In the given figure, two tangents RQ and RP are drawn from an external point R to the circle with centre O.
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
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
$\cos 60^{\circ}=\frac{\mathrm{PR}}{\mathrm{OR}}$
$\Rightarrow \frac{1}{2}=\frac{\mathrm{PR}}{\mathrm{OR}}$
$\Rightarrow \mathrm{OR}=2 \mathrm{PR}$
$\Rightarrow \mathrm{OR}=\mathrm{PR}+\mathrm{PR}$
$\Rightarrow \mathrm{OR}=\mathrm{PR}+\mathrm{RQ}$
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