In the given figure, QS and RS are bisectors of exterior angles Q and R.Then ∠QSR + ∠QPR2 is equal to:
Let, ∠QSR+P2=x ..................(1)
Now, in a △QRS,
∠RQS+∠QRS+∠QSR=180∘ .................. (2)
From (1) and (2),
180∘−∠RQS−∠QRS+∠QPR2=x .................. (3)
In a △PQR,
∠QPR=180∘−∠QRP−∠PQR
Divide the above equation by 2, we get
∠QPR2=90∘−∠QRP2−∠PQR2 .................. (4)
From (3) and (4), we get
270∘−∠RQS−∠QRS−∠QRP2−∠PQR2=x .................. (5)
Now, ∠PQR+2∠RQS=180∘
Dividing by 2 , we get
∠PQR2+∠RQS=90∘ .................. (6) (Linear pair and exterior bisector)
∠QRP+2∠QRS=180∘
Dividing by 2, we get
∠QRP2+∠QRS=90∘ .................. (7) (Linear pair and exterior bisector)
From (5),(6) and (7), we get
270∘−90∘−90∘=x
x=90∘
Hence, option C is the correct answer.