In the below given figure, two tangents RQ and RP are drawn from an external point R to the circle with centre O. If∠PRQ = 120°, then prove that OR = PR + RQ.
Given that m∠PRQ = 120º
We know that the line joining the centre and the external point is the angle bisector between the tangents.
Thus, m∠PRO = m∠QRO =120º/2= 60º
Also we know that lengths of tangents from an external point are equal.
Thus, PR = RQ.
Join OP and OQ.
Since OP and OQ are the radii from the centre O,
OP ⊥ PR and OQ ⊥ RQ.
Thus, Δ OPR and ΔOQR are right angled congruent triangles.
Hence, ∠POR=90º- ∠PRO=90º - 60º= 30º
∠QOR=90°- ∠QRO=90° - 60° =30°
sin ∠QRO = sin30°=1/2
OR =PR + PR
OR =PR + QR