{"id":491,"date":"2020-09-07T06:05:31","date_gmt":"2020-09-07T00:35:31","guid":{"rendered":"https:\/\/kseebsolutions.guru\/?p=491"},"modified":"2021-07-02T15:03:46","modified_gmt":"2021-07-02T09:33:46","slug":"kseeb-solutions-class-9-maths-chapter-3-ex-3-3","status":"publish","type":"post","link":"https:\/\/kseebsolutions.guru\/kseeb-solutions-class-9-maths-chapter-3-ex-3-3\/","title":{"rendered":"KSEEB Solutions for Class 9 Maths Chapter 3 Lines and Angles Ex 3.3"},"content":{"rendered":"

KSEEB Solutions for Class 9 Maths Chapter 3 Lines and Angles Ex 3.3<\/strong> are part of KSEEB Solutions for Class 9 Maths<\/a>. Here we have given Karnataka Board Class 9 Maths Solutions Chapter 3 Lines and Angles Exercise 3.3.<\/p>\n

Karnataka Board Class 9 Maths Chapter 3 Lines and Angles Ex 3.3<\/h2>\n

Question 1.
\nIn Fig. 3.39, sides QP and RQ of \u2206PQR are produced to points S and T respectively. If \u2220SPR = 135\u00b0 and \u2220PQT = 110\u00b0, find \u2220PRQ.
\n\"KSEEB
\nAnswer:
\n\"KSEEB
\nArms of the \u2206PQR QP and RP are produced to S and T and \u2220PQT = 110\u00b0, \u2220SPR = 135\u00b0. \u2220PRQ =?
\nStraight-line PR is on straight line SQ.
\n\u2220SPR and \u2220RPQ are Adjacent angles.
\n\u2234 \u2220SPR + \u2220RPQ = 180\u00b0
\n135 + \u2220RPQ = 180\u00b0
\n\u2234 \u2220RPQ =180 – 135
\n\u2220RPQ = 45\u00b0 (i)
\nSimilarly QP straight line is on straight line TR.
\n\u2220TQP and \u2220PQR are Adjacent angles.
\n\u2234 \u2220RQP + \u2220PQR = 180\u00b0
\n110 + \u2220PQR = 180\u00b0
\n\u2220PQR = 180 – 110
\n\u2234 \u2220PQR = 70\u00b0
\nNow, in \u2206PQR,
\n\u2220QPR + \u2220PQR + \u2220PRQ = 180\u00b0
\n45 + 70 + \u2220PRQ = 180\u00b0
\n115 + \u2220PRQ = 180\u00b0
\n\u2220PRQ = 180 – 115
\n\u2234 \u2220PRQ = 65\u00b0.<\/p>\n

Question 2.
\nIn Fig. 3.40, \u2220X = 62\u00b0. \u2220XYZ = 54\u00b0. If YO and ZO are the bisectors of \u2220XYZ and \u2220XZY respectively of \u2206XYZ, find \u2220OZY and \u2220YOZ
\n\"KSEEB
\nAnswer:
\nIn this figure, \u2220X = 62\u00b0
\n\u2220XYZ = 54\u00b0
\nIn \u2206XYZ, YO and ZO are angular bisectors of \u2220XYZ and \u2220XZY.
\nThen, \u2220OZY =?
\n\u2220OYZ =?
\n\"KSEEB
\nIn \u2206XYZ
\n\u2220X + \u2220Y + \u2220Z = 180\u00b0
\n62 + 54 + \u2220Z= 180
\n116 + \u2220Z= 180
\n\u2220Z= 180- 116
\n\u2234 \u2220Z = 64\u00b0
\nYO is the angular bisector of \u2220Y
\n\u2234 \u2220OYZ = \\(\\frac{54}{2}\\) = 27\u00b0
\nZO is the angular bisector of \u2220Z
\n\u2234 \u2220OZY = \\(\\frac{64}{2}\\) = 32\u00b0
\n\u2234 \u2220OZY = 32\u00b0
\nNow, in \u2206OYZ,
\n\u2220OYZ + \u2220OZY + \u2220YOZ = 180\u00b0
\n27 + 32 + \u2220YOZ = 180
\n59 + \u2220YOZ = 180
\n\u2220YOZ = 180 – 59
\n\u2234\u2220YOZ = 121\u00b0
\n\u2234 \u2220OZY = 32\u00b0
\n\u2220YOZ = 121\u00b0<\/p>\n

Question 3.
\nIn Fig. 3.41, if AB||DE, \u2220BAC = 35\u00b0 and \u2220CDE = 53\u00b0, find \u2220DCE.
\n\"KSEEB
\nAnswer:
\nIf AB || DE, \u2220BAC = 35, \u2220CDE = 53 then \u2220DCE = ?
\nAB || DE, AE is the bisector.
\n\u2234\u2220BAC = \u2220DEC = 35\u00b0 (\u2235 Alternate angles)
\n\u2234\u2220DEC= 35\u00b0
\n\"KSEEB
\nNow in \u2206CDE,
\n\u2220DCE + \u2220CDE + \u2220CED = 180\u00b0
\n\u2220DCE + 53 + 35 = 180
\n\u2220DCE + 88 = 180
\n\u2220DEC = 180 – 88
\n\u2234 \u2220DCE = 92\u00b0.<\/p>\n

Question 4.
\nIn Fig. 3.42, if lines PQ and RS intersect at point T, such that \u2220PRT = 40\u00b0, \u2220RPT = 95\u00b0 and \u2220TSQ = 75\u00b0, find \u2220SQT.
\n\"KSEEB
\nAnswer:
\nPQ and RS straight lines intersect at T.
\nIf \u2220PRT = 40\u00b0, \u2220RPT = 95\u00b0, and \u2220TSQ = 75\u00b0, then \u2220SQT =?
\n\"KSEEB
\nIn \u2206PRT,
\n\u2220RPT + \u2220PRT + \u2220PTR = 180\u00b0
\n95 + 40 + \u2220PTR = 180\u00b0
\n135 + \u2220PTR = 180
\n\u2220PTR = 180 – 135
\n\u2234 \u2220PTR = 45\u00b0
\n\u2220PTR = \u2220STQ = 45\u00b0 (\u2235 Vertically opposite angles)
\nIn \u2206TSQ,
\n\u2220STQ + \u2220TSQ + \u2220SQT =180
\n45 + 75 + \u2220SQT = 180
\n120 + \u2220SQT = 180
\n\u2234\u2220SQT = 180 – 120
\n\u2234 \u2220SQT = 60\u00b0.<\/p>\n

Question 5.
\nIn Fig. 3.43, PQ \u22a5 PS, PQ || SR, \u2220SQR = 28\u00b0 and \u2220QRT = 65\u00b0. then find the value of x and y.
\n\"KSEEB
\nAnswer:
\nPQ \u22a5 PS, PQ \u22a5 SR, \u2220SQR = 28\u00b0, \u2220QRT = 65\u00b0, Then x = ?, y = ?
\n\"KSEEB
\nSolution: \u2220QRT + \u2220QRS = 180\u00b0 (\u2235 Linear pairs)
\n65 + \u2220QRS = 180
\n\u2220QRS = 180 – 65
\n\u2234\u2220QRS =115\u00b0
\nIn \u2206SRQ,
\n\u2220QRS + \u2220SQR + \u2220RSQ = 180\u00b0
\n115 + 28 + \u2220RSQ = 180
\n\u2234\u2220RSQ =180 – 143
\n\u2234\u2220RSQ = 37
\nNow, \u2220RSQ = \u2220PQS
\n37\u00b0 = x
\n\u2234x = 37
\nIn \u2206SPQ,
\n\u2220SPQ + \u2220PSQ + \u2220PQS = 180\u00b0
\n90 + y + 37 = 180
\n\u2234y = 180- 127
\n\u2234y = 53\u00b0.<\/p>\n

Question 6.
\nIn Fig. 3.44, the side QR of \u2206PQR is produced to a point S. If the bisectors of \u2220PQR and \u2220PRS meet at point T, then prove that \u2220QTR = \\(\\frac{1}{2}\\)\u2220QPR.
\n\"KSEEB
\nAnswer:
\nData: Arm QR of \u2206PQR is produced upto S. Angular bisectors of \u2220PQR and \u2220PRS meet at T.
\nTo Prove: \u2220QTR = \\(\\frac{1}{2}\\) \u2220QPR
\n\"KSEEB
\nProof: Let \u2220PQR = 50\u00b0, and \u2220PRS = 120\u00b0
\n\u2220PQT = \u2220TQR = 25\u00b0
\n\u2220PRT = \u2220TRS = 60\u00b0
\nQR arm of \u2206 PQR is produced upto S.
\n\u2234Exterior angle \u2220PRS = \u2220PQR + \u2220QPR
\n120 = 50 + \u2220QPR
\n\u2234 \u2220QPR = 120 – 50
\n\u2220QPR = 70\u00b0
\n\u2234 \u2220PRQ = 60\u00b0
\nNow, in \u2206TRQ,
\n\u2220TQR + \u2220TRQ + \u2220QTR = 180\u00b0
\n25 + 120 + \u2220QTR = 180\u00b0
\n145 + \u2220QTR = 180\u00b0
\n\u2220QTR = 180 – 145
\n\u2234 \u2220QTR = 35\u00b0
\nNow, \u2220QTR = 35\u00b0 \u2220QPR = 70\u00b0
\n\u2220QTR = \\(\\frac{70}{2}\\)
\n\u2234 \u2220QTR = \\(\\frac{10}{2}\\) x \u2220QPR.<\/p>\n

We hope the KSEEB Solutions for Class 9 Maths Chapter 3 Lines and Angles Ex 3.3 help you. If you have any query regarding Karnataka Board Class 9 Maths Chapter 3 Lines and Angles Exercise 3.3, drop a comment below and we will get back to you at the earliest.<\/p>\n","protected":false},"excerpt":{"rendered":"

KSEEB Solutions for Class 9 Maths Chapter 3 Lines and Angles Ex 3.3 are part of KSEEB Solutions for Class 9 Maths. Here we have given Karnataka Board Class 9 Maths Solutions Chapter 3 Lines and Angles Exercise 3.3. Karnataka Board Class 9 Maths Chapter 3 Lines and Angles Ex 3.3 Question 1. In Fig. …<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[5],"tags":[],"jetpack_sharing_enabled":true,"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/491"}],"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\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/comments?post=491"}],"version-history":[{"count":0,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/491\/revisions"}],"wp:attachment":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/media?parent=491"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/categories?post=491"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/tags?post=491"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}