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 = ${120}^o$, then prove that OR = PR + RQ.

Answer 1

∠OPR = ∠OQR = 90° ---- 1
And in ΔOPR and ΔOQR
∠OPR = ∠ OQR = 90° (from equation 1)

OP = OQ (Radii of same circle)
And

OR = OR (common side)

ΔOPR = ΔOQR (ByRHS Congruency)

So, RP = RQ --- 2

And ∠ORP = ∠ORQ --- 3
∠PRQ = ∠ORP + ∠ORQ

Substitute ∠PQR = 120° (given)
And from equation 3 we get

∠ORP + ∠ORP = 120°
2 ∠ORP = 120°
∠ORP = 60°
Now in ΔOPR we get

Cos ∠ORP =

Cos 60° =


(we know cos 60° = ½)


OR = 2PR
OR = PR + PR (substitute value from equation 2 we get)
OR = PR + RQ