Question

in ∆ PQR , PR is greater than PQ and PS bisects angle QPR prove that angle PSR is greater than angle PSQ

Answer 1

sorry for bad handwriting

Answer 2

Hello mate =_=

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Solution:

PR>PQ              (Given)

⇒∠PQR>∠PRQ             ....... (1)

(In any triangle, the angle opposite to the longer side is larger.)

We also have ∠PQR+∠QPS+∠PSQ=180°      (Angle sum property of triangle)     

⇒∠PQR=180°−∠QPS−∠PSQ             ......(2)

And, ∠PRQ+∠RPS+∠PSR=180°               (Angle sum property of triangle)          

⇒∠PRQ=180°−∠PSR−∠RPS            ....... (3)

 Putting (2) and (3) in (1), we get

180°−∠QPS−∠PSQ>180°−∠PSR−∠RPS             

We also have ∠QPS=∠RPS, using this in the above equation, we get

−∠PSQ>−∠PSR

⇒∠PSQ<∠PSR

Hence Proved

hope, this will help you.

Thank you______❤

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