Menu Close

Question-16579




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

Leave a Reply

Your email address will not be published. Required fields are marked *