In Fig. 3, from an external point P, two tangents PT and PS are drawn to a circle with centre O and radius r. If \[OP=2r\], show that \[\angle OTS=\angle OST=30{}^\circ \]. |
Answer:
We have, \[OP=2r\] Let \[\angle TOP=\theta \] In \[\Delta \,OTP,\] \[\cos \theta =\frac{OT}{OP}=\frac{r}{2r}=\frac{1}{2}\] \[\because \] \[\theta =60{}^\circ \] Hence, \[\angle TOS=2\theta =2\times 60{}^\circ =120{}^\circ \] In \[\Delta \,TOS\] \[\angle \,TOS+\angle OTS+\angle OST=180{}^\circ \] \[120{}^\circ +2\angle OTS=180{}^\circ (\because \,\angle OTS=\angle OST)\] \[2\angle OTS=180{}^\circ -120{}^\circ \] \[\angle OTS=30{}^\circ \] Hence, \[\angle OTS=\angle OST=30{}^\circ \] Hence Proved.
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