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Transcript

Ex7.4, 2 (Method 1) In the given figure sides AB and AC of ABC are extended to points P and Q respectively. Also, PBC < QCB. Show that AC > AB. Similarly, Now, PBC < QCB A + ACB < A + ABC ACB < ABC AB < AC Thus, AC > AB Hence proved. Ex7.4, 2 (Method 2) In the given figure sides AB and AC of ABC are extended to points P and Q respectively. Also, PBC < QCB. Show that AC > AB. Similarly, Now, PBC < QCB PBC > QCB 180 PBC > 180 QCB

  1. Chapter 7 Class 9 Triangles
  2. Serial order wise

About the Author

Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science and Computer Science at Teachoo