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Question Number 48763 by mr W last updated on 28/Nov/18

Commented by mr W last updated on 28/Nov/18

Find R in terms of a, b, c.

$${Find}\:{R}\:{in}\:{terms}\:{of}\:{a},\:{b},\:{c}. \\ $$

Answered by ajfour last updated on 28/Nov/18

Answered by ajfour last updated on 28/Nov/18

Rsin θ = (a/2) 2Rcos (θ+α)=(√(b^2 +c^2 )) ⇒ 2R(cos θcos α−sin θsin α)=(√(b^2 +c^2 )) 2R(cos θ×(b/(√(b^2 +c^2 )))−(a/(2R))×(c/(√(b^2 +c^2 ))))=(√(b^2 +c^2 )) ⇒ 2bRcos θ=b^2 +c^2 +ac or (4R^2 −a^2 )b^2 = (b^2 +c^2 +ac)^2 4R^2 = a^2 +(((b^2 +c^2 +ac)^2 )/b^2 ) R = (1/2)(√(a^2 +(((b^2 +c^2 +ac)^2 )/b^2 ))) .

$${R}\mathrm{sin}\:\theta\:=\:\frac{{a}}{\mathrm{2}} \\ $$$$\mathrm{2}{R}\mathrm{cos}\:\left(\theta+\alpha\right)=\sqrt{{b}^{\mathrm{2}} +{c}^{\mathrm{2}} } \\ $$$$\Rightarrow\:\mathrm{2}{R}\left(\mathrm{cos}\:\theta\mathrm{cos}\:\alpha−\mathrm{sin}\:\theta\mathrm{sin}\:\alpha\right)=\sqrt{{b}^{\mathrm{2}} +{c}^{\mathrm{2}} } \\ $$$$\mathrm{2}{R}\left(\mathrm{cos}\:\theta×\frac{{b}}{\sqrt{{b}^{\mathrm{2}} +{c}^{\mathrm{2}} }}−\frac{{a}}{\mathrm{2}{R}}×\frac{{c}}{\sqrt{{b}^{\mathrm{2}} +{c}^{\mathrm{2}} }}\right)=\sqrt{{b}^{\mathrm{2}} +{c}^{\mathrm{2}} } \\ $$$$\Rightarrow\:\mathrm{2}{bR}\mathrm{cos}\:\theta={b}^{\mathrm{2}} +{c}^{\mathrm{2}} +{ac} \\ $$$${or}\:\:\:\:\left(\mathrm{4}{R}^{\mathrm{2}} −{a}^{\mathrm{2}} \right){b}^{\mathrm{2}} \:=\:\left({b}^{\mathrm{2}} +{c}^{\mathrm{2}} +{ac}\right)^{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\mathrm{4}{R}^{\mathrm{2}} \:=\:{a}^{\mathrm{2}} +\frac{\left({b}^{\mathrm{2}} +{c}^{\mathrm{2}} +{ac}\right)^{\mathrm{2}} }{{b}^{\mathrm{2}} } \\ $$$$\:\:\:\:\:{R}\:=\:\frac{\mathrm{1}}{\mathrm{2}}\sqrt{{a}^{\mathrm{2}} +\frac{\left({b}^{\mathrm{2}} +{c}^{\mathrm{2}} +{ac}\right)^{\mathrm{2}} }{{b}^{\mathrm{2}} }}\:. \\ $$

Commented by mr W last updated on 28/Nov/18

correct! thank you sir!

$${correct}!\:{thank}\:{you}\:{sir}! \\ $$

Commented by mr W last updated on 28/Nov/18

alternative way: circumcircle of a triangle with sides: a (√(b^2 +c^2 )) (√(b^2 +(a+c)^2 )) R=((a(√((b^2 +c^2 )(b^2 +(a+c)^2 ))))/(4A)) A=((ab)/2) R=((a(√((b^2 +c^2 )(b^2 +(a+c)^2 ))))/(2ab)) ⇒R=((√((b^2 +c^2 )(b^2 +(a+c)^2 )))/(2b))

$${alternative}\:{way}: \\ $$$${circumcircle}\:{of}\:{a}\:{triangle}\:{with}\:{sides}: \\ $$$${a} \\ $$$$\sqrt{{b}^{\mathrm{2}} +{c}^{\mathrm{2}} } \\ $$$$\sqrt{{b}^{\mathrm{2}} +\left({a}+{c}\right)^{\mathrm{2}} } \\ $$$$ \\ $$$${R}=\frac{{a}\sqrt{\left({b}^{\mathrm{2}} +{c}^{\mathrm{2}} \right)\left({b}^{\mathrm{2}} +\left({a}+{c}\right)^{\mathrm{2}} \right)}}{\mathrm{4}{A}} \\ $$$${A}=\frac{{ab}}{\mathrm{2}} \\ $$$${R}=\frac{{a}\sqrt{\left({b}^{\mathrm{2}} +{c}^{\mathrm{2}} \right)\left({b}^{\mathrm{2}} +\left({a}+{c}\right)^{\mathrm{2}} \right)}}{\mathrm{2}{ab}} \\ $$$$\Rightarrow{R}=\frac{\sqrt{\left({b}^{\mathrm{2}} +{c}^{\mathrm{2}} \right)\left({b}^{\mathrm{2}} +\left({a}+{c}\right)^{\mathrm{2}} \right)}}{\mathrm{2}{b}} \\ $$

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