutions-chapter-11-exercise-112-pdf\" title=\"Download RD Sharma Class 9 Solutions Chapter 11 Exercise 11.2 PDF\">Download RD Sharma Class 9 Solutions Chapter 11 Exercise 11.2 PDF<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-2'><a class=\"ez-toc-link ez-toc-heading-2\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-class-9-solutions-chapter-11-exercise-11-2\/#access-answers-of-rd-sharma-class-9-solutions-chapter-11-exercise-112\" title=\"Access answers of RD Sharma Class 9 Solutions Chapter 11 Exercise 11.2\">Access answers of RD Sharma Class 9 Solutions Chapter 11 Exercise 11.2<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-2'><a class=\"ez-toc-link ez-toc-heading-3\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-class-9-solutions-chapter-11-exercise-11-2\/#faqs-on-rd-sharma-class-9-solutions-chapter-11-exercise-112\" title=\"FAQs on RD Sharma Class 9 Solutions Chapter 11 Exercise 11.2\">FAQs on RD Sharma Class 9 Solutions Chapter 11 Exercise 11.2<\/a><ul class='ez-toc-list-level-3'><li class='ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-4\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-class-9-solutions-chapter-11-exercise-11-2\/#from-where-can-i-download-the-pdf-of-rd-sharma-class-9-solutions-chapter-11-exercise-112\" title=\"From where can I download the PDF of RD Sharma Class 9 Solutions Chapter 11 Exercise 11.2?\">From where can I download the PDF of RD Sharma Class 9 Solutions Chapter 11 Exercise 11.2?<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-5\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-class-9-solutions-chapter-11-exercise-11-2\/#how-much-does-it-cost-to-download-the-pdf-of-rd-sharma-class-9-solutions-chapter-11-exercise-112\" title=\"How much does it cost to download the PDF of RD Sharma Class 9 Solutions Chapter 11 Exercise 11.2?\">How much does it cost to download the PDF of RD Sharma Class 9 Solutions Chapter 11 Exercise 11.2?<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-6\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-class-9-solutions-chapter-11-exercise-11-2\/#can-i-access-the-rd-sharma-class-9-solutions-chapter-11-exercise-112-pdf-offline\" title=\"Can I access the RD Sharma Class 9 Solutions Chapter 11 Exercise 11.2\u00a0PDF offline?\">Can I access the RD Sharma Class 9 Solutions Chapter 11 Exercise 11.2\u00a0PDF offline?<\/a><\/li><\/ul><\/li><\/ul><\/nav><\/div>\n<h2><span class=\"ez-toc-section\" id=\"download-rd-sharma-class-9-solutions-chapter-11-exercise-112-pdf\"><\/span><strong>Download RD Sharma Class 9 Solutions Chapter 11 Exercise 11.2 PDF<\/strong><span class=\"ez-toc-section-end\"><\/span><\/h2>\n<p><a href=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/09\/RD-Sharma-Solutions-Class-9-Chapter-11-Ex-11.2-1.pdf\" target=\"_blank\" rel=\"noopener\">RD Sharma Class 9 Solutions Chapter 11 Exercise 11.2<\/a><\/p>\n<div id=\"example1\" style=\"text-align: justify;\">&nbsp;<\/div>\n<p style=\"text-align: justify;\"><style>\n.pdfobject-container { height: 500px;}<br \/>\n.pdfobject { border: 1px solid #666; }<br \/>\n<\/style><\/p>\n<p style=\"text-align: justify;\"><script src=\"https:\/\/www.kopykitab.com\/_utility\/js\/pdfobject.min.js\"><\/script><br><script>PDFObject.embed(\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/09\/RD-Sharma-Solutions-Class-9-Chapter-11-Ex-11.2-1.pdf\", \"#example1\");<\/script><\/p>\n<h2><span class=\"ez-toc-section\" id=\"access-answers-of-rd-sharma-class-9-solutions-chapter-11-exercise-112\"><\/span><strong>Access answers of RD Sharma Class 9 Solutions Chapter 11 Exercise 11.2<\/strong><span class=\"ez-toc-section-end\"><\/span><\/h2>\n<p>Question 1.<br>The exterior angles obtained on producing the base of a triangle both ways are 104\u00b0 and 136\u00b0. Find all the angles of the triangle.<br>Solution:<br>In \u2206ABC, base BC has produced both ways to D and E respectively forming \u2220ABE = 104\u00b0 and \u2220ACD = 136\u00b0<br><img src=\"https:\/\/farm2.staticflickr.com\/1943\/30703039077_e49a07397c_o.png\" alt=\"Class 9 RD Sharma Solutions Chapter 11 Coordinate Geometry\" width=\"353\" height=\"507\"><\/p>\n<p>Question 2.<br>In the figure, the sides BC, CA, and AB of an \u2206ABC have been produced to D, E, and F respectively. If \u2220ACD = 105\u00b0 and \u2220EAF = 45\u00b0, find all the angles of the \u2206ABC.<br><img src=\"https:\/\/farm2.staticflickr.com\/1946\/45593358102_a3d2c61d79_o.png\" alt=\"Class 9 Maths Chapter 11 Coordinate Geometry RD Sharma Solutions\" width=\"205\" height=\"202\"><br>Solution:<br>In \u2206ABC, sides BC, CA, and BA are produced to D, E, and F respectively.<br>\u2220ACD = 105\u00b0 and \u2220EAF = 45\u00b0<br>\u2220ACD + \u2220ACB = 180\u00b0 (Linear pair)<br>\u21d2 105\u00b0 + \u2220ACB = 180\u00b0<br>\u21d2 \u2220ACB = 180\u00b0- 105\u00b0 = 75\u00b0<br>\u2220BAC = \u2220EAF (Vertically opposite angles)<br>= 45\u00b0<br>But \u2220BAC + \u2220ABC + \u2220ACB = 180\u00b0<br>\u21d2 45\u00b0 + \u2220ABC + 75\u00b0 = 180\u00b0<br>\u21d2 120\u00b0 +\u2220ABC = 180\u00b0<br>\u21d2 \u2220ABC = 180\u00b0- 120\u00b0<br>\u2234 \u2220ABC = 60\u00b0<br>Hence \u2220ABC = 60\u00b0, \u2220BCA = 75\u00b0<br>and \u2220BAC = 45\u00b0<\/p>\n<p>Question 3.<br>Compute the value of x in each of the following figures:<br><img src=\"https:\/\/farm2.staticflickr.com\/1919\/30703038737_b28e3e2e33_o.png\" alt=\"RD Sharma Book Class 9 PDF Free Download Chapter 11 Coordinate Geometry\" width=\"271\" height=\"514\"><br><img src=\"https:\/\/farm2.staticflickr.com\/1939\/45593357912_c8ecf52f5f_o.png\" alt=\"RD Sharma Class 9 Book Chapter 11 Coordinate Geometry\" width=\"205\" height=\"256\"><br>Solution:<br>(i) In \u2206ABC, sides BC and CA are produced to D and E respectively<br><img src=\"https:\/\/farm2.staticflickr.com\/1974\/45643870021_5b1428fdbb_o.png\" alt=\"Coordinate Geometry With Solutions PDF RD Sharma Class 9 Solutions\" width=\"361\" height=\"531\"><br>(ii) In \u2206ABC, side BC is produced to either side of D and E respectively<br>\u2220ABE = 120\u00b0 and \u2220ACD =110\u00b0<br>\u2235 \u2220ABE + \u2220ABC = 180\u00b0 (Linear pair)<br><img src=\"https:\/\/farm2.staticflickr.com\/1938\/45593357622_0e671134b5_o.png\" alt=\"RD Sharma Class 9 Maths Book Questions Chapter 11 Coordinate Geometry\" width=\"340\" height=\"469\"><br>(iii) In the figure, BA || DC<br><img src=\"https:\/\/farm2.staticflickr.com\/1971\/45643869581_4277c744cb_o.png\" alt=\"RD Sharma Mathematics Class 9 Solutions Chapter 11 Coordinate Geometry\" width=\"342\" height=\"449\"><\/p>\n<p>Question 4.<br>In the figure, AC \u22a5 CE and \u2220A: \u2220B: \u2220C = 3:2:1, find the value of \u2220ECD.<br><img src=\"https:\/\/farm2.staticflickr.com\/1916\/45643869171_a057c46844_o.png\" alt=\"Solution Of Rd Sharma Class 9 Chapter 11 Coordinate Geometry\" width=\"268\" height=\"161\"><br>Solution:<br>In \u2206ABC, \u2220A: \u2220B : \u2220C = 3: 2 : 1<br>BC is produced for D and CE \u22a5 AC<br>\u2235 \u2220A + \u2220B + \u2220C = 180\u00b0 (Sum of angles of triangles)<br>Let\u2220A = 3x, then \u2220B = 2x and \u2220C = x<br>\u2234 3x + 2x + x = 180\u00b0 \u21d2 6x = 180\u00b0<br>\u21d2 x =&nbsp;<span id=\"MathJax-Element-24-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-155\" class=\"math\"><span id=\"MathJax-Span-156\" class=\"mrow\"><span id=\"MathJax-Span-157\" class=\"mfrac\"><span id=\"MathJax-Span-158\" class=\"msubsup\"><span id=\"MathJax-Span-159\" class=\"texatom\"><span id=\"MathJax-Span-160\" class=\"mrow\"><span id=\"MathJax-Span-161\" class=\"mn\">180<\/span><\/span><\/span><span id=\"MathJax-Span-162\" class=\"texatom\"><span id=\"MathJax-Span-163\" class=\"mrow\"><span id=\"MathJax-Span-164\" class=\"mo\">\u2218<\/span><\/span><\/span><\/span><span id=\"MathJax-Span-165\" class=\"mn\">6<\/span><\/span><\/span><\/span><\/span>&nbsp; = 30\u00b0<br>\u2234 \u2220A = 3x = 3 x 30\u00b0 = 90\u00b0<br>\u2220B = 2x = 2 x 30\u00b0 = 60\u00b0<br>\u2220C = x = 30\u00b0<br>In \u2206ABC,<br>Ext. \u2220ACD = \u2220A + \u2220B<br>\u21d2 90\u00b0 + \u2220ECD = 90\u00b0 + 60\u00b0 = 150\u00b0<br>\u2234 \u2220ECD = 150\u00b0-90\u00b0 = 60\u00b0<\/p>\n<p>Question 5.<br>In the figure, AB || DE, find \u2220ACD.<br><img src=\"https:\/\/farm2.staticflickr.com\/1957\/45593357032_3d6c2881aa_o.png\" alt=\"RD Sharma Math Solution Class 9 Chapter 11 Coordinate Geometry\" width=\"277\" height=\"221\"><br>Solution:<br>In the figure, AB || DE<br>AE and BD intersect each other at C \u2220BAC = 30\u00b0 and \u2220CDE = 40\u00b0<br>\u2235 AB || DE<br>\u2234 \u2220ABC = \u2220CDE (Alternate angles)<br><img src=\"https:\/\/farm2.staticflickr.com\/1915\/45643869001_99ba566f78_o.png\" alt=\"RD Sharma Class 9 Questions Chapter 11 Coordinate Geometry\" width=\"270\" height=\"221\"><br>\u21d2 \u2220ABC = 40\u00b0<br>In \u2206ABC, BC is produced<br>Ext. \u2220ACD = Int. \u2220A + \u2220B<br>= 30\u00b0 + 40\u00b0 = 70\u00b0<\/p>\n<p>Question 6.<br>Which of the following statements are true (T) and which are false (F):<br>(i) Sum of the three angles of a triangle is 180\u00b0.<br>(ii) A triangle can have two right angles.<br>(iii) All the angles of a triangle can be less than 60\u00b0.<br>(iv) All the angles of a triangle can be greater than 60\u00b0.<br>(v) All the angles of a triangle can be equal to 60\u00b0.<br>(vi) A triangle can have two obtuse angles.<br>(vii) A triangle can have at most one obtuse angle.<br>(viii) If one angle of a triangle is obtuse, then it cannot be a right-angled triangle.<br>(ix) An exterior angle of a triangle is less than either of its interior opposite angles.<br>(x) An exterior angle of a triangle is equal to the sum of the two interior opposite angles.<br>(xi) An exterior angle of a triangle is greater than the opposite interior angles.<br>Solution:<br>(i) True.<br>(ii) False. A right triangle has only one right angle.<br>(iii) False. In this, the sum of three angles will be less than 180\u00b0 which is not true.<br>(iv) False. In this, the sum of three angles will be more than 180\u00b0 which is not true.<br>(v) True. The sum of three angles will be 180\u00b0 which is true.<br>(vi) False. A triangle has only one obtuse angle.<br>(vii) True.<br>(viii)True.<br>(ix) False. An exterior angle of a triangle is always greater than its each interior opposite angles.<br>(x) True.<br>(xi) True.<\/p>\n<p>Question 7.<br>Fill in the blanks to make the following statements true:<br>(i) Sum of the angles of a triangle is \u2026\u2026\u2026<br>(ii) An exterior angle of a triangle is equal to the two \u2026\u2026.. opposite angles.<br>(iii) An exterior angle of a triangle is always \u2026\u2026.. than either of the interior opposite angles.<br>(iv) A triangle cannot have more than \u2026\u2026\u2026. right angles.<br>(v) A triangle cannot have more than \u2026\u2026\u2026 obtuse angles.<br>Solution:<br>(i) Sum of the angles of a triangle is 180\u00b0.<br>(ii) An exterior angle of a triangle is equal to the two interior opposite angles.<br>(iii) An exterior angle of a triangle is always greater than either of the interior opposite angles.<br>(iv) A triangle cannot have more than one right angle.<br>(v) A triangle cannot have more than one obtuse angle.<\/p>\n<p>Question 8.<br>In an \u2206ABC, the internal bisectors of \u2220B and \u2220C meet at P and the external bisectors of \u2220B and \u2220C meet at Q. Prove that \u2220BPC + \u2220BQC = 180\u00b0.<br>Solution:<br>Given: In \u2206ABC, sides AB, and AC are produced to D and E respectively. Bisectors of interior \u2220B and \u2220C meet at P and bisectors of exterior angles B and C meet at Q.<br><img src=\"https:\/\/farm2.staticflickr.com\/1918\/45593356782_846e6279f7_o.png\" alt=\"Maths RD Sharma Class 9 Chapter 11 Coordinate Geometry\" width=\"246\" height=\"256\"><br>To prove : \u2220BPC + \u2220BQC = 180\u00b0<br>Proof: \u2235 PB and PC are the internal bisectors of \u2220B and \u2220C<br>\u2220BPC = 90\u00b0+&nbsp;<span id=\"MathJax-Element-25-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-166\" class=\"math\"><span id=\"MathJax-Span-167\" class=\"mrow\"><span id=\"MathJax-Span-168\" class=\"mfrac\"><span id=\"MathJax-Span-169\" class=\"mn\">1<\/span><span id=\"MathJax-Span-170\" class=\"mn\">2<\/span><\/span><\/span><\/span><\/span>&nbsp;\u2220A \u2026(i)<br>Similarly, QB and QC are the bisectors of exterior angles B and C<br>\u2234 \u2220BQC = 90\u00b0 +&nbsp;<span id=\"MathJax-Element-26-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-171\" class=\"math\"><span id=\"MathJax-Span-172\" class=\"mrow\"><span id=\"MathJax-Span-173\" class=\"mfrac\"><span id=\"MathJax-Span-174\" class=\"mn\">1<\/span><span id=\"MathJax-Span-175\" class=\"mn\">2<\/span><\/span><\/span><\/span><\/span>&nbsp;\u2220A \u2026(ii)<br>Adding (i) and (ii),<br>\u2220BPC + \u2220BQC = 90\u00b0 +&nbsp;<span id=\"MathJax-Element-27-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-176\" class=\"math\"><span id=\"MathJax-Span-177\" class=\"mrow\"><span id=\"MathJax-Span-178\" class=\"mfrac\"><span id=\"MathJax-Span-179\" class=\"mn\">1<\/span><span id=\"MathJax-Span-180\" class=\"mn\">2<\/span><\/span><\/span><\/span><\/span>&nbsp;\u2220A + 90\u00b0 \u2013&nbsp;<span id=\"MathJax-Element-28-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-181\" class=\"math\"><span id=\"MathJax-Span-182\" class=\"mrow\"><span id=\"MathJax-Span-183\" class=\"mfrac\"><span id=\"MathJax-Span-184\" class=\"mn\">1<\/span><span id=\"MathJax-Span-185\" class=\"mn\">2<\/span><\/span><\/span><\/span><\/span>&nbsp;\u2220A<br>= 90\u00b0 + 90\u00b0 = 180\u00b0<br>Hence \u2220BPC + \u2220BQC = 180\u00b0<\/p>\n<p>Question 9.<br>In the figure, compute the value of x.<br><img src=\"https:\/\/farm2.staticflickr.com\/1948\/45593356622_a26904c9aa_o.png\" alt=\"RD Sharma Class 9 Chapter 11 Coordinate Geometry\" width=\"229\" height=\"229\"><br>Solution:<br>In the figure,<br>\u2220ABC = 45\u00b0, \u2220BAD = 35\u00b0 and \u2220BCD = 50\u00b0 Join BD and produce it E<br><img src=\"https:\/\/farm2.staticflickr.com\/1922\/45643868601_2fd5a4eac4_o.png\" alt=\"RD Sharma Class 9 Solutions Chapter 11 Coordinate Geometry\" width=\"340\" height=\"468\"><\/p>\n<p>Question 10.<br>In the figure, AB divides \u2220D AC in the ratio 1 : 3 and AB = DB. Determine the value of x.<br><img src=\"https:\/\/farm2.staticflickr.com\/1964\/45643868361_ff6f3b98d0_o.png\" alt=\"RD Sharma Solutions Class 9 Chapter 11 Coordinate Geometry\" width=\"247\" height=\"173\"><br>Solution:<br>In the figure AB = DB<br><img src=\"https:\/\/farm2.staticflickr.com\/1959\/45593356382_df4f219f84_o.png\" alt=\"RD Sharma Class 9 PDF Chapter 11 Coordinate Geometry\" width=\"259\" height=\"174\"><br><img src=\"https:\/\/farm2.staticflickr.com\/1927\/30703037367_3fb8dd1e90_o.png\" alt=\"Coordinate Geometry Class 9 RD Sharma Solutions\" width=\"339\" height=\"423\"><\/p>\n<p>Question 11.<br>ABC is a triangle. The bisector of the exterior angle at B and the bisector of \u2220C intersect each other at D. Prove that \u2220D =&nbsp;<span id=\"MathJax-Element-29-Frame\" class=\"MathJax\" tabindex=\"0\"><span id=\"MathJax-Span-186\" class=\"math\"><span id=\"MathJax-Span-187\" class=\"mrow\"><span id=\"MathJax-Span-188\" class=\"mfrac\"><span id=\"MathJax-Span-189\" class=\"mn\">1<\/span><span id=\"MathJax-Span-190\" class=\"mn\">2<\/span><\/span><\/span><\/span><\/span>&nbsp;\u2220A.<br>Solution:<br>Given: In \u2220ABC, CB is produced to E bisectors of ext. \u2220ABE and into \u2220ACB meet at D.<br><img src=\"https:\/\/farm2.staticflickr.com\/1968\/45593356172_40bff58f4c_o.png\" alt=\"RD Sharma Class 9 Solution Chapter 11 Coordinate Geometry\" width=\"280\" height=\"371\"><br><img src=\"https:\/\/farm2.staticflickr.com\/1979\/45643868071_8f2185882d_o.png\" alt=\"Class 9 RD Sharma Solutions Chapter 11 Coordinate Geometry\" width=\"357\" height=\"410\"><\/p>\n<p>Question 12.<br>In the figure, AM \u22a5 BC, and AN is the bisector of \u2220A. If \u2220B = 65\u00b0 and \u2220C = 33\u00b0, find \u2220MAN.<br><img src=\"https:\/\/farm2.staticflickr.com\/1958\/45593355992_8103016355_o.png\" alt=\"Class 9 Maths Chapter 11 Coordinate Geometry RD Sharma Solutions\" width=\"232\" height=\"156\"><br>Solution:<br><img src=\"https:\/\/farm2.staticflickr.com\/1928\/30703037067_fb84003d03_o.png\" alt=\"RD Sharma Book Class 9 PDF Free Download Chapter 11 Coordinate Geometry\" width=\"349\" height=\"305\"><br><img src=\"https:\/\/farm2.staticflickr.com\/1927\/45593355872_1d95c9f5bf_o.png\" alt=\"RD Sharma Class 9 Book Chapter 11 Coordinate Geometry\" width=\"348\" height=\"352\"><\/p>\n<p>Question 13.<br>In an AABC, AD bisects \u2220A and \u2220C &gt; \u2220B. Prove that \u2220ADB &gt; \u2220ADC.<br>Solution:<br>Given: In \u2206ABC,<br>\u2220C &gt; \u2220B and AD is the bisector of \u2220A<br><img src=\"https:\/\/farm2.staticflickr.com\/1962\/45593355462_3a93a0243d_o.png\" alt=\"Coordinate Geometry With Solutions PDF RD Sharma Class 9 Solutions\" width=\"231\" height=\"194\"><br>To prove: \u2220ADB &gt; \u2220ADC<br>Proof: In \u2206ABC, AD is the bisector of \u2220A<br>\u2234 \u22201 = \u22202<br>In \u2206ADC,<br>Ext. \u2220ADB = \u2220l+ \u2220C<br>\u21d2 \u2220C = \u2220ADB \u2013 \u22201 \u2026(i)<br>Similarly, in \u2206ABD,<br>Ext. \u2220ADC = \u22202 + \u2220B<br>\u21d2 \u2220B = \u2220ADC \u2013 \u22202 \u2026(ii)<br>From (i) and (ii)<br>\u2235 \u2220C &gt; \u2220B (Given)<br>\u2234 (\u2220ADB \u2013 \u22201) &gt; (\u2220ADC \u2013 \u22202)<br>But \u22201 = \u22202<br>\u2234 \u2220ADB &gt; \u2220ADC<\/p>\n<p>Question 14.<br>In \u2206ABC, BD \u22a5 AC, and CE \u22a5 AB. If BD and CE intersect at O, prove that \u2220BOC = 180\u00b0-\u2220A.<br>Solution:<br>Given: In \u2206ABC, BD \u22a5 AC and CE\u22a5 AB BD and CE intersect each other at O<br><img src=\"https:\/\/farm2.staticflickr.com\/1929\/45593355192_62f1936a30_o.png\" alt=\"RD Sharma Class 9 Maths Book Questions Chapter 11 Coordinate Geometry\" width=\"280\" height=\"197\"><br>To prove : \u2220BOC = 180\u00b0 \u2013 \u2220A<br>Proof: In quadrilateral ADOE<br>\u2220A + \u2220D + \u2220DOE + \u2220E = 360\u00b0 (Sum of angles of quadrilateral)<br>\u21d2 \u2220A + 90\u00b0 + \u2220DOE + 90\u00b0 = 360\u00b0<br>\u2220A + \u2220DOE = 360\u00b0 \u2013 90\u00b0 \u2013 90\u00b0 = 180\u00b0<br>But \u2220BOC = \u2220DOE (Vertically opposite angles)<br>\u21d2 \u2220A + \u2220BOC = 180\u00b0<br>\u2234 \u2220BOC = 180\u00b0 \u2013 \u2220A<\/p>\n<p>Question 15.<br>In the figure, AE bisects \u2220CAD and \u2220B = \u2220C. Prove that AE || BC.<br><img src=\"https:\/\/farm2.staticflickr.com\/1943\/45593354882_d870a95ebe_o.png\" alt=\"RD Sharma Mathematics Class 9 Solutions Chapter 11 Coordinate Geometry\" width=\"207\" height=\"208\"><br>Solution:<br>Given: In AABC, BA is produced and AE is the bisector of \u2220CAD<br>\u2220B = \u2220C<br><img src=\"https:\/\/farm2.staticflickr.com\/1979\/45593354722_ae04496774_o.png\" alt=\"Solution Of Rd Sharma Class 9 Chapter 11 Coordinate Geometry\" width=\"202\" height=\"200\"><br>To prove: AE || BC<br>Proof: In \u2206ABC, BA is produced<br>\u2234 Ext. \u2220CAD = \u2220B + \u2220C<br>\u21d2 2\u2220EAC = \u2220C + \u2220C (\u2235 AE is the bisector of \u2220CAE) (\u2235 \u2220B = \u2220C)<br>\u21d2 2\u2220EAC = 2\u2220C<br>\u21d2 \u2220EAC = \u2220C<br>But there are alternate angles<br>\u2234 AE || BC<\/p>\n<p>This is the complete blog on RD Sharma Class 9 Solutions Chapter 11 Exercise 11.2. To know more about the <a href=\"https:\/\/www.cbse.gov.in\/\" target=\"_blank\" rel=\"noopener\">CBSE<\/a> Class 11 Maths exam, ask in the comments.&nbsp;<\/p>\n<h2><span class=\"ez-toc-section\" id=\"faqs-on-rd-sharma-class-9-solutions-chapter-11-exercise-112\"><\/span><strong>FAQs on RD Sharma Class 9 Solutions Chapter 11 Exercise 11.2<\/strong><span class=\"ez-toc-section-end\"><\/span><\/h2>\n\n\n<div id=\"rank-math-faq\" class=\"rank-math-block\">\n<div class=\"rank-math-list \">\n<div id=\"faq-question-1631284248674\" class=\"rank-math-list-item\">\n<h3 class=\"rank-math-question \"><span class=\"ez-toc-section\" id=\"from-where-can-i-download-the-pdf-of-rd-sharma-class-9-solutions-chapter-11-exercise-112\"><\/span>From where can I download the PDF of RD Sharma Class 9 Solutions Chapter 11 Exercise 11.2?<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<div class=\"rank-math-answer \">\n\n<p>You can find the download link from the above blog.<\/p>\n\n<\/div>\n<\/div>\n<div id=\"faq-question-1631284318159\" class=\"rank-math-list-item\">\n<h3 class=\"rank-math-question \"><span class=\"ez-toc-section\" id=\"how-much-does-it-cost-to-download-the-pdf-of-rd-sharma-class-9-solutions-chapter-11-exercise-112\"><\/span>How much does it cost to download the PDF of RD Sharma Class 9 Solutions Chapter 11 Exercise 11.2?<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<div class=\"rank-math-answer \">\n\n<p>You can download it for free.<\/p>\n\n<\/div>\n<\/div>\n<div id=\"faq-question-1631284363454\" class=\"rank-math-list-item\">\n<h3 class=\"rank-math-question \"><span class=\"ez-toc-section\" id=\"can-i-access-the-rd-sharma-class-9-solutions-chapter-11-exercise-112-pdf-offline\"><\/span>Can I access the RD Sharma Class 9 Solutions Chapter 11 Exercise 11.2\u00a0PDF offline?<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<div class=\"rank-math-answer \">\n\n<p>Once you have downloaded the PDF online, you can access it offline as well.<\/p>\n\n<\/div>\n<\/div>\n<\/div>\n<\/div>","protected":false},"excerpt":{"rendered":"<p>RD Sharma Class 9 Solutions Chapter 11 Exercise 11.2:&nbsp;Worried about your upcoming maths test or the piled-up maths assignments? Don&#8217;t worry about them and start studying the RD Sharma Solutions Class 9 Maths. You will find solutions to all your problems at once. The solutions of RD Sharma Class 9 Solutions Chapter 11 Exercise 11.2 &#8230; <a title=\"RD Sharma Class 9 Solutions Chapter 11 Exercise 11.2 (Updated for 2024)\" class=\"read-more\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-class-9-solutions-chapter-11-exercise-11-2\/\" aria-label=\"More on RD Sharma Class 9 Solutions Chapter 11 Exercise 11.2 (Updated for 2024)\">Read more<\/a><\/p>\n","protected":false},"author":243,"featured_media":126667,"comment_status":"open","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"fifu_image_url":"","fifu_image_alt":""},"categories":[73411],"tags":[3086,4388],"amp_enabled":true,"_links":{"self":[{"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/posts\/126664"}],"collection":[{"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/users\/243"}],"replies":[{"embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/comments?post=126664"}],"version-history":[{"count":4,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/posts\/126664\/revisions"}],"predecessor-version":[{"id":499621,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/posts\/126664\/revisions\/499621"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/media\/126667"}],"wp:attachment":[{"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/media?parent=126664"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/categories?post=126664"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/tags?post=126664"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}