Question

1. In the fig:QS is the bisector of < PQR . Also QS=QR.If <QRS =50^ Find the measure of <PQR PSRQ​

Answer 1

Step-by-step explanation:

In given △PAB,

QS, RS are the angle bisector of ∠AQR and ∠BRQ.

So, ∠AQS = ∠SQR and ∠QRS = ∠SRB --- (1)

Side PQ and PR of △PQR are produced to A and B respectively.

∴ Exterior of ∠AQR = ∠P + ∠R --- (2)

and Exterior of ∠BRQ = ∠P + ∠Q --- (3)

Adding (2) and (3) we het,

∠AQR + ∠BRQ = ∠P + ∠R + ∠P + ∠Q

∠AQR + ∠BRQ = 2∠P + ∠R + ∠Q

2∠SQR + 2∠QRS = ∠P + 180

∘

∠SQR + ∠QRS =

2

1

∠P + 90

∘

--- (4)

But in △QSR,

∠SQR + ∠QRS + ∠QSR = 180

∘

--- (5)

From equatiomn (4) and (5) we het,

2

1

∠P + 90

∘

+ ∠QSR = 180

∘

∴ ∠QSR +

2

1

∠ P = 90

∘