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Question-54022

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Question Number 54022 by ajfour last updated on 28/Jan/19
Commented by ajfour last updated on 28/Jan/19
Given a and b, find θ and R in terms of a and b.
$${Given}\:{a}\:{and}\:{b},\:{find}\:\theta\:{and}\:{R}\:{in}\:{terms} \\ $$$${of}\:{a}\:{and}\:{b}.\:\:\:\: \\ $$
Commented by tanmay.chaudhury50@gmail.com last updated on 28/Jan/19
Commented by ajfour last updated on 28/Jan/19
Commented by ajfour last updated on 28/Jan/19
AE=AD+DE (b/(sin θ))= acot (θ/2)+(√((a+b)^2 −a^2 )) (b/(2sin (θ/2)cos (θ/2)))−((acos (θ/2))/(sin (θ/2)))=(√(b(b+2a))) ((b−a(1+cos θ))/(sin θ))= (√(b(b+2a))) ⇒ b^2 +a^2 (1+cos θ)^2 −2ab(1+cos θ) =b(b+2a)sin^2 θ b^2 +a^2 +2a^2 cos θ+a^2 cos^2 θ−2ab −2abcos θ=b^2 +2ab−b^2 cos^2 θ−2abcos^2 θ let cos θ = s ⇒ (a+b)^2 s^2 +2a(a−b)s−a(4b−a)=0 s=((a(a−b)+(√(a^2 (a−b)^2 +a(a+b)^2 (4b−a))))/((a+b)^2 )) a and R both are roots of (x+b)^2 s^2 +2x(x−b)s−x(4b−x)=0 ⇒ x^2 (1+s)^2 +2bx(s^2 −s−2)+b^2 s^2 =0 ⇒ aR = ((b^2 s^2 )/((1+s)^2 )) hence R = ((b^2 s^2 )/(a(1+s)^2 )) .
$${AE}={AD}+{DE} \\ $$$$\frac{{b}}{\mathrm{sin}\:\theta}=\:{a}\mathrm{cot}\:\frac{\theta}{\mathrm{2}}+\sqrt{\left({a}+{b}\right)^{\mathrm{2}} −{a}^{\mathrm{2}} } \\ $$$$\frac{{b}}{\mathrm{2sin}\:\frac{\theta}{\mathrm{2}}\mathrm{cos}\:\frac{\theta}{\mathrm{2}}}−\frac{{a}\mathrm{cos}\:\frac{\theta}{\mathrm{2}}}{\mathrm{sin}\:\frac{\theta}{\mathrm{2}}}=\sqrt{{b}\left({b}+\mathrm{2}{a}\right)} \\ $$$$\frac{{b}−{a}\left(\mathrm{1}+\mathrm{cos}\:\theta\right)}{\mathrm{sin}\:\theta}=\:\sqrt{{b}\left({b}+\mathrm{2}{a}\right)} \\ $$$$\Rightarrow\:{b}^{\mathrm{2}} +{a}^{\mathrm{2}} \left(\mathrm{1}+\mathrm{cos}\:\theta\right)^{\mathrm{2}} −\mathrm{2}{ab}\left(\mathrm{1}+\mathrm{cos}\:\theta\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:={b}\left({b}+\mathrm{2}{a}\right)\mathrm{sin}\:^{\mathrm{2}} \theta \\ $$$${b}^{\mathrm{2}} +{a}^{\mathrm{2}} +\mathrm{2}{a}^{\mathrm{2}} \mathrm{cos}\:\theta+{a}^{\mathrm{2}} \mathrm{cos}\:^{\mathrm{2}} \theta−\mathrm{2}{ab} \\ $$$$\:\:−\mathrm{2}{ab}\mathrm{cos}\:\theta={b}^{\mathrm{2}} +\mathrm{2}{ab}−{b}^{\mathrm{2}} \mathrm{cos}\:^{\mathrm{2}} \theta−\mathrm{2}{ab}\mathrm{cos}\:^{\mathrm{2}} \theta \\ $$$${let}\:\mathrm{cos}\:\theta\:=\:{s} \\ $$$$\Rightarrow\:\left({a}+{b}\right)^{\mathrm{2}} \boldsymbol{{s}}^{\mathrm{2}} +\mathrm{2}{a}\left({a}−{b}\right)\boldsymbol{{s}}−{a}\left(\mathrm{4}{b}−{a}\right)=\mathrm{0} \\ $$$$\boldsymbol{{s}}=\frac{{a}\left({a}−{b}\right)+\sqrt{{a}^{\mathrm{2}} \left({a}−{b}\right)^{\mathrm{2}} +{a}\left({a}+{b}\right)^{\mathrm{2}} \left(\mathrm{4}{b}−{a}\right)}}{\left({a}+{b}\right)^{\mathrm{2}} } \\ $$$${a}\:{and}\:{R}\:{both}\:{are}\:{roots}\:{of} \\ $$$$\:\left({x}+{b}\right)^{\mathrm{2}} \boldsymbol{{s}}^{\mathrm{2}} +\mathrm{2}{x}\left({x}−{b}\right)\boldsymbol{{s}}−{x}\left(\mathrm{4}{b}−{x}\right)=\mathrm{0} \\ $$$$\Rightarrow\:\boldsymbol{{x}}^{\mathrm{2}} \left(\mathrm{1}+\boldsymbol{{s}}\right)^{\mathrm{2}} +\mathrm{2}{b}\boldsymbol{{x}}\left(\boldsymbol{{s}}^{\mathrm{2}} −\boldsymbol{{s}}−\mathrm{2}\right)+{b}^{\mathrm{2}} \boldsymbol{{s}}^{\mathrm{2}} =\mathrm{0} \\ $$$$\Rightarrow\:\boldsymbol{{aR}}\:=\:\:\frac{{b}^{\mathrm{2}} \boldsymbol{{s}}^{\mathrm{2}} }{\left(\mathrm{1}+\boldsymbol{{s}}\right)^{\mathrm{2}} } \\ $$$${hence}\:\:\:\boldsymbol{{R}}\:=\:\frac{{b}^{\mathrm{2}} \boldsymbol{{s}}^{\mathrm{2}} }{{a}\left(\mathrm{1}+\boldsymbol{{s}}\right)^{\mathrm{2}} }\:. \\ $$

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