{"id":125887,"date":"2023-09-09T16:23:00","date_gmt":"2023-09-09T10:53:00","guid":{"rendered":"https:\/\/www.kopykitab.com\/blog\/?p=125887"},"modified":"2023-11-28T10:20:42","modified_gmt":"2023-11-28T04:50:42","slug":"rd-sharma-class-10-solutions-chapter-4-exercise-4-3","status":"publish","type":"post","link":"https:\/\/www.kopykitab.com\/blog\/rd-sharma-class-10-solutions-chapter-4-exercise-4-3\/","title":{"rendered":"RD Sharma Class 10 Solutions Chapter 4 Exercise 4.3 (Updated for 2024)"},"content":{"rendered":"\n<p><img class=\"alignnone size-full wp-image-125910\" src=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/09\/RD-Sharma-Class-10-Solutions-Chapter-4-Exercise-4.3-scaled.jpg\" alt=\"RD Sharma Class 10 Solutions Chapter 4 Exercise 4.3\" width=\"2048\" height=\"1152\" srcset=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/09\/RD-Sharma-Class-10-Solutions-Chapter-4-Exercise-4.3-scaled.jpg 2048w, https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/09\/RD-Sharma-Class-10-Solutions-Chapter-4-Exercise-4.3-768x432.jpg 768w, https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/09\/RD-Sharma-Class-10-Solutions-Chapter-4-Exercise-4.3-1536x864.jpg 1536w\" sizes=\"(max-width: 2048px) 100vw, 2048px\" \/><\/p>\n<p><strong>RD Sharma Class 10 Solutions Chapter 4 Exercise 4.3:\u00a0<\/strong>Students will learn about the internal and external bisectors of a triangle angle from this exercise. The chapter-by-chapter <a href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-class-10-solutions-for-maths\/\"><strong>RD Sharma Class 10 Solutions<\/strong><\/a> created by Kopykitab experts is a very beneficial material for students to refer to and confidently prepare for their exams. Students can also use the links below to download <a href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-class-10-solutions-chapter-4-triangles\/\"><strong>RD Sharma Solutions for Class 10 Maths Chapter 4 Triangles<\/strong><\/a> Exercise 4.3 PDF.<\/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-69e77c0a03952\" 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-69e77c0a03952\"><\/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-10-solutions-chapter-4-exercise-4-3\/#download-rd-sharma-class-10-solutions-chapter-4-exercise-43-free-pdf\" title=\"Download RD Sharma Class 10 Solutions Chapter 4 Exercise 4.3 Free PDF\">Download RD Sharma Class 10 Solutions Chapter 4 Exercise 4.3 Free PDF<\/a><ul class='ez-toc-list-level-3'><li class='ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-2\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-class-10-solutions-chapter-4-exercise-4-3\/#access-answers-to-rd-sharma-solutions-class-10-maths-chapter-4-exercise-43-important-question-with-answers\" title=\"Access answers to RD Sharma Solutions Class 10 Maths Chapter 4 Exercise 4.3- Important Question with Answers\">Access answers to RD Sharma Solutions Class 10 Maths Chapter 4 Exercise 4.3- Important Question with Answers<\/a><\/li><\/ul><\/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-10-solutions-chapter-4-exercise-4-3\/#faqs-on-rd-sharma-class-10-solutions-chapter-4-exercise-43\" title=\"FAQs on RD Sharma Class 10 Solutions Chapter 4 Exercise 4.3\">FAQs on RD Sharma Class 10 Solutions Chapter 4 Exercise 4.3<\/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-10-solutions-chapter-4-exercise-4-3\/#is-it-important-to-study-all-of-the-concepts-included-in-rd-sharma-class-10-solutions-chapter-4-exercise-43\" title=\"Is it important to study all of the concepts included in RD Sharma Class 10 Solutions Chapter 4 Exercise 4.3?\">Is it important to study all of the concepts included in RD Sharma Class 10 Solutions Chapter 4 Exercise 4.3?<\/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-10-solutions-chapter-4-exercise-4-3\/#where-can-i-download-rd-sharma-class-10-solutions-chapter-4-exercise-43-free-pdf\" title=\"Where can I download RD Sharma Class 10 Solutions Chapter 4 Exercise 4.3 free PDF?\">Where can I download RD Sharma Class 10 Solutions Chapter 4 Exercise 4.3 free PDF?<\/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-10-solutions-chapter-4-exercise-4-3\/#what-are-the-benefits-of-using-rd-sharma-class-10-solutions-chapter-4-exercise-43\" title=\"What are the benefits of using RD Sharma Class 10 Solutions Chapter 4 Exercise 4.3?\">What are the benefits of using RD Sharma Class 10 Solutions Chapter 4 Exercise 4.3?<\/a><\/li><\/ul><\/li><\/ul><\/nav><\/div>\n<h2><span class=\"ez-toc-section\" id=\"download-rd-sharma-class-10-solutions-chapter-4-exercise-43-free-pdf\"><\/span>Download RD Sharma Class 10 Solutions Chapter 4 Exercise 4.3 Free PDF<span class=\"ez-toc-section-end\"><\/span><\/h2>\n<div id=\"example1\" style=\"text-align: justify;\">\u00a0<\/div>\n<p style=\"text-align: justify;\"><style>\n.pdfobject-container { height: 800px;}<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-Solution-Class-10-Chapter-4-Exercise-4.3.pdf\", \"#example1\");<\/script><\/p>\n<p><a href=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/09\/RD-Sharma-Solution-Class-10-Chapter-4-Exercise-4.3.pdf\">RD Sharma Class 10 Solutions Chapter 4 Exercise 4.3<\/a><\/p>\n<h3><span class=\"ez-toc-section\" id=\"access-answers-to-rd-sharma-solutions-class-10-maths-chapter-4-exercise-43-important-question-with-answers\"><\/span>Access answers to RD Sharma Solutions Class 10 Maths Chapter 4 Exercise 4.3- Important Question with Answers<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<p><strong>1. In a\u00a0\u0394\u00a0ABC, AD is the bisector of\u00a0\u2220\u00a0A, meeting side BC at D.<\/strong><\/p>\n<p><strong>(i) if BD = 2.5 cm, AB = 5 cm, and AC = 4.2 cm, find DC.<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Given: \u0394\u00a0ABC and AD bisects\u00a0\u2220A, meeting side BC at D. And BD = 2.5 cm, AB = 5 cm, and AC = 4.2 cm.<\/p>\n<p>Required to find: DC<\/p>\n<p><img class=\"\" title=\"R D Sharma Solutions For Class 10 Maths Chapter 4 Triangles ex 4.3 - 1\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2019\/10\/r-d-sharma-solutions-for-class-10-maths-chapter-4.png\" alt=\"R D Sharma Solutions For Class 10 Maths Chapter 4 Triangles ex 4.3 - 1\" width=\"417\" height=\"195\" \/><\/p>\n<p>Since AD is the bisector of \u2220 A meeting side BC at D in \u0394 ABC<\/p>\n<p>\u21d2 AB\/ AC = BD\/ DC<\/p>\n<p>5\/ 4.2 = 2.5\/ DC<\/p>\n<p>5DC = 2.5 x 4.2<\/p>\n<p>\u2234 DC = 2.1 cm<\/p>\n<p><strong>(ii) if BD = 2 cm, AB = 5 cm, and DC = 3 cm, find AC.<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Given: \u0394\u00a0ABC and AD bisects\u00a0\u2220A, meeting side BC at D. And BD = 2 cm, AB = 5 cm, and DC = 3 cm.<\/p>\n<p>Required to find: AC<\/p>\n<p><img class=\"\" title=\"R D Sharma Solutions For Class 10 Maths Chapter 4 Triangles ex 4.3 - 2\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2019\/10\/r-d-sharma-solutions-for-class-10-maths-chapter-4-1.png\" alt=\"R D Sharma Solutions For Class 10 Maths Chapter 4 Triangles ex 4.3 - 2\" width=\"442\" height=\"207\" \/><\/p>\n<p>Since AD is the bisector of \u2220 A meeting side BC at D in \u0394 ABC<\/p>\n<p>\u21d2 AB\/ AC = BD\/ DC<\/p>\n<p>5\/ AC = 2\/ 3<\/p>\n<p>2AC = 5 x 3<\/p>\n<p>\u2234 AC = 7.5 cm<\/p>\n<p><strong>(iii) if AB = 3.5 cm, AC = 4.2 cm, and DC = 2.8 cm, find BD.<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Given: \u0394\u00a0ABC and AD bisects\u00a0\u2220A, meeting side BC at D. And AB = 3.5 cm, AC = 4.2 cm, and DC = 2.8 cm.<\/p>\n<p>Required to find: BD<\/p>\n<p><img class=\"\" title=\"R D Sharma Solutions For Class 10 Maths Chapter 4 Triangles ex 4.3 - 3\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2019\/10\/r-d-sharma-solutions-for-class-10-maths-chapter-4-2.png\" alt=\"R D Sharma Solutions For Class 10 Maths Chapter 4 Triangles ex 4.3 - 3\" width=\"406\" height=\"190\" \/><\/p>\n<p>Since AD is the bisector of \u2220 A meeting side BC at D in \u0394 ABC<\/p>\n<p>\u21d2 AB\/ AC = BD\/ DC<\/p>\n<p>3.5\/ 4.2 = BD\/ 2.8<\/p>\n<p>4.2 x BD = 3.5 x 2.8<\/p>\n<p>BD = 7\/3<\/p>\n<p>\u2234 BD = 2.3 cm<\/p>\n<p><strong>(iv) if AB = 10 cm, AC = 14 cm, and BC = 6 cm, find BD and DC.<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Given: In\u00a0\u0394\u00a0ABC, AD is the bisector of\u00a0\u2220A meeting side BC at D. And, AB = 10 cm, AC = 14 cm, and BC = 6 cm<\/p>\n<p>Required to find: BD and DC.<\/p>\n<p><img class=\"\" title=\"R D Sharma Solutions For Class 10 Maths Chapter 4 Triangles ex 4.3 - 4\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2019\/10\/r-d-sharma-solutions-for-class-10-maths-chapter-4-3.png\" alt=\"R D Sharma Solutions For Class 10 Maths Chapter 4 Triangles ex 4.3 - 4\" width=\"417\" height=\"195\" \/><\/p>\n<p>Since AD is the bisector of \u2220A<\/p>\n<p>We have,<\/p>\n<p>AB\/AC = BD\/DC (AD is bisector of\u00a0\u2220 A and side BC)<\/p>\n<p>Then, 10\/ 14 = x\/ (6 \u2013 x)<\/p>\n<p>14x = 60 \u2013 6x<\/p>\n<p>20x = 60<\/p>\n<p>x = 60\/20<\/p>\n<p>\u2234 BD = 3 cm and DC = (6 \u2013 3) = 3 cm.<\/p>\n<p><strong>(v) if AC = 4.2 cm, DC = 6 cm, and BC = 10 cm, find AB.<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Given: \u0394\u00a0ABC and AD bisects\u00a0\u2220A, meeting side BC at D. And AC = 4.2 cm, DC = 6 cm, and BC = 10 cm.<\/p>\n<p>Required to find: AB<\/p>\n<p><img class=\"\" title=\"R D Sharma Solutions For Class 10 Maths Chapter 4 Triangles ex 4.3 - 5\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2019\/10\/r-d-sharma-solutions-for-class-10-maths-chapter-4-4.png\" alt=\"R D Sharma Solutions For Class 10 Maths Chapter 4 Triangles ex 4.3 - 5\" width=\"460\" height=\"215\" \/><\/p>\n<p>Since AD is the bisector of \u2220 A meeting side BC at D in \u0394 ABC<\/p>\n<p>\u21d2 AB\/ AC = BD\/ DC<\/p>\n<p>AB\/ 4.2 = BD\/ 6<\/p>\n<p>We know that,<\/p>\n<p>BD = BC \u2013 DC = 10 \u2013 6 = 4 cm<\/p>\n<p>\u21d2 AB\/ 4.2 = 4\/ 6<\/p>\n<p>AB = (2 x 4.2)\/ 3<\/p>\n<p>\u2234 AB = 2.8 cm<\/p>\n<p><strong>(vi) if AB = 5.6 cm, AC = 6 cm, and DC = 3 cm, find BC.<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Given: \u0394\u00a0ABC and AD bisects\u00a0\u2220A, meeting side BC at D. And AB = 5.6 cm, AC = 6 cm, and DC = 3 cm.<\/p>\n<p>Required to find: BC<\/p>\n<p><img class=\"\" title=\"R D Sharma Solutions For Class 10 Maths Chapter 4 Triangles ex 4.3 - 6\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2019\/10\/r-d-sharma-solutions-for-class-10-maths-chapter-4-5.png\" alt=\"R D Sharma Solutions For Class 10 Maths Chapter 4 Triangles ex 4.3 - 6\" width=\"436\" height=\"204\" \/><\/p>\n<p>Since AD is the bisector of \u2220 A meeting side BC at D in \u0394 ABC<\/p>\n<p>\u21d2 AB\/ AC = BD\/ DC<\/p>\n<p>5.6\/ 6 = BD\/ 3<\/p>\n<p>BD = 5.6\/ 2 = 2.8cm<\/p>\n<p>And we know that,<\/p>\n<p>BD = BC \u2013 DC<\/p>\n<p>2.8 = BC \u2013 3<\/p>\n<p>\u2234 BC = 5.8 cm<\/p>\n<p><strong>(vii) if AB = 5.6 cm, BC = 6 cm, and BD = 3.2 cm, find AC.<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Given: \u0394\u00a0ABC and AD bisects\u00a0\u2220A, meeting side BC at D. And AB = 5.6 cm, BC = 6 cm, and BD = 3.2 cm.<\/p>\n<p>Required to find: AC<\/p>\n<p><img class=\"\" title=\"R D Sharma Solutions For Class 10 Maths Chapter 4 Triangles ex 4.3 - 7\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2019\/10\/r-d-sharma-solutions-for-class-10-maths-chapter-4-6.png\" alt=\"R D Sharma Solutions For Class 10 Maths Chapter 4 Triangles ex 4.3 - 7\" width=\"442\" height=\"206\" \/><\/p>\n<p>Since AD is the bisector of \u2220 A meeting side BC at D in \u0394 ABC<\/p>\n<p>\u21d2 AB\/ AC = BD\/ DC<\/p>\n<p>5.6\/ AC = 3.2\/ DC<\/p>\n<p>And we know that<\/p>\n<p>BD = BC \u2013 DC<\/p>\n<p>3.2 = 6 \u2013 DC<\/p>\n<p>\u2234 DC = 2.8 cm<\/p>\n<p>\u21d2 5.6\/ AC = 3.2\/ 2.8<\/p>\n<p>AC = (5.6 x 2.8)\/ 3.2<\/p>\n<p>\u2234 AC = 4.9 cm<\/p>\n<p><strong>(viii) if AB = 10 cm, AC = 6 cm, and BC = 12 cm, find BD and DC.<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Given: \u0394\u00a0ABC and AD bisects\u00a0\u2220A, meeting side BC at D. AB = 10 cm, AC = 6 cm, and BC = 12 cm.<\/p>\n<p>Required to find: DC<\/p>\n<p><img class=\"\" title=\"R D Sharma Solutions For Class 10 Maths Chapter 4 Triangles ex 4.3 - 8\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2019\/10\/c-users-tnluser-downloads-rd-sharma-class-10-chap.png\" alt=\"C:\\Users\\Tnluser\\Downloads\\RD-sharma-class-10-chapter-4.3---1.8.png\" width=\"474\" height=\"221\" \/><\/p>\n<p>Since AD is the bisector of \u2220 A meeting side BC at D in \u0394 ABC<\/p>\n<p>\u21d2 AB\/ AC = BD\/ DC<\/p>\n<p>10\/ 6 = BD\/ DC \u2026\u2026.. (i)<\/p>\n<p>And we know that<\/p>\n<p>BD = BC \u2013 DC = 12 \u2013 DC<\/p>\n<p>Let BD = x,<\/p>\n<p>\u21d2 DC = 12 \u2013 x<\/p>\n<p>Thus (i) becomes,<\/p>\n<p>10\/ 6 = x\/ (12 \u2013 x)<\/p>\n<p>5(12 \u2013 x) = 3x<\/p>\n<p>60 -5x = 3x<\/p>\n<p>\u2234 x = 60\/8 = 7.5<\/p>\n<p>Hence, DC = 12 \u2013 7.5 = 4.5cm and BD = 7.5 cm<\/p>\n<p><strong>2. In Figure 4.57, AE is the bisector of the exterior \u2220CAD meeting BC produced in E. If AB = 10 cm, AC = 6 cm, and BC = 12 cm, find CE.<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Given: AE is the bisector of the exterior\u00a0<strong>\u2220<\/strong>CAD and AB = 10 cm, AC = 6 cm, and BC = 12 cm.<\/p>\n<p>Required to find: CE<\/p>\n<p><img class=\"\" title=\"R D Sharma Solutions For Class 10 Maths Chapter 4 Triangles ex 4.3 - 9\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2019\/10\/r-d-sharma-solutions-for-class-10-maths-chapter-4-7.png\" alt=\"R D Sharma Solutions For Class 10 Maths Chapter 4 Triangles ex 4.3 - 9\" width=\"387\" height=\"181\" \/><\/p>\n<p>Since AE is the bisector of the exterior\u00a0<strong>\u2220<\/strong>CAD.<\/p>\n<p>BE \/ CE = AB \/ AC<\/p>\n<p>Let\u2019s take CE as x.<\/p>\n<p>So, we have<\/p>\n<p>BE\/ CE = AB\/ AC<\/p>\n<p>(12+x)\/ x = 10\/ 6<\/p>\n<p>6x + 72 = 10x<\/p>\n<p>10x \u2013 6x = 72<\/p>\n<p>4x = 72<\/p>\n<p>\u2234 x = 18<\/p>\n<p>Therefore, CE = 18 cm.<\/p>\n<p><strong>3. \u00a0In fig. 4.58,\u00a0\u0394 ABC\u00a0is a triangle such that\u00a0AB\/AC = BD\/DC,\u00a0\u2220B=70<sup>o<\/sup>,\u00a0\u2220C = 50<sup>o<\/sup>, find\u00a0\u2220BAD.<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Given:\u00a0\u0394 ABC<strong>\u00a0<\/strong>such that AB\/AC = BD\/DC,\u00a0\u2220B = 70<sup>o<\/sup>\u00a0and\u00a0\u2220C = 50<sup>o<\/sup><\/p>\n<p>Required to find:\u00a0\u2220BAD<\/p>\n<p><img class=\"\" title=\"R D Sharma Solutions For Class 10 Maths Chapter 4 Triangles ex 4.3 - 10\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2019\/10\/r-d-sharma-solutions-for-class-10-maths-chapter-4-8.png\" alt=\"R D Sharma Solutions For Class 10 Maths Chapter 4 Triangles ex 4.3 - 10\" width=\"321\" height=\"150\" \/><\/p>\n<p>We know that,<\/p>\n<p>In \u0394ABC,<\/p>\n<p>\u2220A\u00a0= 180 \u2013 (70 + 50) [Angle sum property of a triangle]<\/p>\n<p>= 180 \u2013 120<\/p>\n<p>= 60<sup>o<\/sup><\/p>\n<p>Since,<\/p>\n<p>AB\/AC = BD\/DC,<\/p>\n<p>AD is the angle bisector of angle \u2220A.<\/p>\n<p>Thus,<\/p>\n<p>\u2220BAD = \u2220A\/2 = 60\/2 = 30<sup>o<\/sup><\/p>\n<p>We have provided complete details of RD Sharma Class 10 Solutions Chapter 4 Exercise 4.3. If you have any queries related to <a href=\"https:\/\/www.cbse.gov.in\/\" target=\"_blank\" rel=\"noopener\"><strong>CBSE<\/strong><\/a>\u00a0Class 10, feel free to ask us in the comment section below.<\/p>\n<h2><span class=\"ez-toc-section\" id=\"faqs-on-rd-sharma-class-10-solutions-chapter-4-exercise-43\"><\/span>FAQs on RD Sharma Class 10 Solutions Chapter 4 Exercise 4.3<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-1631178064020\" class=\"rank-math-list-item\">\n<h3 class=\"rank-math-question \"><span class=\"ez-toc-section\" id=\"is-it-important-to-study-all-of-the-concepts-included-in-rd-sharma-class-10-solutions-chapter-4-exercise-43\"><\/span>Is it important to study all of the concepts included in RD Sharma Class 10 Solutions Chapter 4 Exercise 4.3?<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<div class=\"rank-math-answer \">\n\n<p>Yes, learning all of the concepts included in RD Sharma Solutions for Class 10 Maths Chapter 4 Exercise 4.3 is required in order to achieve high scores on the Class 10 board exams. These solutions were created by subject matter specialists who compiled model questions that covered all of the textbook&#8217;s exercise questions.<\/p>\n\n<\/div>\n<\/div>\n<div id=\"faq-question-1631178203860\" class=\"rank-math-list-item\">\n<h3 class=\"rank-math-question \"><span class=\"ez-toc-section\" id=\"where-can-i-download-rd-sharma-class-10-solutions-chapter-4-exercise-43-free-pdf\"><\/span>Where can I download RD Sharma Class 10 Solutions Chapter 4 Exercise 4.3 free PDF?<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<div class=\"rank-math-answer \">\n\n<p>You can download RD Sharma Class 10 Solutions Chapter 4 Exercise 4.3 free PDF from the above article.<\/p>\n\n<\/div>\n<\/div>\n<div id=\"faq-question-1631178333530\" class=\"rank-math-list-item\">\n<h3 class=\"rank-math-question \"><span class=\"ez-toc-section\" id=\"what-are-the-benefits-of-using-rd-sharma-class-10-solutions-chapter-4-exercise-43\"><\/span>What are the benefits of using RD Sharma Class 10 Solutions Chapter 4 Exercise 4.3?<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<div class=\"rank-math-answer \">\n\n<p>1. Correct answers according to the latest CBSE guidelines and syllabus.<br \/>2. The RD Sharma Class 10 Solutions Chapter 4 Exercise 4.3 is written in simple language to assist students in their board examination, &amp; competitive examination preparation.<\/p>\n\n<\/div>\n<\/div>\n<\/div>\n<\/div>","protected":false},"excerpt":{"rendered":"<p>RD Sharma Class 10 Solutions Chapter 4 Exercise 4.3:\u00a0Students will learn about the internal and external bisectors of a triangle angle from this exercise. The chapter-by-chapter RD Sharma Class 10 Solutions created by Kopykitab experts is a very beneficial material for students to refer to and confidently prepare for their exams. Students can also use &#8230; <a title=\"RD Sharma Class 10 Solutions Chapter 4 Exercise 4.3 (Updated for 2024)\" class=\"read-more\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-class-10-solutions-chapter-4-exercise-4-3\/\" aria-label=\"More on RD Sharma Class 10 Solutions Chapter 4 Exercise 4.3 (Updated for 2024)\">Read more<\/a><\/p>\n","protected":false},"author":238,"featured_media":125910,"comment_status":"open","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"fifu_image_url":"","fifu_image_alt":""},"categories":[73411,2985,73410],"tags":[3243,9206,73520,4388],"amp_enabled":true,"_links":{"self":[{"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/posts\/125887"}],"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\/238"}],"replies":[{"embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/comments?post=125887"}],"version-history":[{"count":4,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/posts\/125887\/revisions"}],"predecessor-version":[{"id":513204,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/posts\/125887\/revisions\/513204"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/media\/125910"}],"wp:attachment":[{"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/media?parent=125887"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/categories?post=125887"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/tags?post=125887"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}