Question-205588

Question Number 205588 by cortano12 last updated on 25/Mar/24

Answered by A5T last updated on 25/Mar/24

Commented by A5T last updated on 25/Mar/24

BC=r(√2);BD^2 =r^2 +r^2 −2r^2 cos60⇒BD=r 6^2 +AB^2 −6(√3)AB=r^2 ...(i)⇒2r^2 =72+2AB^2 −12(√3)AB 2r^2 =BC^2 =96+AB^2 −8(√3)AB=72+2AB^2 −12(√3)AB ⇒AB^2 +4(√3)AB−24=0⇒AB=2((√3)+3)≈9.464

$${BC}={r}\sqrt{\mathrm{2}};{BD}^{\mathrm{2}} ={r}^{\mathrm{2}} +{r}^{\mathrm{2}} −\mathrm{2}{r}^{\mathrm{2}} {cos}\mathrm{60}\Rightarrow{BD}={r} \\ $$$$\mathrm{6}^{\mathrm{2}} +{AB}^{\mathrm{2}} −\mathrm{6}\sqrt{\mathrm{3}}{AB}={r}^{\mathrm{2}} …\left({i}\right)\Rightarrow\mathrm{2}{r}^{\mathrm{2}} =\mathrm{72}+\mathrm{2}{AB}^{\mathrm{2}} −\mathrm{12}\sqrt{\mathrm{3}}{AB} \\ $$$$\mathrm{2}{r}^{\mathrm{2}} ={BC}^{\mathrm{2}} =\mathrm{96}+{AB}^{\mathrm{2}} −\mathrm{8}\sqrt{\mathrm{3}}{AB}=\mathrm{72}+\mathrm{2}{AB}^{\mathrm{2}} −\mathrm{12}\sqrt{\mathrm{3}}{AB} \\ $$$$\Rightarrow{AB}^{\mathrm{2}} +\mathrm{4}\sqrt{\mathrm{3}}{AB}−\mathrm{24}=\mathrm{0}\Rightarrow{AB}=\mathrm{2}\left(\sqrt{\mathrm{3}}+\mathrm{3}\right)\approx\mathrm{9}.\mathrm{464} \\ $$

Answered by mr W last updated on 25/Mar/24

Commented by mr W last updated on 25/Mar/24

CD^2 =6^2 +(4(√6))^2 −2×6×4(√6)×cos 75° ⇒CD=2(√(15+6(√3))) ((4(√6))/(sin α))=((2(√(15+6(√3))))/(sin 75°))=2R sin α=((4(√6)×sin 75°)/(2(√(15+6(√3)))))=((1+(√3))/( (√(5+2(√3))))) ⇒cos α=(1/( (√(5+2(√3))))) ((AB)/(sin (α+45°)))=2R ⇒AB=((2(√(15+6(√3))))/(sin 75°))×((sin α+cos α)/( (√2))) =2(3+(√3)) ✓

$${CD}^{\mathrm{2}} =\mathrm{6}^{\mathrm{2}} +\left(\mathrm{4}\sqrt{\mathrm{6}}\right)^{\mathrm{2}} −\mathrm{2}×\mathrm{6}×\mathrm{4}\sqrt{\mathrm{6}}×\mathrm{cos}\:\mathrm{75}° \\ $$$$\Rightarrow{CD}=\mathrm{2}\sqrt{\mathrm{15}+\mathrm{6}\sqrt{\mathrm{3}}} \\ $$$$\frac{\mathrm{4}\sqrt{\mathrm{6}}}{\mathrm{sin}\:\alpha}=\frac{\mathrm{2}\sqrt{\mathrm{15}+\mathrm{6}\sqrt{\mathrm{3}}}}{\mathrm{sin}\:\mathrm{75}°}=\mathrm{2}{R} \\ $$$$\mathrm{sin}\:\alpha=\frac{\mathrm{4}\sqrt{\mathrm{6}}×\mathrm{sin}\:\mathrm{75}°}{\mathrm{2}\sqrt{\mathrm{15}+\mathrm{6}\sqrt{\mathrm{3}}}}=\frac{\mathrm{1}+\sqrt{\mathrm{3}}}{\:\sqrt{\mathrm{5}+\mathrm{2}\sqrt{\mathrm{3}}}} \\ $$$$\Rightarrow\mathrm{cos}\:\alpha=\frac{\mathrm{1}}{\:\sqrt{\mathrm{5}+\mathrm{2}\sqrt{\mathrm{3}}}} \\ $$$$\frac{{AB}}{\mathrm{sin}\:\left(\alpha+\mathrm{45}°\right)}=\mathrm{2}{R} \\ $$$$\Rightarrow{AB}=\frac{\mathrm{2}\sqrt{\mathrm{15}+\mathrm{6}\sqrt{\mathrm{3}}}}{\mathrm{sin}\:\mathrm{75}°}×\frac{\mathrm{sin}\:\alpha+\mathrm{cos}\:\alpha}{\:\sqrt{\mathrm{2}}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{2}\left(\mathrm{3}+\sqrt{\mathrm{3}}\right)\:\checkmark \\ $$

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