Question

In fig 7.51,
PR>PQ and PS bisects angle QPR. prove that angle PSR >angle PSQ

Answer 1

$<b>Given:</b>$ ▪PR>PQ ▪angle QPS = angle RPS
$<b> To Prove: </b>$angle PSR > angle PSQ
$<b>Proof:</b>$PR>PQ
=angle PQR > angle PRQ
=angle PQR + angle QPS > angle PRQ + angle RPS......(i).......
[°•° angle QPS = angle RPS(given)]

✴Also by Exterior Angle Property
angle PSR = angle PQS + angle QPS ....(ii)....
angle PSQ = angle RPS + angle PRS .....(iii)....

✴Using (ii) and (iii) in (i)
angle PSR > angle PSQ



HOPE IT HELPS !
PLS MARK BRAINLIEST !
DON'T FORGET TO FOLLOW !
✨☺✨

Answer 2

Hello mate =_=

____________________________

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______❤

_____________________________❤