Question-53145

Question Number 53145 by ajfour last updated on 18/Jan/19

Commented by ajfour last updated on 18/Jan/19

If entire string length is l, find θ in terms of l,a, r, and R. Or find l from a,r,R, and θ.

$${If}\:{entire}\:{string}\:{length}\:{is}\:{l},\:{find}\:\theta \\ $$$${in}\:{terms}\:{of}\:{l},{a},\:{r},\:{and}\:{R}. \\ $$$${Or}\:{find}\:{l}\:{from}\:{a},{r},{R},\:{and}\:\theta. \\ $$

Commented by tanmay.chaudhury50@gmail.com last updated on 18/Jan/19

Answered by ajfour last updated on 18/Jan/19

l=a+R(π+θ+2cot (θ/2)) +r((π/2)−(θ/2)+tan (θ/2)−(2/(sin θ))) (my answer).

$${l}={a}+{R}\left(\pi+\theta+\mathrm{2cot}\:\frac{\theta}{\mathrm{2}}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:+{r}\left(\frac{\pi}{\mathrm{2}}−\frac{\theta}{\mathrm{2}}+\mathrm{tan}\:\frac{\theta}{\mathrm{2}}−\frac{\mathrm{2}}{\mathrm{sin}\:\theta}\right)\: \\ $$$$\left({my}\:{answer}\right). \\ $$

Answered by mr W last updated on 18/Jan/19

Commented by mr W last updated on 18/Jan/19

AB=a BC^(⌢) =r((π/2)−(θ/2)) CD=HD−HC=(R/(tan (θ/2)))−(r/(tan θ)) DE_(⌣) =R(π+θ) EF=HE−HF=(R/(tan (θ/2)))−(r/(sin θ)) ⇒l=a+r((π/2)−(θ/2))+(R/(tan (θ/2)))−(r/(tan θ))+R(π+θ)+(R/(tan (θ/2)))−(r/(sin θ)) ⇒l=a+r((π/2)−(θ/2)−(1/(tan θ))−(1/(sin θ)))+R(π+θ+(2/(tan (θ/2))))

$${AB}={a} \\ $$$$\overset{\frown} {{BC}}={r}\left(\frac{\pi}{\mathrm{2}}−\frac{\theta}{\mathrm{2}}\right) \\ $$$${CD}={HD}−{HC}=\frac{{R}}{\mathrm{tan}\:\frac{\theta}{\mathrm{2}}}−\frac{{r}}{\mathrm{tan}\:\theta} \\ $$$$\underset{\smile} {{DE}}={R}\left(\pi+\theta\right) \\ $$$${EF}={HE}−{HF}=\frac{{R}}{\mathrm{tan}\:\frac{\theta}{\mathrm{2}}}−\frac{{r}}{\mathrm{sin}\:\theta} \\ $$$$\Rightarrow{l}={a}+{r}\left(\frac{\pi}{\mathrm{2}}−\frac{\theta}{\mathrm{2}}\right)+\frac{{R}}{\mathrm{tan}\:\frac{\theta}{\mathrm{2}}}−\frac{{r}}{\mathrm{tan}\:\theta}+{R}\left(\pi+\theta\right)+\frac{{R}}{\mathrm{tan}\:\frac{\theta}{\mathrm{2}}}−\frac{{r}}{\mathrm{sin}\:\theta} \\ $$$$\Rightarrow{l}={a}+{r}\left(\frac{\pi}{\mathrm{2}}−\frac{\theta}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{tan}\:\theta}−\frac{\mathrm{1}}{\mathrm{sin}\:\theta}\right)+{R}\left(\pi+\theta+\frac{\mathrm{2}}{\mathrm{tan}\:\frac{\theta}{\mathrm{2}}}\right) \\ $$

Commented by Otchere Abdullai last updated on 18/Jan/19

$${thus}\:{my}\:{prof}\:{W} \\ $$

Commented by ajfour last updated on 18/Jan/19

$${Thank}\:{you}\:{very}\:{much}\:{Sir}.\: \\ $$

Commented by mr W last updated on 18/Jan/19

$${we}\:{had}\:{some}\:{difference}\:{in}\:{our}\:{results}. \\ $$$${do}\:{you}\:{know}\:{the}\:{reason}\:{sir}? \\ $$

Commented by ajfour last updated on 18/Jan/19

yes, they are the same, tan (θ/2)−(1/(2sin (θ/2)cos (θ/2))) = −(1/(tan θ)) hence tan (θ/2)−(2/(sin θ)) = −(1/(tan θ))−(1/(sin θ))^.

$${yes},\:{they}\:{are}\:{the}\:{same}, \\ $$$$\mathrm{tan}\:\frac{\theta}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{2sin}\:\frac{\theta}{\mathrm{2}}\mathrm{cos}\:\frac{\theta}{\mathrm{2}}}\:=\:−\frac{\mathrm{1}}{\mathrm{tan}\:\theta} \\ $$$${hence}\:\:\mathrm{tan}\:\frac{\theta}{\mathrm{2}}−\frac{\mathrm{2}}{\mathrm{sin}\:\theta}\:=\:−\frac{\mathrm{1}}{\mathrm{tan}\:\theta}−\frac{\mathrm{1}}{\mathrm{sin}\:\theta}\:^{.} \\ $$

Commented by mr W last updated on 18/Jan/19

$${thanks}\:{sir}\:{for}\:{the}\:{explanation}! \\ $$

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