Question

Hey guys solve this question..........​

Answer 1

Step-by-step explanation:

PR > PQ & PS bisects ∠QPR

To prove:

∠PSR > ∠PSQ

Proof:

∠PQR > ∠PRQ — (i) (PR > PQ as angle opposite to larger side is larger.)

∠QPS = ∠RPS — (ii) (PS bisects ∠QPR)

∠PSR = ∠PQR +∠QPS — (iii)

(exterior angle of a triangle equals to the sum of opposite interior angles)

∠PSQ = ∠PRQ + ∠RPS — (iv)

(exterior angle of a triangle equals to the sum of opposite interior angles)

Adding (i) and (ii)

∠PQR + ∠QPS > ∠PRQ + ∠RPS

⇒ ∠PSR > ∠PSQ [from (i), (ii), (iii) and (iv)]

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Hope this will help you.......

Answer 2

Step-by-step explanation:

5..Ans. In PQR, PR > PQ [Given]

PQR > PRQ …..(i) [Angle opposite to longer side is greater]

Again 1 = 2 …..(ii) [ PS is the bisector of P]

PQR + 1 > PRQ + 2 ……….(iii)

But PQS + 1 + PSQ = PRS + 2 + PSR = [Angle sum property]

PQR + 1 + PSQ = PRQ + 2 + PSR ………(iv)

[PRS = PRQ and PQS = PQR]

From eq. (iii) and (iv),

PSQ < PSR

Or PSR > PSQ