#### Question

In the following figure, PQ = QR, `∠`RQP = 68° , PC and CQ are tangents to the circle with centre O

(i) `∠`QOP

(ii) `∠`QCP

#### Solution

Consider two concentric circles with centres at O. Let AB and CD be two chords of the outer circle which touch the inner circle at the points M and N respectively.

To prove the given question, it is sufficient to prove AB = CD.

For this join OM, ON, OB and OD.

Let the radius of outer and inner circles be R and r respectively.

AB touches the inner circle at M.

AB is a tangent to the inner circle

∴ OM ⊥ AB

⇒ BM =`1/2 `AB

⇒ AB = 2BM

Similarly ON ⊥ CD, and CD = 2DN

Using Pythagoras theorem in ΔOMB and ΔOND

`OB^2 = OM^2 + BM^2 , OD^2 = ON^2 + DM^2`

⇒ `BM= sqrt(R^2 - r^2) ,DN = sqrt(R^2 - r^2`

Now,

`AB = 2BM = 2 sqrt(R^2 -r^2) ,CD = 2DN =2 sqrt(R^2 -r^2)`

∴ AB = CD

Hence proved.