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Question Number 16579 by Joel577 last updated on 24/Jun/17

Commented by Joel577 last updated on 24/Jun/17

$$\mathrm{Mr}.\mathrm{Ajfour},\mathrm{why}\mathrm{\angle }P\mathrm{is}2\theta ?$$

Commented by Joel577 last updated on 24/Jun/17

$$\mathrm{pls}\mathrm{explain}\mathrm{sir}$$

Commented by ajfour last updated on 24/Jun/17

$$\mathrm{let}\mathrm{point}\mathrm{above}\mathrm{P}\mathrm{on}\mathrm{CD}\mathrm{is}\mathrm{Q}.$$$$\mathrm{I}\mathrm{have}\mathrm{let}\mathrm{\angle }\mathrm{DPQ}=\theta$$$$\mathrm{PQ}=\mathrm{DQ},\mathrm{so}\mathrm{\angle }\mathrm{PDQ}=\mathrm{\angle }\mathrm{DPQ}=\theta$$$$\mathrm{\angle }\mathrm{CQP}=\mathrm{\angle }\mathrm{PDQ}+\mathrm{\angle }\mathrm{DPQ}=2\theta$$$$\mathrm{Now}\mathrm{BP}=\mathrm{radius}\mathbf{r}$$$$\mathrm{Also}\mathrm{CP}=\mathrm{r}\mathrm{and}\mathrm{CQ}=\mathrm{BP}=\mathrm{r}$$$$\mathrm{hence}\mathrm{CP}=\mathrm{CQ}$$$$\mathrm{so}\mathrm{\angle }\mathrm{CPQ}=\mathrm{\angle }\mathrm{CQP}=2\theta .$$

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