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Question Number 17524 by 786 last updated on 07/Jul/17

$$\mathrm{The}\mathrm{circle}\omega \mathrm{touches}\mathrm{the}\mathrm{circle}\mathrm{\Omega }$$$$\mathrm{internally}\mathrm{at}P.\mathrm{The}\mathrm{centre}O\mathrm{of}\mathrm{\Omega }\mathrm{is}$$$$\mathrm{outside}\omega .\mathrm{Let}XY\mathrm{be}\mathrm{a}\mathrm{diameter}\mathrm{of}\mathrm{\Omega }$$$$\mathrm{which}\mathrm{is}\mathrm{also}\mathrm{tangent}\mathrm{to}\omega .\mathrm{Assume}$$$$PY>PX.\mathrm{Let}PY\mathrm{intersect}\omega \mathrm{at}Z.\mathrm{If}$$$$YZ=2PZ,\mathrm{what}\mathrm{is}\mathrm{the}\mathrm{magnitude}\mathrm{of}$$$$\mathrm{\angle }PYX\mathrm{in}\mathrm{degrees}?$$

Answered by ajfour last updated on 07/Jul/17

$$\mathrm{\angle }\mathrm{PYX}=15°.$$

Commented by ajfour last updated on 07/Jul/17

Commented by ajfour last updated on 08/Jul/17

$$\mathrm{YZ}=2\mathrm{PZ}\mathrm{or}3\mathrm{PZ}=\mathrm{PY}$$$$3\left(2\mathrm{rcos}\theta \right)=2\mathrm{Rcos}\theta$$$$\Rightarrow \mathrm{R}=3\mathrm{r}$$$$\mathrm{In}△\mathrm{OCN},\sin 2\theta =\frac{\mathrm{r}}{\mathrm{R}-\mathrm{r}}=\frac{\mathrm{r}}{3\mathrm{r}-\mathrm{r}}$$$$\sin 2\theta =\frac{1}{2}\mathrm{or}\theta =15°.$$

Commented by Tinkutara last updated on 07/Jul/17

$$\mathrm{Thanks}\mathrm{Sir}!$$

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