Question

In the given figure, two tangents RQ and RP are drawn from an external point R to the circle with centre O. If angle PRQ= 120° , then prove that OR= PR+RQ

Answer 1

as angle PRQ is 120
In triangle PRO n QRO
PR=QR
OR=OR
OP=OQ
BOTH THE TRIANGLES ARE CONGRUENT
therefore angles PRO n QRO
therefore angles PRO N QRO IS HALF OF ANGLE PRQ i.e. 60
NOW IN TRIANGLE PRO
COS 60= PR/OR
1/2=PR/OR
2PR=OR---(1)
NOW IN TRIANGLE QRO
COS 60=QR/OR
1/2=QR/OR
2QR=OR---(2)
ADD (1) AND (2)
2(PR+QR)=2OR
PR+QR=OR

Answer 2

Here , RQ and PR are the tangents from an external point R
Join OP and OQ
In ∆OPR and ∆OQR ,
OPR=OQR (each 90°)
OR = OR ( common )
OP=OQ ( radii )
thus , ∆ OPR ≈ ∆ OQR
PRO = QRO (cpct)
PRO = 1/2×120 = 60°
Now in rt. ∆ OPR ,
RP/OR = cos60°=1/2
RP/OR=1/2
OR=2RP
OR = RP + RP
OR = RP + RQ ( RP = RQ )
Hence , OR = RP + RQ. ( proved )