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Question-77651




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

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