Question

In Fig 5.51, PR > PQ and PS bisects angle QPR. Prove
that PSR> angle PSQ❓❓❓❓❓❓❓❓.​

Answer 1

Step-by-step explanation:

given PR>PQ

ANGLE PSR > ANGLE PSQ ( angle PSR is opposite to PR in ∆PRS and angle PSQ is opposite to PQ in ∆PQS)

Answer 2

Hello Mate,

Here is your answer,

Given: PR > PQ

Angles that are opposite to the longer sides are greater.

Angle opposite to PR => PQR (= PQS)

Angle opposite to PQ => PRQ (= PRS)

So, we can say that PQR > PRQ

(Or)

PQS > PRS

From Triangle PQS,

Angles (PQS + SPQ + PSQ = 180°) - - - (1)

(angle sum property of triangle)

Similarly from triangle PRS,

Angles (PRS + PSR + SPR = 180°) - - - (2)

Equate (1) and (2),

(PQS + SPQ + PSQ = PRS + PSR + SPR)

- - - (4)

Given: PS is bisector

So, SPR = SPQ

Let SPR = SPQ = x

Now substituting 'x' in (4), we get,

PQS + PSQ + x = PRS + PSR + x

So, 'x' gets canceled,

PQS + PSQ = PRS + PSR

and we know PQS > PRS

PQS + PSQ > PRS + PSR

So, PSQ > PSR

Hence, Proved.