Question

In the 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

Answer 1

Construction: Join PO and OQ
In △POR and △QOR
OP = OQ (Radii)
RP = RQ (Tangents from the external point are congruent)
OR = OR (Common)
By SSS congruency, △POR ≅ △QOR
∠PRO = ∠QRO (C.P.C.T)
Now, ∠PRO + ∠QRO = ∠PRQ
⇒ 2∠PRO = 120^∘
⇒ ∠PRO = 60^∘
Now, In △POR
$$\begin{gathered} \cos {60}^{°}=\frac{\mathrm{PR}}{\mathrm{OR}} \\ \Rightarrow \frac{1}{2}=\frac{\mathrm{PR}}{\mathrm{OR}} \\ \Rightarrow \mathrm{OR}=2\mathrm{PR} \\ \Rightarrow \mathrm{OR}=\mathrm{PR}+\mathrm{PR} \\ \Rightarrow \mathrm{OR}=\mathrm{PR}+\mathrm{RQ} \end{gathered}$$