Question

In adjacent figure, PR > PQ and PS bisects ∠QPR. Prove that ∠PSR > ∠PSQ.

Answer 1

Answer:

Step-by-step explanation:

Given:

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)]

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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