Question

"Question 5 In the given figure, PR > PQ and PS bisects ∠QPR. Prove that ∠PSR >∠PSQ.

Class 9 - Math - Triangles Page 132"

Answer 1

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)]

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

Answer 2

⭐ I am useless User ⭐

Hi

here is ur answer

Prof :-
_______

In ∆ PQR

PR > PQ ( Given )

.°. < PQR > <PRQ ---------(1)

( A triangles longest sides of opposite angle is big )

°.° PS , < QPRs a bisect.

.°. < QPS = < RPS

in ∆ PQR ------------(2)

< PQR + < QPS + < PSQ = 180° ----------(3)

( all angles of a triangle equal is 180° )

In ∆ PRS

< PRS + < SPR + < PSR = 180° -------(4)

( all angles of a triangle equal is 180° )

From (3) and (4) ,

< PQR + < QPS + < PSQ = < PRS + < SPR + < PSR

=> < PQR + < PSQ = < PRS + < PSR

=> < PRS + < PSR = < PQR + < PSQ

=> < PRS + < PSR > < PRQ + < PSQ ( From (1) )

=> < PRQ + < PSR > < PRS + < PSQ ( °.° < PRQ = < PRS

=> < PSR > < PSQ

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