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Example 4 - Chapter 6 Class 9 Lines and Angles
Last updated at April 16, 2024 by Teachoo
Parallel lines and traversal - Problems
Chapter 6 Class 9 Lines and Angles
Concept wise
- Basic terms and defination
- Angles - Properties
- Angles - Problems
- Parallel lines and traversal - Properties
-
Parallel lines and traversal - Problems
- Triangle - Properties
- Triangle - Problems
Transcript
Example 4 In figure, if PQ || RS, MXQ = 135 and MYR = 40 , find XMY. Let AB || PQ, drawn through point M Since PQ || RS, AB || RS Since AB || PQ and MX is the transversal XMB + QXM = 180 XMB + 135 = 180 XMB = 180 135 XMB = 45 Similarly, Since AB || RS and MY is the transversal BMY = RYM BMY = 40 Now, XMY = XMB + MBY XMY = 45 + 40 XMY = 85
