Question

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Answer 1

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∠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 0 = 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

$\huge\blue{Thank\:You}$

Answer 2

Step-by-step explanation:

∠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 0 = 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