Question

Two tangents RQ and RP are drawn from an external point R to the circle with centre O. If angle PRQ=120, 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° 

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

Answer 2

Answer:

Step-by-step explanation:

given: PQ and RP are tangent from an external point R to the circle with center O such that angle PRQ=120

TO PROVE: PR+RQ=OR

CONSTRUCTION: Join OP and OR

PROOF: ∠OPR=∠OQR=90 (PR AND PQ ARE TANGENT)

NOW,

in ΔPRR

cos60=PR/OR

⇒1/2=PR/OR

⇒PR=1/2OR .......................... (1)

similarly,

in triangle QRO

RQ=1/2OR ....................(2)

adding (1) and (2)

we get

PR+PQ=  1/2OR +1/2OR

PR++PQ=OR

HENCE PROVED