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

Commented by Joel577 last updated on 24/Jun/17

Mr. Ajfour, why ∠P is 2θ?

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

Commented by Joel577 last updated on 24/Jun/17

pls explain sir

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

Commented by ajfour last updated on 24/Jun/17

let point above P on CD is Q. I have let ∠DPQ=θ PQ=DQ, so ∠PDQ=∠DPQ=θ ∠CQP =∠PDQ+∠DPQ=2θ Now BP=radius r Also CP= r and CQ=BP =r hence CP=CQ so ∠CPQ = ∠CQP = 2θ .

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

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