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Question-226486

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Question Number 226486 by ajfour last updated on 30/Nov/25
Commented by ajfour last updated on 30/Nov/25
yes
Commented by ajfour last updated on 30/Nov/25
Length OB=θ as well.
$${Length}\:{OB}=\theta\:\:{as}\:{well}. \\ $$
Commented by fantastic2 last updated on 30/Nov/25
in OB θ represents length and in AOC it represents angle?
$${in}\:{OB}\:\theta\:{represents}\:{length} \\ $$$${and}\:{in}\:{AOC}\:{it}\:{represents}\:{angle}? \\ $$
Answered by fantastic2 last updated on 30/Nov/25
Commented by fantastic2 last updated on 30/Nov/25
△ACE≊△BDC BO=θ=a BD=asin (90^0 −θ)=acos θ AE=sin θ DC=OC−asin θ=p−asin θ CE=cos θ−p ((AE)/(BD))=((CE)/(DC)) ((tan θ)/a)=((cos θ−p)/(p−asin θ)) ptan θ−asin θtan θ=acos θ−ap p(tan θ+a)=a(cos θ+sin θtan θ) ⇒p=((a(cos θ+sin θtan θ))/((tan θ+a))) p=((θ(cos θ+sin θtan θ))/((tan θ+θ))) ✓ example : θ=0.5 and angle θ=60^0 p≈0.448018 θ=.3 and angle θ=25^0 p≈0.431958 ((a(cos θ+sin θtan θ))/((tan θ+a))) =((atan θ(((cos^2 θ)/(sin θ))+sin θ))/((tan θ+a))) =((atan θ)/(sin θ(tan θ+a))) =((asin θ)/(asin θcos θ(((sin θ)/(acos θ))+1))) =(1/(cos θ+((sin θ)/a))) ∴p=((θ(cos θ+sin θtan θ))/((tan θ+θ)))=(1/(cos θ+((sin θ)/θ)))
$$\bigtriangleup{ACE}\approxeq\bigtriangleup{BDC} \\ $$$${BO}=\theta={a} \\ $$$${BD}={a}\mathrm{sin}\:\left(\mathrm{90}^{\mathrm{0}} −\theta\right)={a}\mathrm{cos}\:\theta \\ $$$${AE}=\mathrm{sin}\:\theta \\ $$$${DC}={OC}−{a}\mathrm{sin}\:\theta={p}−{a}\mathrm{sin}\:\theta \\ $$$${CE}=\mathrm{cos}\:\theta−{p} \\ $$$$\frac{{AE}}{{BD}}=\frac{{CE}}{{DC}} \\ $$$$\frac{\mathrm{tan}\:\theta}{{a}}=\frac{\mathrm{cos}\:\theta−{p}}{{p}−{a}\mathrm{sin}\:\theta} \\ $$$${p}\mathrm{tan}\:\theta−{a}\mathrm{sin}\:\theta\mathrm{tan}\:\theta={a}\mathrm{cos}\:\theta−{ap} \\ $$$${p}\left(\mathrm{tan}\:\theta+{a}\right)={a}\left(\mathrm{cos}\:\theta+\mathrm{sin}\:\theta\mathrm{tan}\:\theta\right) \\ $$$$\Rightarrow{p}=\frac{{a}\left(\mathrm{cos}\:\theta+\mathrm{sin}\:\theta\mathrm{tan}\:\theta\right)}{\left(\mathrm{tan}\:\theta+{a}\right)} \\ $$$${p}=\frac{\theta\left(\mathrm{cos}\:\theta+\mathrm{sin}\:\theta\mathrm{tan}\:\theta\right)}{\left(\mathrm{tan}\:\theta+\theta\right)}\:\checkmark \\ $$$${example}\:: \\ $$$$\theta=\mathrm{0}.\mathrm{5}\:{and}\:{angle}\:\theta=\mathrm{60}^{\mathrm{0}} \\ $$$${p}\approx\mathrm{0}.\mathrm{448018} \\ $$$$\theta=.\mathrm{3}\:{and}\:{angle}\:\theta=\mathrm{25}^{\mathrm{0}} \\ $$$${p}\approx\mathrm{0}.\mathrm{431958} \\ $$$$\frac{{a}\left(\mathrm{cos}\:\theta+\mathrm{sin}\:\theta\mathrm{tan}\:\theta\right)}{\left(\mathrm{tan}\:\theta+{a}\right)} \\ $$$$=\frac{{a}\mathrm{tan}\:\theta\left(\frac{\mathrm{cos}\:^{\mathrm{2}} \theta}{\mathrm{sin}\:\theta}+\mathrm{sin}\:\theta\right)}{\left(\mathrm{tan}\:\theta+{a}\right)} \\ $$$$=\frac{{a}\mathrm{tan}\:\theta}{\mathrm{sin}\:\theta\left(\mathrm{tan}\:\theta+{a}\right)} \\ $$$$=\frac{{a}\mathrm{sin}\:\theta}{{a}\mathrm{sin}\:\theta\mathrm{cos}\:\theta\left(\frac{\mathrm{sin}\:\theta}{{a}\mathrm{cos}\:\theta}+\mathrm{1}\right)} \\ $$$$=\frac{\mathrm{1}}{\mathrm{cos}\:\theta+\frac{\mathrm{sin}\:\theta}{{a}}} \\ $$$$\therefore{p}=\frac{\theta\left(\mathrm{cos}\:\theta+\mathrm{sin}\:\theta\mathrm{tan}\:\theta\right)}{\left(\mathrm{tan}\:\theta+\theta\right)}=\frac{\mathrm{1}}{\mathrm{cos}\:\theta+\frac{\mathrm{sin}\:\theta}{\theta}} \\ $$
Commented by fantastic2 last updated on 30/Nov/25
can we find c_(max) ? [0<θ≤1 ]
$${can}\:{we}\:{find}\:{c}_{{max}} \:?\:\left[\mathrm{0}<\theta\leqslant\mathrm{1}\:\right] \\ $$
Commented by mr W last updated on 30/Nov/25
why do you think θ≤1? following figure with θ>1 is clearly also possible. if C should lie between A and C, then 0<θ<(π/2).
$${why}\:{do}\:{you}\:{think}\:\theta\leqslant\mathrm{1}? \\ $$$${following}\:{figure}\:{with}\:\theta>\mathrm{1}\:{is}\:{clearly}\: \\ $$$${also}\:{possible}. \\ $$$${if}\:{C}\:{should}\:{lie}\:{between}\:{A}\:{and}\:{C}, \\ $$$${then}\:\mathrm{0}<\theta<\frac{\pi}{\mathrm{2}}. \\ $$
Commented by mr W last updated on 30/Nov/25
Commented by fantastic2 last updated on 30/Nov/25
i know θ>1 is possible but in the main q posted by ajfour sir, θ<1 [although he didnt say anything about the domain of θ,so you have a point]
$${i}\:{know}\:\theta>\mathrm{1}\:{is}\:{possible} \\ $$$${but}\:{in}\:{the}\:{main}\:{q}\:{posted}\:{by} \\ $$$${ajfour}\:{sir},\:\theta<\mathrm{1} \\ $$$$\left[{although}\:{he}\:{didnt}\:{say}\:{anything}\right. \\ $$$${about}\:{the}\:{domain}\:{of}\:\theta,{so}\:{you} \\ $$$$\left.{have}\:{a}\:{point}\right] \\ $$
Commented by fantastic2 last updated on 30/Nov/25
if we take θ relies in the circle p_(max) at θ=(π/2) & θ_(length) =1⇒p_(max) =1[when 0<θ_(length) ≤1]
$${if}\:{we}\:{take}\:\theta\:{relies}\:{in}\:{the}\:{circle} \\ $$$${p}_{{max}} \:\:{at}\:\theta=\frac{\pi}{\mathrm{2}}\:\&\:\theta_{{length}} =\mathrm{1}\Rightarrow{p}_{{max}} =\mathrm{1}\left[{when}\:\mathrm{0}<\theta_{{length}} \leqslant\mathrm{1}\right] \\ $$
Commented by mr W last updated on 30/Nov/25
if 0<θ≤(π/2): p_(max) =(π/2) if 0<θ≤1: p_(max) =(1/(cos 1+sin 1))≈0.7237
$${if}\:\mathrm{0}<\theta\leqslant\frac{\pi}{\mathrm{2}}:\:{p}_{{max}} =\frac{\pi}{\mathrm{2}} \\ $$$${if}\:\mathrm{0}<\theta\leqslant\mathrm{1}:\:\:{p}_{{max}} =\frac{\mathrm{1}}{\mathrm{cos}\:\mathrm{1}+\mathrm{sin}\:\mathrm{1}}\approx\mathrm{0}.\mathrm{7237} \\ $$
Answered by mr W last updated on 30/Nov/25
say ∠OAB=φ tan φ=θ ((OC)/(sin φ))=((OA)/(sin (φ+θ))) (p/(sin φ))=(1/(sin (φ+θ)))=(1/(sin φ cos θ+cos φ sin θ)) p=(1/(cos θ+((sin θ)/(tan φ))))=(1/(cos θ+((sin θ)/θ))) ✓
$${say}\:\angle{OAB}=\phi \\ $$$$\mathrm{tan}\:\phi=\theta \\ $$$$\frac{{OC}}{\mathrm{sin}\:\phi}=\frac{{OA}}{\mathrm{sin}\:\left(\phi+\theta\right)} \\ $$$$\frac{{p}}{\mathrm{sin}\:\phi}=\frac{\mathrm{1}}{\mathrm{sin}\:\left(\phi+\theta\right)}=\frac{\mathrm{1}}{\mathrm{sin}\:\phi\:\mathrm{cos}\:\theta+\mathrm{cos}\:\phi\:\mathrm{sin}\:\theta} \\ $$$${p}=\frac{\mathrm{1}}{\mathrm{cos}\:\theta+\frac{\mathrm{sin}\:\theta}{\mathrm{tan}\:\phi}}=\frac{\mathrm{1}}{\mathrm{cos}\:\theta+\frac{\mathrm{sin}\:\theta}{\theta}}\:\checkmark \\ $$
Commented by ajfour last updated on 30/Nov/25
thanks sir, i did it same way.
$${thanks}\:{sir},\:{i}\:{did}\:{it}\:{same}\:{way}. \\ $$

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