Question-171574

Question Number 171574 by cortano1 last updated on 18/Jun/22

Answered by mr W last updated on 18/Jun/22

Commented by mr W last updated on 18/Jun/22

sin β=(b/(2r))=(b/d) sin α=(a/(2r))=(a/d) α+β=θ cos (α+β)=cos θ (√((1−(a^2 /d^2 ))(1−(b^2 /d^2 ))))−(a/d)×(b/d)=cos θ (d^2 −a^2 )(d^2 −b^2 )=(ab+d^2 cos θ)^2 sin^2 θ d^2 =a^2 +b^2 +2ab cos θ d=((√(a^2 +b^2 +2ab cos θ))/(sin θ)) ⇒r=((√(a^2 +b^2 +2ab cos θ))/(2 sin θ)) =((√(3^2 +5^2 +2×3×5 cos 60°))/(2 sin 60°))=((7(√3))/( 3))

$$\mathrm{sin}\:\beta=\frac{{b}}{\mathrm{2}{r}}=\frac{{b}}{{d}} \\ $$$$\mathrm{sin}\:\alpha=\frac{{a}}{\mathrm{2}{r}}=\frac{{a}}{{d}} \\ $$$$\alpha+\beta=\theta \\ $$$$\mathrm{cos}\:\left(\alpha+\beta\right)=\mathrm{cos}\:\theta \\ $$$$\sqrt{\left(\mathrm{1}−\frac{{a}^{\mathrm{2}} }{{d}^{\mathrm{2}} }\right)\left(\mathrm{1}−\frac{{b}^{\mathrm{2}} }{{d}^{\mathrm{2}} }\right)}−\frac{{a}}{{d}}×\frac{{b}}{{d}}=\mathrm{cos}\:\theta \\ $$$$\left({d}^{\mathrm{2}} −{a}^{\mathrm{2}} \right)\left({d}^{\mathrm{2}} −{b}^{\mathrm{2}} \right)=\left({ab}+{d}^{\mathrm{2}} \:\mathrm{cos}\:\theta\right)^{\mathrm{2}} \\ $$$$\mathrm{sin}^{\mathrm{2}} \:\theta\:{d}^{\mathrm{2}} ={a}^{\mathrm{2}} +{b}^{\mathrm{2}} +\mathrm{2}{ab}\:\mathrm{cos}\:\theta \\ $$$${d}=\frac{\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} +\mathrm{2}{ab}\:\mathrm{cos}\:\theta}}{\mathrm{sin}\:\theta} \\ $$$$\Rightarrow{r}=\frac{\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} +\mathrm{2}{ab}\:\mathrm{cos}\:\theta}}{\mathrm{2}\:\mathrm{sin}\:\theta} \\ $$$$\:\:\:\:\:\:\:=\frac{\sqrt{\mathrm{3}^{\mathrm{2}} +\mathrm{5}^{\mathrm{2}} +\mathrm{2}×\mathrm{3}×\mathrm{5}\:\mathrm{cos}\:\mathrm{60}°}}{\mathrm{2}\:\mathrm{sin}\:\mathrm{60}°}=\frac{\mathrm{7}\sqrt{\mathrm{3}}}{\:\mathrm{3}} \\ $$

Commented by infinityaction last updated on 18/Jun/22

$${nice}\:{sir} \\ $$

Commented by Tawa11 last updated on 18/Jun/22

$$\mathrm{Great}\:\mathrm{sir} \\ $$

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