{"id":4919,"date":"2020-09-09T11:37:12","date_gmt":"2020-09-09T06:07:12","guid":{"rendered":"https:\/\/kseebsolutions.guru\/?p=4919"},"modified":"2021-07-02T15:03:35","modified_gmt":"2021-07-02T09:33:35","slug":"kseeb-solutions-class-8-maths-chapter-6-ex-6-3","status":"publish","type":"post","link":"https:\/\/kseebsolutions.guru\/kseeb-solutions-class-8-maths-chapter-6-ex-6-3\/","title":{"rendered":"KSEEB Solutions for Class 8 Maths Chapter 6 Theorems on Triangles Ex 6.3"},"content":{"rendered":"

Students can Download Maths Chapter 6 Theorems on Triangles Ex 6.3 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 6.3<\/h2>\n

Question 1.
\nThe exterior angles obtained on producing the base of a triangle both ways are 104\u00b0 and 136\u00b0. Find the angles of the triangle.
\nAnswer:
\nIn triangle ABC, BC is produced on either sides.
\nLet \u2220ABC = 140\u00b0 and \u2220ACE = 136\u00b0
\n\u2220ABD +\u2220ABC = 180\u00b0 [ Linear pair]
\n104 + \u2220ABC = 180\u00b0
\n\u2220ABC = 180 – 104
\n\u2220 ABC = 76\u00b0
\n\"KSEEB
\n\u2220ACB + \u2220ACE = 180\u00b0 [Linearpair]
\n\u2220ACB + 136\u00b0 = 180\u00b0
\n\u2220ACB = 180 – 136 \u2220ACB = 44\u00b0
\n\u2220ABD + \u2220ACB + \u2220BAC = 180\u00b0
\n[Sum of the angles of a triangle 180\u00b0 ]
\n76 + 44 + \u2220BAC = 180\u00b0
\n120 + \u2220BAC = 180\u00b0
\n\u2220BAC = 180 – 120 \u2220BAC = 60\u00b0
\nThree angles are 76\u00b0, 44\u00b0 & 60\u00b0<\/p>\n

Question 2.
\nSides BC, CA and AB of a triangle ABC are produced in order, forming exterior angles \u2220ACD, \u2220BAE, and\u2220CBF.
\nShow that \u2220ACD + \u2220BAE + \u2220CBF = 360\u00b0.
\nAnswer:
\n\"KSEEB
\nABC + \u2220CBF = 180\u00b0 [Linear pair]\u2026\u2026\u2026..(i)
\n\u2220ACB + \u2220ACD = 180\u00b0 [Linear pair]\u2026\u2026(ii)
\n\u2220BAC + \u2220BAE = 180\u00b0 [Linear pair]\u2026\u2026.(iii)
\nByadding(i), (ii) and (iii)
\n\u2220ABC + \u2220CBF + \u2220ACB + \u2220ACD+
\n\u2220BAC + \u2220BAE = 180 + 180 + 180
\n\u2220ABC + \u2220ACB + \u2220BAC +
\n\u2220CBF + \u2220ACD + \u2220BAE = 540\u00b0
\n180 + \u2220CBF + \u2220ACD + \u2220BAE = 540\u00b0
\n[Sum of the angles of a triangle 180\u00b0]
\n\u2220CBF + \u2220ACD + \u2220BAE = 360\u00b0<\/p>\n

Question 3
\nCompute the value of x in each of the following figures
\nAnswer:
\ni.
\n\"KSEEB
\nAB = AC
\n\u2234 \u2220ABC = \u2220ACB = 50\u00b0
\n\u2220ACB + \u2220ACD = 180\u00b0 [Linear pair]
\n50 + x = 180\u00b0
\nx = 180 – 50 = 130\u00b0<\/p>\n

ii.
\n\"KSEEB
\n\u2220ABC + \u2220CBF = 180\u00b0 [Linear pair].
\n106\u00b0+\u2220ABC = 180\u00b0
\n\u2220ABC = 180 – 106 = 74\u00b0
\n\u2220EAC = \u2220ABC + \u2220ACB
\n130\u00b0 =74 +x
\n130 – 74 = x
\n\u2234x = 56\u00b0<\/p>\n

iii.
\n\"KSEEB
\n\u2220QPR = \u2220TPU = 65\u00b0
\n[Vertically opposite angles]
\n\u2220PRS = \u2220PQR + \u2220QPR [Exterior angle=Sum of interior opposire angle]
\n100 = 65 + x
\n100 – 65 = x
\n35 = x
\n\u2234 x = 35\u00b0.
\niv.
\n\"KSEEB
\n\u2220BAE + \u2220BAC = 180\u00b0 [Linear pair]
\n120\u00b0 + \u2220BAC = 180\u00b0
\n\u2220BAC = 180 – 120
\n\u2220BAC = 60\u00b0
\n\u2220ACD = \u2220BAC + \u2220ABC
\n[Exterior angle = Sum of interior opposuite angle]
\n112\u00b0 = 60+x
\n112 – 60 = x
\n52 = x
\nx = 52\u00b0
\nv.
\n\"KSEEB
\nIn \u0394 ABC, BA = BC (data)
\n\u2234 \u2220BAC = \u2220BCA = 20\u00b0
\n[Base angles of an isosceles traingle]
\n\u2220ABD = \u2220BAC + \u2220BCA
\n[Exterior angle=Sum of interior opposite angle]
\nx = 20 + 20
\nx=40\u00b0<\/p>\n

Question 4.
\nIn figure QT \u22a5 PR, \u2220TOR = 40\u00b0 and \u2220SPR = 30\u00b0 find \u2220TRS and \u2220PSQ.
\n\"KSEEB
\nAnswer:
\nIn \u0394TQR,
\n\u2220TQR + \u2220QTR + \u2220TRQ = 180\u00b0
\n40 + 90 + \u2220TRQ = 180\u00b0
\n130 + \u2220TRQ = 180
\n\u2220 TRQ = 180 \u2013 130
\n\u2220 TRQ = 50\u00b0
\n\u2220 TRS = 50\u00b0 [\u2220PRS is same as \u2220TRS ]
\nIn \u0394 PRS, RS is produced to Q
\n\u2234Exterior\u2220PSQ = \u2220SPR + \u2220PRS = 30 + 50
\n\u2220PSQ = 80\u00b0<\/p>\n

Question 5.
\nAn exterior angle of a triangle is 120\u00b0 and one of the interior opposite angle is 30 \u00b0. Find the other angles of the triangle.
\nAnswer:
\n\"KSEEB
\nIn \u0394 ABC, BC is produced to D. Let
\n\u2220ACD = 120\u00b0 and \u2220ABC = 30\u00b0
\nExterior \u2220ACD = \u2220BAC + \u2220ABC
\n120 = \u2220BAC + 30
\n120 \u2013 30 = \u2220BAC
\n90\u00b0 = \u2220BAC
\n\u2234\u2220BAC = 90
\n\u2220ACB +\u2220ACD = 180\u00b0 [Linear pair]
\n\u2220ACB + 120 = 180\u00b0
\n\u2220ACB = 180 \u2013 120
\n\u2220ACB = 60\u00b0
\nOther two angles are 90\u00b0 & 60\u00b0<\/p>\n","protected":false},"excerpt":{"rendered":"

Students can Download Maths Chapter 6 Theorems on Triangles Ex 6.3 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 6.3 Question 1. The exterior …<\/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\/4919"}],"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=4919"}],"version-history":[{"count":0,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/4919\/revisions"}],"wp:attachment":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/media?parent=4919"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/categories?post=4919"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/tags?post=4919"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}