Question

^QPS =2^SPR, ^Q=^R+40? and ^PSR=120? find ^QPR​

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

Proof :-

_______

In ∆ PQR

°.° 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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