Question Number 77651 by ajfour last updated on 08/Jan/20
Commented by ajfour last updated on 08/Jan/20
$${Wish}\:{to}\:{know}\:{s}\:{in}\:{terms}\:{of}\:{x},{R}. \\ $$$${s}={AC},\:\:{x}={AB},\:{BP}=\mathrm{1}. \\ $$
Commented by mr W last updated on 08/Jan/20
$${DC}\bot{AP}\:,\:{sir}? \\ $$
Commented by ajfour last updated on 08/Jan/20
$${yes}\:{sir}. \\ $$
Commented by key of knowledge last updated on 08/Jan/20
$$\mathrm{how}\:\mathrm{wanderfullllll} \\ $$
Answered by mr W last updated on 08/Jan/20
$${OB}={R}−{x} \\ $$$${DB}=\sqrt{{R}^{\mathrm{2}} −\left({R}−{x}\right)^{\mathrm{2}} }=\sqrt{\left(\mathrm{2}{R}−{x}\right){x}} \\ $$$$\frac{{BC}}{{DB}}=\frac{{BP}}{{AB}} \\ $$$$\frac{{x}−{s}}{\sqrt{\left(\mathrm{2}{R}−{x}\right){x}}}=\frac{\mathrm{1}}{{x}} \\ $$$$\Rightarrow{s}={x}−\sqrt{\frac{\mathrm{2}{R}}{{x}}−\mathrm{1}} \\ $$
Commented by ajfour last updated on 08/Jan/20
$${And}\:{as}\:{s}\:{approaches}\:{zero},\: \\ $$$${x}\:{can}\:{be}\:{found}\:{from}\:{x}^{\mathrm{3}} +{x}=\mathrm{2}{R}. \\ $$$${Thanks}\:{Sir},\:{want}\:{to}\:{discuss} \\ $$$${with}\:{you}\:{this}\:{question}\:{at}\:{length} \\ $$$$...{please}\:{cooperate}.. \\ $$
Commented by mr W last updated on 08/Jan/20
$${with}\:{pleasure}\:{sir}! \\ $$
Commented by ajfour last updated on 08/Jan/20
$${I}\:{think}\:{we}\:{should}\:{try}\:{taking} \\ $$$${limitfor}\:{finding}\:\:\:\mathrm{cos}\:\theta=\underset{{s}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\left({s}\mathrm{cos}\:\theta\right)\left({x}\right)}{{s}\left({x}\right)} \\ $$$${where}\:{i}\:{mean}\:{we}\:{have}\:{found} \\ $$$${s}\left({x}\right)\:,\:{we}\:{should}\:{find}\:{side} \\ $$$${s}\mathrm{cos}\:\theta\:{in}\:{terms}\:{of}\:{x}\:{without} \\ $$$${using}\:\mathrm{cos}\:\theta.. \\ $$
Commented by Tawa11 last updated on 29/Dec/21
$$\mathrm{Grest}\:\mathrm{sir} \\ $$