{"id":4914,"date":"2020-09-09T11:16:30","date_gmt":"2020-09-09T05:46:30","guid":{"rendered":"https:\/\/kseebsolutions.guru\/?p=4914"},"modified":"2021-07-02T15:03:35","modified_gmt":"2021-07-02T09:33:35","slug":"kseeb-solutions-class-8-maths-chapter-6-ex-6-2","status":"publish","type":"post","link":"https:\/\/kseebsolutions.guru\/kseeb-solutions-class-8-maths-chapter-6-ex-6-2\/","title":{"rendered":"KSEEB Solutions for Class 8 Maths Chapter 6 Theorems on Triangles Ex 6.2"},"content":{"rendered":"

Students can Download Maths Chapter 6 Theorems on Triangles Ex 6.2 Questions and Answers, Notes Pdf, KSEEB Solutions for Class 8 Maths<\/a> helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations.<\/p>\n

Karnataka Board Class 8 Maths Chapter 6 Theorems on Triangles Ex 6.2<\/h2>\n

Question 1.
\nIn a triangle ABC, if \u2220A= 55\u00b0 and\u2220B.= 40\u00b0 find \u2220C
\nAnswer:
\n\u2220A +\u2220B + \u2220C = 180\u00b0
\n[Sum of the angles of a triangle 180\u00b0]
\n55 + 40 + \u2220C = 180\u00b0
\n95 + \u2220C = 180\u201c
\n\u2220C = 180 – 95
\n\u2220C = 85\u00b0<\/p>\n

Question 2.
\nIn a right-angled triangle, if one of the other two angles is 35\u00b0, find the remaining angle.
\nAnswer:
\nLet the angles be \u2220A, \u2220B and \u2220C
\nThen \u2220A = 90\u00b0,\u2220B = 35\u00b0 and \u2220C = ?
\n\u2220A + \u2220B + \u2220C = 180\u00b0
\n90 + 35 + \u2220C = 180\u00b0
\n125 + \u2220C = 180
\n\u2220C = 180 – 125 = 55\u00b0
\n\u2220C = 55\u00b0<\/p>\n

Question 3.
\nIf the vertex angle of an isosceles triangle is 50\u00b0 find the other two angles.
\nAnswer:
\nIn an isosceles triangle, the base angles are equal, Let the each base angle be \u2018x\u2019
\n\u2220A + \u2220B + \u2220C = l80\u00b0
\n50 + x + x = 180\u00b0
\n[Sum of the angles of a traingle 180\u00b0]
\n50 + 2x = 180
\n2x = 180 – 50
\n2x = 130
\nx = \\(\\frac{130}{2}\\)
\nx = 65\u00b0
\n\u2234 The other two angles are equal to 65\u00b0 & 65\u00b0<\/p>\n

Question 4.
\nThe angles of a triangle are in the ratio 1: 2 : 3. Determine the three angles.
\nAnswer:
\nLet the common ratio be \u2018x\u2019
\nThe three angles are x, 2x and 3x
\nx + 2x + 3x = 180\u00b0
\n[Sum of the angles of a triangle 180\u00b0]
\n6x = 180\u00b0
\nx = \\(\\frac{180}{6}\\)\u00a0= 30
\nx = 30\u00b0
\n2x = 2 \u00d7 30 = 60\u00b0
\n3x = 3 \u00d7 30 = 90\u00b0
\n\u2234 The angles are 30\u00b0, 60\u00b0 and 90\u00b0<\/p>\n

Question 5.
\nIn the adjacent triangle ABC, find the value of x and calculate the measure of all the angles of the triangle.
\nAnswer:
\n\u2220A + \u2220B + \u2220C = 180\u00b0
\n[Sum of the angles of a triangle 180\u00b0]
\nx + 15 + x – 15 + x + 30 = 180\u00b0
\n3x + 30 = 180\u00b0
\n3x = 180\u00b0 \u2013 30
\n3x = 150
\nx = \\(\\frac{150}{3}\\)\u00a0= 50\u00b0
\n\u2220A = x +15 = 50 + 15 = 65\u00b0
\n\u2220B = x – 15 = 50 – 15 = 35\u00b0
\n\u2220C = x + 30 = 50 + 30 = 80\u00b0
\n\u2234 The angles are 65\u00b0, 3 5\u00b0 and 80\u00b0<\/p>\n

Question 6.
\nThe angles of a triangle are arranged in ascending order of their magnitude. If the difference between two consecutive angles is 10\u00b0 find the three angles.
\nAnswer:
\nLet the first angle be x, second angle is x + 10\u00b0 and third angle is x + 20
\nx + x + 10 + x + 20 = 180\u00b0
\n[Sum of the angles of a triangle 180\u00b0]
\n3x + 30 = 180\u00b0
\n3x = 180 – 30
\n3x = 150\u00b0
\nx = \\(\\frac{150}{3}\\)\u00a0= 50\u00b0
\nx = 50\u00b0
\n\u2234 First angle x = 50\u00b0
\nSecond angle = x + 10 = 50 + 10 = 60\u00b0
\nThird angle = x + 20 = 50 + 20 = 70\u00b0
\nThree angles are 50\u00b0, 60\u00b0 & 70\u00b0<\/p>\n","protected":false},"excerpt":{"rendered":"

Students can Download Maths Chapter 6 Theorems on Triangles Ex 6.2 Questions and Answers, Notes Pdf, KSEEB Solutions for Class 8 Maths helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations. Karnataka Board Class 8 Maths Chapter 6 Theorems on Triangles Ex 6.2 Question 1. In a …<\/p>\n","protected":false},"author":5,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[6],"tags":[],"jetpack_sharing_enabled":true,"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/4914"}],"collection":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/users\/5"}],"replies":[{"embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/comments?post=4914"}],"version-history":[{"count":0,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/4914\/revisions"}],"wp:attachment":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/media?parent=4914"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/categories?post=4914"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/tags?post=4914"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}