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Question Number 166036 by mr W last updated on 12/Feb/22

Commented by mr W last updated on 12/Feb/22

if the chords on a quater circle have the length a, b, c respectively, find the radius R of the quater circle in terms of a, b, c.

$${if}\:{the}\:{chords}\:{on}\:{a}\:{quater}\:{circle}\:{have} \\ $$$${the}\:{length}\:{a},\:{b},\:{c}\:{respectively},\:{find} \\ $$$${the}\:{radius}\:{R}\:{of}\:{the}\:{quater}\:{circle}\:{in} \\ $$$${terms}\:{of}\:{a},\:{b},\:{c}. \\ $$

Commented by ajfour last updated on 13/Feb/22

great!

$${great}! \\ $$

Answered by Rasheed.Sindhi last updated on 12/Feb/22

A Try a^2 =R^2 +R^2 −2.R.R.cos α { ((a^2 =2R^2 (1−cos α))),((b^2 =2R^2 (1−cos β))),((c^2 =2R^2 (1−cos γ))) :} ; α+β+γ=90 R^2 =(a^2 /(2(1−cos α)))=(b^2 /(2(1−cos β)))=(c^2 /(2(1−cos γ))) 3R^2 =((a^2 +b^2 +c^2 )/(2(3−(cos α+cos β+cos γ)))) 3R^2 =((a^2 +b^2 +c^2 )/(2(3−(cos α+cos β+cos(90−(α+β))))) 3R^2 =((a^2 +b^2 +c^2 )/(2(3−(cos α+cos β+sin(α+β))))) R^2 =((a^2 +b^2 +c^2 )/(6(3−(cos α+cos β+sin(α+β))))) I think R is also dependant on α,β

$${A}\:\mathcal{T}{ry} \\ $$$${a}^{\mathrm{2}} =\mathrm{R}^{\mathrm{2}} +\mathrm{R}^{\mathrm{2}} −\mathrm{2}.\mathrm{R}.\mathrm{R}.\mathrm{cos}\:\alpha \\ $$$$\begin{cases}{{a}^{\mathrm{2}} =\mathrm{2R}^{\mathrm{2}} \left(\mathrm{1}−\mathrm{cos}\:\alpha\right)}\\{{b}^{\mathrm{2}} =\mathrm{2R}^{\mathrm{2}} \left(\mathrm{1}−\mathrm{cos}\:\beta\right)}\\{{c}^{\mathrm{2}} =\mathrm{2R}^{\mathrm{2}} \left(\mathrm{1}−\mathrm{cos}\:\gamma\right)}\end{cases}\:;\:\alpha+\beta+\gamma=\mathrm{90} \\ $$$$\mathrm{R}^{\mathrm{2}} =\frac{{a}^{\mathrm{2}} }{\mathrm{2}\left(\mathrm{1}−\mathrm{cos}\:\alpha\right)}=\frac{{b}^{\mathrm{2}} }{\mathrm{2}\left(\mathrm{1}−\mathrm{cos}\:\beta\right)}=\frac{{c}^{\mathrm{2}} }{\mathrm{2}\left(\mathrm{1}−\mathrm{cos}\:\gamma\right)} \\ $$$$\mathrm{3R}^{\mathrm{2}} =\frac{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} }{\mathrm{2}\left(\mathrm{3}−\left(\mathrm{cos}\:\alpha+\mathrm{cos}\:\beta+\mathrm{cos}\:\gamma\right)\right)} \\ $$$$\mathrm{3R}^{\mathrm{2}} =\frac{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} }{\mathrm{2}\left(\mathrm{3}−\left(\mathrm{cos}\:\alpha+\mathrm{cos}\:\beta+\mathrm{cos}\left(\mathrm{90}−\left(\alpha+\beta\right)\right)\right)\right.} \\ $$$$\mathrm{3R}^{\mathrm{2}} =\frac{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} }{\mathrm{2}\left(\mathrm{3}−\left(\mathrm{cos}\:\alpha+\mathrm{cos}\:\beta+\mathrm{sin}\left(\alpha+\beta\right)\right)\right)} \\ $$$$\mathrm{R}^{\mathrm{2}} =\frac{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} }{\mathrm{6}\left(\mathrm{3}−\left(\mathrm{cos}\:\alpha+\mathrm{cos}\:\beta+\mathrm{sin}\left(\alpha+\beta\right)\right)\right)} \\ $$$${I}\:{think}\:{R}\:{is}\:{also}\:{dependant}\:{on}\:\alpha,\beta\: \\ $$

Commented by mr W last updated on 12/Feb/22

you have sin (α/2)=(a/(2R)) ⇒(α/2)=sin^(−1) (a/(2R)) sin (β/2)=(b/(2R))⇒(β/2)=sin^(−1) (b/(2R)) sin (γ/2)=(c/(2R))⇒(γ/2)=sin^(−1) (c/(2R)) (α/2)+(β/2)+(γ/2)=(π/4) ⇒sin^(−1) (a/(2R))+sin^(−1) (b/(2R))+sin^(−1) (c/(2R))=(π/4) there is an unique solution for R, but we can not solve it in exact form in terms of a, b, c.

$${you}\:{have} \\ $$$$\mathrm{sin}\:\frac{\alpha}{\mathrm{2}}=\frac{{a}}{\mathrm{2}{R}}\:\Rightarrow\frac{\alpha}{\mathrm{2}}=\mathrm{sin}^{−\mathrm{1}} \frac{{a}}{\mathrm{2}{R}} \\ $$$$\mathrm{sin}\:\frac{\beta}{\mathrm{2}}=\frac{{b}}{\mathrm{2}{R}}\Rightarrow\frac{\beta}{\mathrm{2}}=\mathrm{sin}^{−\mathrm{1}} \frac{{b}}{\mathrm{2}{R}} \\ $$$$\mathrm{sin}\:\frac{\gamma}{\mathrm{2}}=\frac{{c}}{\mathrm{2}{R}}\Rightarrow\frac{\gamma}{\mathrm{2}}=\mathrm{sin}^{−\mathrm{1}} \frac{{c}}{\mathrm{2}{R}} \\ $$$$\frac{\alpha}{\mathrm{2}}+\frac{\beta}{\mathrm{2}}+\frac{\gamma}{\mathrm{2}}=\frac{\pi}{\mathrm{4}} \\ $$$$\Rightarrow\mathrm{sin}^{−\mathrm{1}} \frac{{a}}{\mathrm{2}{R}}+\mathrm{sin}^{−\mathrm{1}} \frac{{b}}{\mathrm{2}{R}}+\mathrm{sin}^{−\mathrm{1}} \frac{{c}}{\mathrm{2}{R}}=\frac{\pi}{\mathrm{4}} \\ $$$${there}\:{is}\:{an}\:{unique}\:{solution}\:{for}\:{R}, \\ $$$${but}\:{we}\:{can}\:{not}\:{solve}\:{it}\:{in}\:{exact}\:{form} \\ $$$${in}\:{terms}\:{of}\:{a},\:{b},\:{c}. \\ $$

Commented by mahdipoor last updated on 12/Feb/22

⇒0=cos90=cos(α+β+λ)= cosα.cosβ.cosλ−sinβ.sinα.cosλ− sinλ.sinα.cosβ−sinλ.sinβ.cosα , u^2 =2R^2 (1−cosv) ⇒ { ((cosv=1−(u/( (√2)R)))),((sinv=(√((√2)(u/R)−(u^2 /(2R^2 )))))) :}

$$\Rightarrow\mathrm{0}={cos}\mathrm{90}={cos}\left(\alpha+\beta+\lambda\right)= \\ $$$${cos}\alpha.{cos}\beta.{cos}\lambda−{sin}\beta.{sin}\alpha.{cos}\lambda− \\ $$$${sin}\lambda.{sin}\alpha.{cos}\beta−{sin}\lambda.{sin}\beta.{cos}\alpha \\ $$$$, \\ $$$${u}^{\mathrm{2}} =\mathrm{2}{R}^{\mathrm{2}} \left(\mathrm{1}−{cosv}\right) \\ $$$$\Rightarrow\begin{cases}{{cosv}=\mathrm{1}−\frac{{u}}{\:\sqrt{\mathrm{2}}{R}}}\\{{sinv}=\sqrt{\sqrt{\mathrm{2}}\frac{{u}}{{R}}−\frac{{u}^{\mathrm{2}} }{\mathrm{2}{R}^{\mathrm{2}} }}}\end{cases} \\ $$

Commented by Rasheed.Sindhi last updated on 12/Feb/22

You′re right mr W sir, Iconcluded so becsuse I saw that cos α+cos β+sin(α+β) is not constant! But inspite this it may be dependant upon a,b & c.

$${You}'{re}\:{right}\:{mr}\:{W}\:{sir},\:{Iconcluded} \\ $$$${so}\:{becsuse}\:{I}\:{saw}\:{that}\:\mathrm{cos}\:\alpha+\mathrm{cos}\:\beta+\mathrm{sin}\left(\alpha+\beta\right) \\ $$$${is}\:{not}\:{constant}!\:{But}\:{inspite}\:{this} \\ $$$${it}\:{may}\:{be}\:{dependant}\:{upon}\:{a},{b}\:\&\:{c}. \\ $$

Commented by mr W last updated on 13/Feb/22

we can transform the equation to: 4R^6 −4(a^2 +b^2 +c^2 )R^4 −6(√2)abcR^3 +(a^4 +b^4 +c^4 )R^2 +(√2)abc(a^2 +b^2 +c^2 )R+a^2 b^2 c^2 =0 but still not solvable.

$${we}\:{can}\:{transform}\:{the}\:{equation}\:{to}: \\ $$$$\mathrm{4}{R}^{\mathrm{6}} −\mathrm{4}\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} \right){R}^{\mathrm{4}} −\mathrm{6}\sqrt{\mathrm{2}}{abcR}^{\mathrm{3}} +\left({a}^{\mathrm{4}} +{b}^{\mathrm{4}} +{c}^{\mathrm{4}} \right){R}^{\mathrm{2}} +\sqrt{\mathrm{2}}{abc}\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} \right){R}+{a}^{\mathrm{2}} {b}^{\mathrm{2}} {c}^{\mathrm{2}} =\mathrm{0} \\ $$$${but}\:{still}\:{not}\:{solvable}. \\ $$

Commented by Rasheed.Sindhi last updated on 12/Feb/22

Nice relation sir!

$${Nice}\:{relation}\:{sir}! \\ $$

Commented by ajfour last updated on 13/Feb/22

θ+ψ=(π/4)−φ cos (θ+ψ)=(1/( (√2))){cos φ+sin φ} (√(1−p^2 ))(√(1−r^2 ))−pr=(1/( (√2))){(√(1−q^2 ))+q} (1−p^2 )(1−r^2 )+((1−q^2 )/2) −(√2)(√(1−q^2 ))(√(1−p^2 ))(√(1−r^2 )) =p^2 r^2 +(q^2 /2)+(√2)pqr ⇒ { (3/2)−(p^2 +r^2 +q^2 )−(√2)pqr}^2 =2(1−p^2 )(1−q^2 )(1−r^2 ) ⇒ (9/4)+(p^2 +q^2 +r^2 )^2 +2p^2 q^2 r^2 −3(p^2 +r^2 +q^2 )−3(√2)pqr +2(√2)pqr(p^2 +q^2 +r^2 ) =2−p^2 q^2 r^2 +2(p^2 q^2 +q^2 r^2 +r^2 p^2 ) ⇒ p^4 +q^4 +r^4 +(1/4)+3p^2 q^2 r^2 +2(√2)pqr(p^2 +q^2 +r^2 ) −3(p^2 +q^2 +r^2 )=0 say (1/(2R))=t ⇒ (3a^2 b^2 c^2 )t^6 +2(√2)abc(a^2 +b^2 +c^2 )t^5 +(a^4 +b^4 +c^4 )t^4 −3(a^2 +b^2 +c^2 )t^2 +(1/4)=0

$$\theta+\psi=\frac{\pi}{\mathrm{4}}−\phi \\ $$$$\mathrm{cos}\:\left(\theta+\psi\right)=\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\left\{\mathrm{cos}\:\phi+\mathrm{sin}\:\phi\right\} \\ $$$$\sqrt{\mathrm{1}−{p}^{\mathrm{2}} }\sqrt{\mathrm{1}−{r}^{\mathrm{2}} }−{pr}=\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\left\{\sqrt{\mathrm{1}−{q}^{\mathrm{2}} }+{q}\right\} \\ $$$$\left(\mathrm{1}−{p}^{\mathrm{2}} \right)\left(\mathrm{1}−{r}^{\mathrm{2}} \right)+\frac{\mathrm{1}−{q}^{\mathrm{2}} }{\mathrm{2}} \\ $$$$\:\:\:−\sqrt{\mathrm{2}}\sqrt{\mathrm{1}−{q}^{\mathrm{2}} }\sqrt{\mathrm{1}−{p}^{\mathrm{2}} }\sqrt{\mathrm{1}−{r}^{\mathrm{2}} } \\ $$$$\:\:\:\:={p}^{\mathrm{2}} {r}^{\mathrm{2}} +\frac{{q}^{\mathrm{2}} }{\mathrm{2}}+\sqrt{\mathrm{2}}{pqr} \\ $$$$\Rightarrow\:\left\{\:\frac{\mathrm{3}}{\mathrm{2}}−\left({p}^{\mathrm{2}} +{r}^{\mathrm{2}} +{q}^{\mathrm{2}} \right)−\sqrt{\mathrm{2}}{pqr}\right\}^{\mathrm{2}} \\ $$$$\:\:\:\:=\mathrm{2}\left(\mathrm{1}−{p}^{\mathrm{2}} \right)\left(\mathrm{1}−{q}^{\mathrm{2}} \right)\left(\mathrm{1}−{r}^{\mathrm{2}} \right) \\ $$$$\Rightarrow\:\frac{\mathrm{9}}{\mathrm{4}}+\left({p}^{\mathrm{2}} +{q}^{\mathrm{2}} +{r}^{\mathrm{2}} \right)^{\mathrm{2}} +\mathrm{2}{p}^{\mathrm{2}} {q}^{\mathrm{2}} {r}^{\mathrm{2}} \\ $$$$−\mathrm{3}\left({p}^{\mathrm{2}} +{r}^{\mathrm{2}} +{q}^{\mathrm{2}} \right)−\mathrm{3}\sqrt{\mathrm{2}}{pqr} \\ $$$$+\mathrm{2}\sqrt{\mathrm{2}}{pqr}\left({p}^{\mathrm{2}} +{q}^{\mathrm{2}} +{r}^{\mathrm{2}} \right) \\ $$$$\:\:=\mathrm{2}−{p}^{\mathrm{2}} {q}^{\mathrm{2}} {r}^{\mathrm{2}} +\mathrm{2}\left({p}^{\mathrm{2}} {q}^{\mathrm{2}} +{q}^{\mathrm{2}} {r}^{\mathrm{2}} +{r}^{\mathrm{2}} {p}^{\mathrm{2}} \right) \\ $$$$\Rightarrow \\ $$$$\:\:{p}^{\mathrm{4}} +{q}^{\mathrm{4}} +{r}^{\mathrm{4}} +\frac{\mathrm{1}}{\mathrm{4}}+\mathrm{3}{p}^{\mathrm{2}} {q}^{\mathrm{2}} {r}^{\mathrm{2}} \\ $$$$\:\:+\mathrm{2}\sqrt{\mathrm{2}}{pqr}\left({p}^{\mathrm{2}} +{q}^{\mathrm{2}} +{r}^{\mathrm{2}} \right) \\ $$$$\:\:\:−\mathrm{3}\left({p}^{\mathrm{2}} +{q}^{\mathrm{2}} +{r}^{\mathrm{2}} \right)=\mathrm{0} \\ $$$${say}\:\:\frac{\mathrm{1}}{\mathrm{2}{R}}={t}\:\:\Rightarrow \\ $$$$\left(\mathrm{3}{a}^{\mathrm{2}} {b}^{\mathrm{2}} {c}^{\mathrm{2}} \right){t}^{\mathrm{6}} +\mathrm{2}\sqrt{\mathrm{2}}{abc}\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} \right){t}^{\mathrm{5}} \\ $$$$\:\:+\left({a}^{\mathrm{4}} +{b}^{\mathrm{4}} +{c}^{\mathrm{4}} \right){t}^{\mathrm{4}} \\ $$$$\:\:−\mathrm{3}\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} \right){t}^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{4}}=\mathrm{0} \\ $$$$ \\ $$

Answered by ajfour last updated on 15/Feb/22

((abc)/(2R))=casin β=absin γ=bcsin α=p let ((abc)/(2r))=q q=casin 2β=absin 2γ=bcsin 2α α+β+γ=(π/4) sin 2β=cos (2α+2γ) (q/(ac))=(√(1−((q/(bc)))^2 ))(√(1−((q/(ab)))^2 )) −(q^2 /(b^2 ac)) ⇒ (q^2 /(a^2 c^2 ))+(q^4 /(b^4 a^2 c^2 ))+((2q^3 )/(a^2 b^2 c^2 )) =1−(q^2 /(a^2 b^2 c^2 ))(a^2 +c^2 )+(q^4 /(b^4 a^2 c^2 )) ⇒ 2q^3 +(a^2 +b^2 +c^2 )q^2 −a^2 b^2 c^2 =0 say we obtained q. Now (q/p)=2cos β ⇒ q^2 =4p^2 (1−(b^2 /(4R^2 )))=4p^2 (1−(p^2 /(a^2 c^2 ))) ⇒ p^4 −a^2 c^2 p^2 +((a^2 c^2 q^2 )/4)=0 p^2 =(((abc)/(2R)))^2 =((a^2 c^2 )/2)±(√(((a^2 c^2 )/4)(a^2 c^2 −q^2 ))) ⇒ R^2 =(b^2 /2)((q/(ac)))^2 {1∓(√(1−((q/(ac)))^2 ))}

$$\frac{{abc}}{\mathrm{2}{R}}={ca}\mathrm{sin}\:\beta={ab}\mathrm{sin}\:\gamma={bc}\mathrm{sin}\:\alpha={p} \\ $$$${let}\:\frac{{abc}}{\mathrm{2}{r}}={q} \\ $$$$\:\:{q}={ca}\mathrm{sin}\:\mathrm{2}\beta={ab}\mathrm{sin}\:\mathrm{2}\gamma={bc}\mathrm{sin}\:\mathrm{2}\alpha \\ $$$$\alpha+\beta+\gamma=\frac{\pi}{\mathrm{4}} \\ $$$$\mathrm{sin}\:\mathrm{2}\beta=\mathrm{cos}\:\left(\mathrm{2}\alpha+\mathrm{2}\gamma\right) \\ $$$$\frac{{q}}{{ac}}=\sqrt{\mathrm{1}−\left(\frac{{q}}{{bc}}\right)^{\mathrm{2}} }\sqrt{\mathrm{1}−\left(\frac{{q}}{{ab}}\right)^{\mathrm{2}} } \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:−\frac{{q}^{\mathrm{2}} }{{b}^{\mathrm{2}} {ac}} \\ $$$$\Rightarrow\:\frac{{q}^{\mathrm{2}} }{{a}^{\mathrm{2}} {c}^{\mathrm{2}} }+\cancel{\frac{{q}^{\mathrm{4}} }{{b}^{\mathrm{4}} {a}^{\mathrm{2}} {c}^{\mathrm{2}} }}+\frac{\mathrm{2}{q}^{\mathrm{3}} }{{a}^{\mathrm{2}} {b}^{\mathrm{2}} {c}^{\mathrm{2}} } \\ $$$$\:\:\:\:\:\:\:\:\:=\mathrm{1}−\frac{{q}^{\mathrm{2}} }{{a}^{\mathrm{2}} {b}^{\mathrm{2}} {c}^{\mathrm{2}} }\left({a}^{\mathrm{2}} +{c}^{\mathrm{2}} \right)+\cancel{\frac{{q}^{\mathrm{4}} }{{b}^{\mathrm{4}} {a}^{\mathrm{2}} {c}^{\mathrm{2}} }} \\ $$$$\Rightarrow\:\:\mathrm{2}{q}^{\mathrm{3}} +\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} \right){q}^{\mathrm{2}} −{a}^{\mathrm{2}} {b}^{\mathrm{2}} {c}^{\mathrm{2}} =\mathrm{0} \\ $$$${say}\:{we}\:{obtained}\:{q}.\:\:\: \\ $$$${Now}\:\:\frac{{q}}{{p}}=\mathrm{2cos}\:\beta \\ $$$$\Rightarrow\:\:\:{q}^{\mathrm{2}} =\mathrm{4}{p}^{\mathrm{2}} \left(\mathrm{1}−\frac{{b}^{\mathrm{2}} }{\mathrm{4}{R}^{\mathrm{2}} }\right)=\mathrm{4}{p}^{\mathrm{2}} \left(\mathrm{1}−\frac{{p}^{\mathrm{2}} }{{a}^{\mathrm{2}} {c}^{\mathrm{2}} }\right) \\ $$$$\Rightarrow\:{p}^{\mathrm{4}} −{a}^{\mathrm{2}} {c}^{\mathrm{2}} {p}^{\mathrm{2}} +\frac{{a}^{\mathrm{2}} {c}^{\mathrm{2}} {q}^{\mathrm{2}} }{\mathrm{4}}=\mathrm{0} \\ $$$$\:\:{p}^{\mathrm{2}} =\left(\frac{{abc}}{\mathrm{2}{R}}\right)^{\mathrm{2}} =\frac{{a}^{\mathrm{2}} {c}^{\mathrm{2}} }{\mathrm{2}}\pm\sqrt{\frac{{a}^{\mathrm{2}} {c}^{\mathrm{2}} }{\mathrm{4}}\left({a}^{\mathrm{2}} {c}^{\mathrm{2}} −{q}^{\mathrm{2}} \right)} \\ $$$$\Rightarrow\:\:{R}^{\mathrm{2}} =\frac{{b}^{\mathrm{2}} }{\mathrm{2}}\left(\frac{{q}}{{ac}}\right)^{\mathrm{2}} \left\{\mathrm{1}\mp\sqrt{\mathrm{1}−\left(\frac{{q}}{{ac}}\right)^{\mathrm{2}} }\right\} \\ $$$$ \\ $$

Commented by ajfour last updated on 15/Feb/22

Commented by mr W last updated on 15/Feb/22

thanks sir! can you please show in a diagram what are α,β,γ?

$${thanks}\:{sir}! \\ $$$${can}\:{you}\:{please}\:{show}\:{in}\:{a}\:{diagram} \\ $$$${what}\:{are}\:\alpha,\beta,\gamma? \\ $$

Commented by mr W last updated on 15/Feb/22

Commented by mr W last updated on 15/Feb/22

are you sure sir? i have doubt if this way is correct. bc sin α=ca sin β=ab sin γ=((abc)/(2R)) ✓ with 2α+2β+2γ=(π/2) ✓ bc sin θ=ca sin ϕ=ab sin φ=((abc)/(2r)) ✓ with 2θ+2ϕ+2φ=π ✓ we can say θ+ϕ+φ=2(α+β+γ). but i think we can not assume (set) θ=2α, ϕ=2β, φ=2γ. actually sin θ=(a/(2r)), sin α=(a/(2R)) if θ=2α, then sin^2 θ= sin^2 2α=1−cos^2 2α=1−(1−2 sin^2 α)^2 ((a/(2r)))^2 =1−(1−(a^2 /(2R^2 )))^2 ...(I) similarly, from ϕ=2β and φ=2γ we′ll get ((b/(2r)))^2 =1−(1−(b^2 /(2R^2 )))^2 ...(II) ((c/(2r)))^2 =1−(1−(c^2 /(2R^2 )))^2 ...(III) (I), (II), (III) are not consistent. that means we can not set θ=2α, ϕ=2β, φ=2γ at the same time. bc sin α=ca sin β=ab sin γ=((abc)/(2R)) ✓ ⇏bc sin 2α=^? ca sin 2β=^? ab sin 2γ=^? ((abc)/(2r))

$${are}\:{you}\:{sure}\:{sir}? \\ $$$${i}\:{have}\:{doubt}\:{if}\:{this}\:{way}\:{is}\:{correct}. \\ $$$$ \\ $$$${bc}\:\mathrm{sin}\:\alpha={ca}\:\mathrm{sin}\:\beta={ab}\:\mathrm{sin}\:\gamma=\frac{{abc}}{\mathrm{2}{R}}\:\checkmark \\ $$$${with}\:\mathrm{2}\alpha+\mathrm{2}\beta+\mathrm{2}\gamma=\frac{\pi}{\mathrm{2}}\:\checkmark \\ $$$${bc}\:\mathrm{sin}\:\theta={ca}\:\mathrm{sin}\:\varphi={ab}\:\mathrm{sin}\:\phi=\frac{{abc}}{\mathrm{2}{r}}\:\checkmark \\ $$$${with}\:\mathrm{2}\theta+\mathrm{2}\varphi+\mathrm{2}\phi=\pi\:\checkmark \\ $$$$ \\ $$$${we}\:{can}\:{say}\:\theta+\varphi+\phi=\mathrm{2}\left(\alpha+\beta+\gamma\right). \\ $$$${but}\:{i}\:{think}\:{we}\:{can}\:{not}\:{assume}\:\left({set}\right) \\ $$$$\theta=\mathrm{2}\alpha,\:\varphi=\mathrm{2}\beta,\:\phi=\mathrm{2}\gamma. \\ $$$$ \\ $$$${actually}\:\mathrm{sin}\:\theta=\frac{{a}}{\mathrm{2}{r}},\:\mathrm{sin}\:\alpha=\frac{{a}}{\mathrm{2}{R}} \\ $$$${if}\:\theta=\mathrm{2}\alpha,\:{then} \\ $$$$\mathrm{sin}^{\mathrm{2}} \:\theta=\:\mathrm{sin}^{\mathrm{2}} \:\mathrm{2}\alpha=\mathrm{1}−\mathrm{cos}^{\mathrm{2}} \:\mathrm{2}\alpha=\mathrm{1}−\left(\mathrm{1}−\mathrm{2}\:\mathrm{sin}^{\mathrm{2}} \:\alpha\right)^{\mathrm{2}} \\ $$$$\left(\frac{{a}}{\mathrm{2}{r}}\right)^{\mathrm{2}} =\mathrm{1}−\left(\mathrm{1}−\frac{{a}^{\mathrm{2}} }{\mathrm{2}{R}^{\mathrm{2}} }\right)^{\mathrm{2}} \:\:\:\:\:...\left({I}\right) \\ $$$${similarly},\:{from}\:\varphi=\mathrm{2}\beta\:{and}\:\phi=\mathrm{2}\gamma \\ $$$${we}'{ll}\:{get} \\ $$$$\left(\frac{{b}}{\mathrm{2}{r}}\right)^{\mathrm{2}} =\mathrm{1}−\left(\mathrm{1}−\frac{{b}^{\mathrm{2}} }{\mathrm{2}{R}^{\mathrm{2}} }\right)^{\mathrm{2}} \:\:\:\:\:...\left({II}\right) \\ $$$$\left(\frac{{c}}{\mathrm{2}{r}}\right)^{\mathrm{2}} =\mathrm{1}−\left(\mathrm{1}−\frac{{c}^{\mathrm{2}} }{\mathrm{2}{R}^{\mathrm{2}} }\right)^{\mathrm{2}} \:\:\:\:\:...\left({III}\right) \\ $$$$\left({I}\right),\:\left({II}\right),\:\left({III}\right)\:{are}\:{not}\:{consistent}. \\ $$$${that}\:{means}\:{we}\:{can}\:{not}\:{set} \\ $$$$\theta=\mathrm{2}\alpha,\:\varphi=\mathrm{2}\beta,\:\phi=\mathrm{2}\gamma\:{at}\:{the}\:{same}\:{time}. \\ $$$$ \\ $$$${bc}\:\mathrm{sin}\:\alpha={ca}\:\mathrm{sin}\:\beta={ab}\:\mathrm{sin}\:\gamma=\frac{{abc}}{\mathrm{2}{R}}\:\checkmark \\ $$$$\nRightarrow{bc}\:\mathrm{sin}\:\mathrm{2}\alpha\overset{?} {=}{ca}\:\mathrm{sin}\:\mathrm{2}\beta\overset{?} {=}{ab}\:\mathrm{sin}\:\mathrm{2}\gamma\overset{?} {=}\frac{{abc}}{\mathrm{2}{r}}\: \\ $$

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