Question

Two tangents rq and rp are drawn from an external point r to the circle with centre ).if angle rpq=120 the 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°

And we know cos Ф = Adjacent/hypotenuse

So in ΔOPR we get 

Cos ∠ORP = PR/OR

Cos 60° = PR/OR

½ = PR/OR (we know cos 60° = ½)

OR = 2PR

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

OR = PR  + RQ

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