NCERT Solutions for Class 9 Maths<\/a> Chapter 6 Lines and Angles Ex 6.3 Questions and Answers are prepared by our highly skilled subject experts.<\/p>\nNCERT Solutions for Class 9 Maths Chapter 6 Lines and Angles Exercise 6.3<\/h2>\n Question 1. \nIn Fig. 6.39, sides QP and RQ of \u2206PQR are produced to points S and T respectively. If \u2220SPR = 135 and \u2220PQT = 110\u00b0, Find \u2220PRQ. \n \nSolution: \nIn \u2206PQR \n\u2220SPR + \u2220RPQ = 180\u00b0 (Linear pair) \nor, \u2220RPQ = 180\u00b0 – 135\u00b0 (\u2235 \u2220SPR = 135\u00b0) \nor, \u2220RPQ = 45\u00b0 …….(i) \nAgain, \u2220PQT + \u2220PQR = 180\u00b0 (Linear pair) \nor, 110\u00b0 + \u2220PQR = 180\u00b0 (\u2235 \u2220PQT = 110\u00b0) \n\u2220PQR = 180\u00b0 – 110\u00b0 \nor, \u2220PQR = 70\u00b0 ……(ii) \nNow, we know that sum of all angles of a \u2206 is 180\u00b0 \nTherefore, \n\u2220RPQ + \u2220PQR + \u2220PRQ = 180\u00b0 \nor, 45\u00b0 + 70\u00b0 + \u2220RPQ = 180\u00b0 [From equ. (i) and (ii)] \n\u2220RPQ = 180\u00b0 – 45\u00b0 – 70\u00b0 = 65\u00b0 \nTherefore, \u2220RPQ = 65\u00b0<\/p>\n
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Question 2. \nIn Fig. 6.40, \u2220X = 62\u00b0, \u2220XYZ = 54\u00b0. If YO and ZO are the bisector of \u2220XYZ and \u2220XZY respectively of \u2206XYZ, find \u2220OZY and \u2220YOZ. \n \nSolution: \nIn \u2206XYZ, \n\u2220X + \u2220XYZ + \u2220XZY = 180\u00b0 (Angle sum property of \u2206) \nor, 62\u00b0 + 54\u00b0 + XZY = 180\u00b0 (\u2234 \u2220X = 65\u00b0 & \u2220XYZ = 54\u00b0) \nor, \u2220XYZ = 180\u00b0 – 116\u00b0 \nor, \u2220XZY = 64\u00b0 ….(i) \nAgain, \u2220XZY = \u2220XZO + \u2220OZY (\u2234 OZ is the bisector of \u2220XZY) \nor, 2\u2220OZY = 64\u00b0 \nor, \u2220OZY = 32\u00b0 \n\u2220OYZ + \u2220OZY + \u2220YOZ = 180\u00b0 (Angle sum property of \u2206) \n27\u00b0 + 32\u00b0 + \u2220YOZ = 180\u00b0 \nor, \u2220YOZ = 121\u00b0<\/p>\n
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Question 3. \nIn Fig. 6.41, if AB || DE, \u2220BAC = 35\u00b0 and \u2220CDE = 53\u00b0, find \u2220DCE. \n \nSolution: \nWe have given AB || DE \n\u2220BAC = \u2220DEC (Pair of alternate interior angles) \nor, \u2220DEC = 35\u00b0 (\u2235 \u2220BAC = 35\u00b0) \nNow, In \u2206CDE \n\u2220CDE + \u2220CED + \u2220DCE = 180\u00b0 (Angle sum property of \u2206) \n\u21d2 53 + 35 + \u2220DCE = 180\u00b0 \n\u21d2 \u2220DCE = 180\u00b0 – 53\u00b0 – 35\u00b0 \n\u21d2 \u2220DCE = 92\u00b0<\/p>\n
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Question 4. \nIn Fig. 6.42, if lines PQ and RS intersect at point T, such that \u2220PRT = 40\u00b0, \u2220RPT = 95\u00b0 and \u2220TSQ = 75\u00b0, find \u2220SQT. \n \nSolution: \nIn \u2206PRT, \n\u2220RPT + \u2220PRT + \u2220PTR = 180\u00b0 (Angle sum property of \u2206) \n95 + 40 + \u2220PTR = 180\u00b0 \nor, \u2220PTR = 180\u00b0 – 95\u00b0 – 40\u00b0 \nor, \u2220PTR = 45\u00b0 \nAgain, \u2220PTR = \u2220QTS (Vertically opposite angles) \n\u2234 \u2220QTS = 45\u00b0 \nNow, In \u2206QTS, \n\u2220QTS + \u2220TSQ + \u2220SQT = 180\u00b0 (Angle sum property of \u2206) \n45\u00b0 + 75\u00b0 + \u2220SQT = 180\u00b0 (\u2235 \u2220QTS = 45\u00b0 and \u2220TSQ = 75\u00b0) \nor, \u2220SQT = 180\u00b0 – 45\u00b0 – 75\u00b0 \n\u2234 \u2220SQT = 60\u00b0<\/p>\n
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Question 5. \nIn Fig. 6.43, if PQ \u22a5 PS, PQ || SR, \u2220SQR = 28\u00b0 and \u2220QRT = 65\u00b0, then find the value of x and y. \n \nSolution: \nWe have given PQ || SR \n\u2234 \u2220PQR = \u2220QRT (Pair of alternate interior angles) \n\u21d2 x + 28\u00b0 = 65\u00b0 \n\u21d2 x = 65\u00b0 – 28\u00b0 \n\u21d2 x = 37\u00b0 \nNow, in \u2206PQS \n\u2220QPS + x + y = 180\u00b0 (Angle sum property of \u2206) \n\u21d2 90\u00b0 + 37\u00b0 + y = 180\u00b0 (\u2234 PQ \u22a5 PS and x = 37) \n\u21d2 y = 180\u00b0 – 90\u00b0 – 37\u00b0 \n\u21d2 y = 53\u00b0<\/p>\n
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Question 6. \nIn Fig. 6.44, 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 \nSolution: \nIn \u2206PQR, side QR is produced to S, so by exterior angle property, \n\u2220PRS = \u2220P + \u2220PQR \n\u21d2 \\(\\frac {1}{2}\\) \u2220PRS = \\(\\frac {1}{2}\\) \u2220P + \\(\\frac {1}{2}\\) \u2220PQR \n\u21d2 \u2220TRS = \\(\\frac {1}{2}\\) \u2220P + \u2220TQR \u2026…(1) \n[\u2235 QT and RT are bisectors of \u2220PQR and \u2220PRS respectively.] \nNow, in \u2206QRT, we have \n\u2220TRS = \u2220TQR + \u2220T \u2026…(2) \n[Exterior angle property of a triangle] \nFrom (1) and (2), \nwe have \u2220TQR + \\(\\frac {1}{2}\\) \u2220P = \u2220TQR + \u2220T \n\u21d2 \\(\\frac {1}{2}\\) \u2220P = \u2220T \n\u21d2 \\(\\frac {1}{2}\\) \u2220QPR = \u2220QTR or \u2220QTR = \\(\\frac {1}{2}\\) \u2220QPR<\/p>\n","protected":false},"excerpt":{"rendered":"
These NCERT Solutions for Class 9 Maths Chapter 6 Lines and Angles Ex 6.3 Questions and Answers are prepared by our highly skilled subject experts. NCERT Solutions for Class 9 Maths Chapter 6 Lines and Angles Exercise 6.3 Question 1. In Fig. 6.39, sides QP and RQ of \u2206PQR are produced to points S and …<\/p>\n
NCERT Solutions for Class 9 Maths Chapter 6 Lines and Angles Ex 6.3<\/span> Read More »<\/a><\/p>\n","protected":false},"author":9,"featured_media":0,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"site-sidebar-layout":"default","site-content-layout":"default","ast-main-header-display":"","ast-hfb-above-header-display":"","ast-hfb-below-header-display":"","ast-hfb-mobile-header-display":"","site-post-title":"","ast-breadcrumbs-content":"","ast-featured-img":"","footer-sml-layout":"","theme-transparent-header-meta":"default","adv-header-id-meta":"","stick-header-meta":"default","header-above-stick-meta":"","header-main-stick-meta":"","header-below-stick-meta":"","spay_email":""},"categories":[6],"tags":[],"yoast_head":"\nNCERT Solutions for Class 9 Maths Chapter 6 Lines and Angles Ex 6.3 - MCQ Questions<\/title>\n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n\t \n\t \n\t \n