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Question-198913

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Question Number 198913 by mr W last updated on 25/Oct/23
Commented by mr W last updated on 25/Oct/23
find θ in terms of a, b, c.
$${find}\:\theta\:{in}\:{terms}\:{of}\:{a},\:{b},\:{c}. \\ $$
Answered by ajfour last updated on 25/Oct/23
sin 2θ=((a^2 +b^2 +c^2 )/(2Rabc((1/a^2 )+(1/b^2 )+(1/c^2 ))))
$$\mathrm{sin}\:\mathrm{2}\theta=\frac{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} }{\mathrm{2}{Rabc}\left(\frac{\mathrm{1}}{{a}^{\mathrm{2}} }+\frac{\mathrm{1}}{{b}^{\mathrm{2}} }+\frac{\mathrm{1}}{{c}^{\mathrm{2}} }\right)} \\ $$
Commented by mr W last updated on 26/Oct/23
i got tan θ=((4Δ)/(a^2 +b^2 +c^2 ))=((abc)/(R(a^2 +b^2 +c^2 ))) do they match?
$${i}\:{got}\:\mathrm{tan}\:\theta=\frac{\mathrm{4}\Delta}{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} }=\frac{{abc}}{{R}\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} \right)} \\ $$$${do}\:{they}\:{match}? \\ $$
Answered by mr W last updated on 26/Oct/23
Commented by mr W last updated on 26/Oct/23
Δ=area of ΔABC ((pc sin θ)/2)+((qa sin θ)/2)+((rb sin θ)/2)=Δ ⇒(pc+qa+rb) sin θ=2Δ ...(i) q^2 =c^2 +p^2 −2pc cos θ r^2 =a^2 +q^2 −2qa cos θ p^2 =b^2 +r^2 −2rb cos θ Σ: q^2 +r^2 +p^2 =c^2 +p^2 +a^2 +q^2 +b^2 +r^2 −2(pc+qa+rb)cos θ ⇒(pc+qa+rb)cos θ=((a^2 +b^2 +c^2 )/2) ...(ii) (i)/(ii): ⇒tan θ=((4Δ)/(a^2 +b^2 +c^2 )) or ⇒tan θ=((√((a+b+c)(−a+b+c)(a−b+c)(a+b−c)))/(a^2 +b^2 +c^2 )) ■
$$\Delta={area}\:{of}\:\Delta{ABC} \\ $$$$\frac{{pc}\:\mathrm{sin}\:\theta}{\mathrm{2}}+\frac{{qa}\:\mathrm{sin}\:\theta}{\mathrm{2}}+\frac{{rb}\:\mathrm{sin}\:\theta}{\mathrm{2}}=\Delta \\ $$$$\Rightarrow\left({pc}+{qa}+{rb}\right)\:\mathrm{sin}\:\theta=\mathrm{2}\Delta\:\:\:…\left({i}\right) \\ $$$${q}^{\mathrm{2}} ={c}^{\mathrm{2}} +{p}^{\mathrm{2}} −\mathrm{2}{pc}\:\mathrm{cos}\:\theta \\ $$$${r}^{\mathrm{2}} ={a}^{\mathrm{2}} +{q}^{\mathrm{2}} −\mathrm{2}{qa}\:\mathrm{cos}\:\theta \\ $$$${p}^{\mathrm{2}} ={b}^{\mathrm{2}} +{r}^{\mathrm{2}} −\mathrm{2}{rb}\:\mathrm{cos}\:\theta \\ $$$$\Sigma:\:{q}^{\mathrm{2}} +{r}^{\mathrm{2}} +{p}^{\mathrm{2}} ={c}^{\mathrm{2}} +{p}^{\mathrm{2}} +{a}^{\mathrm{2}} +{q}^{\mathrm{2}} +{b}^{\mathrm{2}} +{r}^{\mathrm{2}} −\mathrm{2}\left({pc}+{qa}+{rb}\right)\mathrm{cos}\:\theta \\ $$$$\Rightarrow\left({pc}+{qa}+{rb}\right)\mathrm{cos}\:\theta=\frac{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} }{\mathrm{2}}\:\:\:…\left({ii}\right) \\ $$$$\left({i}\right)/\left({ii}\right): \\ $$$$\Rightarrow\mathrm{tan}\:\theta=\frac{\mathrm{4}\Delta}{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} } \\ $$$${or} \\ $$$$\Rightarrow\mathrm{tan}\:\theta=\frac{\sqrt{\left({a}+{b}+{c}\right)\left(−{a}+{b}+{c}\right)\left({a}−{b}+{c}\right)\left({a}+{b}−{c}\right)}}{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} } \\ $$$$\blacksquare \\ $$
Answered by ajfour last updated on 26/Oct/23
((sin (θ+α))/a)=((sin (θ+β))/b)=((sin (θ+γ))/c)=(1/(2R)) ((sin (θ+α))/b)=((sin θ)/p)=((sin α)/r) from △ACO ⇒ (1/b)((a/(2R)))=((sin θ)/p) (1/c)((b/(2R)))=((sin θ)/q) (1/a)((c/(2R)))=((sin θ)/r) ......set (I) 2aqcos θ=a^2 +q^2 −r^2 2brcos θ=b^2 +r^2 −p^2 2cpcos θ=c^2 +p^2 −q^2 2cos θ(aq+br+cp)=a^2 +b^2 +c^2 ...(II) from ..set(I) p=(2Rsin θ)(b/a) q=(2Rsin θ)(c/b) r=(2Rsin θ)(a/c) Now from (II) 4R^2 (sin θcos θ)(((ca)/b)+((ab)/c)+((bc)/a))=a^2 +b^2 +c^2 (2abcRsin 2θ)((1/b^2 )+(1/c^2 )+(1/a^2 ))=a^2 +b^2 +c^2 sin 2θ=((a^2 +b^2 +c^2 )/(2abcR((1/a^2 )+(1/b^2 )+(1/c^2 ))))
$$\frac{\mathrm{sin}\:\left(\theta+\alpha\right)}{{a}}=\frac{\mathrm{sin}\:\left(\theta+\beta\right)}{{b}}=\frac{\mathrm{sin}\:\left(\theta+\gamma\right)}{{c}}=\frac{\mathrm{1}}{\mathrm{2}{R}} \\ $$$$\frac{\mathrm{sin}\:\left(\theta+\alpha\right)}{{b}}=\frac{\mathrm{sin}\:\theta}{{p}}=\frac{\mathrm{sin}\:\alpha}{{r}}\:\:{from}\:\bigtriangleup{ACO} \\ $$$$\Rightarrow\:\:\frac{\mathrm{1}}{{b}}\left(\frac{{a}}{\mathrm{2}{R}}\right)=\frac{\mathrm{sin}\:\theta}{{p}}\: \\ $$$$\:\:\:\:\:\:\:\frac{\mathrm{1}}{{c}}\left(\frac{{b}}{\mathrm{2}{R}}\right)=\frac{\mathrm{sin}\:\theta}{{q}} \\ $$$$\:\:\:\:\:\:\:\frac{\mathrm{1}}{{a}}\left(\frac{{c}}{\mathrm{2}{R}}\right)=\frac{\mathrm{sin}\:\theta}{{r}}\:\:\:\:\:\:……{set}\:\left({I}\right) \\ $$$$\:\:\:\:\:\:\:\: \\ $$$$\mathrm{2}{aq}\mathrm{cos}\:\theta={a}^{\mathrm{2}} +{q}^{\mathrm{2}} −{r}^{\mathrm{2}} \\ $$$$\mathrm{2}{br}\mathrm{cos}\:\theta={b}^{\mathrm{2}} +{r}^{\mathrm{2}} −{p}^{\mathrm{2}} \\ $$$$\mathrm{2}{cp}\mathrm{cos}\:\theta={c}^{\mathrm{2}} +{p}^{\mathrm{2}} −{q}^{\mathrm{2}} \\ $$$$\mathrm{2cos}\:\theta\left({aq}+{br}+{cp}\right)={a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} \:\:\:…\left({II}\right) \\ $$$${from}\:..{set}\left({I}\right) \\ $$$${p}=\left(\mathrm{2}{R}\mathrm{sin}\:\theta\right)\frac{{b}}{{a}} \\ $$$${q}=\left(\mathrm{2}{R}\mathrm{sin}\:\theta\right)\frac{{c}}{{b}} \\ $$$${r}=\left(\mathrm{2}{R}\mathrm{sin}\:\theta\right)\frac{{a}}{{c}} \\ $$$${Now}\:{from}\:\left({II}\right) \\ $$$$\mathrm{4}{R}^{\mathrm{2}} \left(\mathrm{sin}\:\theta\mathrm{cos}\:\theta\right)\left(\frac{{ca}}{{b}}+\frac{{ab}}{{c}}+\frac{{bc}}{{a}}\right)={a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} \\ $$$$\left(\mathrm{2}{abcR}\mathrm{sin}\:\mathrm{2}\theta\right)\left(\frac{\mathrm{1}}{{b}^{\mathrm{2}} }+\frac{\mathrm{1}}{{c}^{\mathrm{2}} }+\frac{\mathrm{1}}{{a}^{\mathrm{2}} }\right)={a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} \\ $$$$\:\:\:\mathrm{sin}\:\mathrm{2}\theta=\frac{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} }{\mathrm{2}{abcR}\left(\frac{\mathrm{1}}{{a}^{\mathrm{2}} }+\frac{\mathrm{1}}{{b}^{\mathrm{2}} }+\frac{\mathrm{1}}{{c}^{\mathrm{2}} }\right)} \\ $$
Commented by mr W last updated on 26/Oct/23
nice solution sir!
$${nice}\:{solution}\:{sir}! \\ $$

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