Q : 3     In Fig. \small 10.37 ,  \small \angle PQR=100^{\circ}, where P, Q and R are points on a circle with centre O.  Find \small \angle OPR.

             

Answers (1)

Construction: Join PS and RS.

PQRS is a cyclic quadrilateral.

So, \anglePSR + \anglePQR = 180 \degree

\Rightarrow \angle PSR+100 \degree=180 \degree

\Rightarrow \angle PSR=180 \degree-100 \degree=80 \degree

Here, \anglePOR = 2 \anglePSR

\Rightarrow \angle POR=2\times 80 \degree=160 \degree

In \triangleOPR , 

          OP=OR    (Radii )

          \angleORP = \angleOPR    (the angles opposite to equal sides)

In \triangleOPR , 

       \angleOPR+\angleORP+\anglePOR=180 \degree

\Rightarrow 2\angle OPR+160 \degree= 180 \degree

\Rightarrow 2\angle OPR=180 \degree- 160 \degree

\Rightarrow 2\angle OPR=20 \degree

\Rightarrow \angle OPR=10 \degree

 

 

 

 

 

 

 

 

 

 

 

 

 

 

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