In figure 3, two tangents RQ and RP are drawn from an external point R to the circle with centre O. If \[\angle PRQ=120{}^\circ \], then prove that, \[OR=PR+RQ\]. |
Answer:
O is the centre of the circle and \[\angle PRQ=120{}^\circ \] Construction: Join \[OP,OQ\] To prove: \[OP=PR+RQ\] Proof: We know that, Tangent to a circle is perpendicular to the radius at the point of tangent i.e., \[OP\bot RP\] and \[OQ\bot RQ\]. \[\therefore \angle OPR=\angle OQR=90{}^\circ \] Now, in \[\Delta \,OPR\] and \[\Delta \,OQR\], \[OP=OQ\] [Radius of circle] \[OR=OR\] [Common] \[\angle OPR=\angle OQR=90{}^\circ \] [Each\[90{}^\circ \]] \[\therefore \Delta \,OPR\cong \Delta \,OQR\] [By SSA congruence] So, \[PR=QR\] [By c.p.c.t.] and \[\angle ORP=\angle ORQ\] \[=\frac{120{}^\circ }{2}=60{}^\circ \] Now, in \[\Delta \,OPR\] \[\cos \,\,60{}^\circ =\frac{PR}{OR}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left[ \because \,\cos \theta =\frac{\text{Base}}{\text{Hypotenuse}} \right]\] \[\frac{1}{2}=\frac{PR}{OQ}\] \[OR=2PR\] \[OR=PR+PR\] \[OR=PR+RQ\] \[[\because \,PR=RQ]\] Hence, \[OR=PR+RQ\] Hence Proved.
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