Question and Answers Forum

All Questions      Topic List

Coordinate Geometry Questions

Previous in All Question      Next in All Question      

Previous in Coordinate Geometry      Next in Coordinate Geometry      

Question Number 183631 by a.lgnaoui last updated on 27/Dec/22

determiner r? AB=6 AE=5

$${determiner}\:\:\:\boldsymbol{{r}}? \\ $$$${AB}=\mathrm{6}\:\:\:\:\:{AE}=\mathrm{5} \\ $$

Commented by a.lgnaoui last updated on 27/Dec/22

Answered by mr W last updated on 28/Dec/22

Commented by mr W last updated on 28/Dec/22

BE^2 =5^2 +6^2 −2×5×6 cos 30° ⇒BE=(√(61−30(√3))) R=((√(61−30(√3)))/(2 sin 30°))=(√(61−30(√3))) cos α=(6/(2(√(61−30(√3)))))=(3/( (√(61−30(√3))))) ⇒sin α=((√(52−30(√3)))/( (√(61−30(√3))))) AF=(r/(sin 15°))=((4r)/( (√6)−(√2)))=((√6)+(√2))r cos (α+15°)=((R^2 +r^2 ((√6)+(√2))^2 −(R−r)^2 )/(2Rr((√6)+(√2)))) cos α cos 15°−sin α sin 15°=(((7+4(√3))r+2R)/(2R((√6)+(√2)))) (3/( (√(61−30(√3)))))×(((√6)+(√2))/4)−((√(52−30(√3)))/( (√(61−30(√3)))))×(((√6)−(√2))/4)=(((7+4(√3))r+2R)/(2R((√6)+(√2)))) 3(2+(√3))−(√(52−30(√3)))=(((7+4(√3))r)/2)+R ⇒r=2(7−4(√3))(6+3(√3)−(√(52−30(√3)))−(√(61−30(√3))))≈1.1478

$${BE}^{\mathrm{2}} =\mathrm{5}^{\mathrm{2}} +\mathrm{6}^{\mathrm{2}} −\mathrm{2}×\mathrm{5}×\mathrm{6}\:\mathrm{cos}\:\mathrm{30}° \\ $$$$\Rightarrow{BE}=\sqrt{\mathrm{61}−\mathrm{30}\sqrt{\mathrm{3}}} \\ $$$${R}=\frac{\sqrt{\mathrm{61}−\mathrm{30}\sqrt{\mathrm{3}}}}{\mathrm{2}\:\mathrm{sin}\:\mathrm{30}°}=\sqrt{\mathrm{61}−\mathrm{30}\sqrt{\mathrm{3}}} \\ $$$$\mathrm{cos}\:\alpha=\frac{\mathrm{6}}{\mathrm{2}\sqrt{\mathrm{61}−\mathrm{30}\sqrt{\mathrm{3}}}}=\frac{\mathrm{3}}{\:\sqrt{\mathrm{61}−\mathrm{30}\sqrt{\mathrm{3}}}} \\ $$$$\Rightarrow\mathrm{sin}\:\alpha=\frac{\sqrt{\mathrm{52}−\mathrm{30}\sqrt{\mathrm{3}}}}{\:\sqrt{\mathrm{61}−\mathrm{30}\sqrt{\mathrm{3}}}} \\ $$$${AF}=\frac{{r}}{\mathrm{sin}\:\mathrm{15}°}=\frac{\mathrm{4}{r}}{\:\sqrt{\mathrm{6}}−\sqrt{\mathrm{2}}}=\left(\sqrt{\mathrm{6}}+\sqrt{\mathrm{2}}\right){r} \\ $$$$\mathrm{cos}\:\left(\alpha+\mathrm{15}°\right)=\frac{{R}^{\mathrm{2}} +{r}^{\mathrm{2}} \left(\sqrt{\mathrm{6}}+\sqrt{\mathrm{2}}\right)^{\mathrm{2}} −\left({R}−{r}\right)^{\mathrm{2}} }{\mathrm{2}{Rr}\left(\sqrt{\mathrm{6}}+\sqrt{\mathrm{2}}\right)} \\ $$$$\mathrm{cos}\:\alpha\:\mathrm{cos}\:\mathrm{15}°−\mathrm{sin}\:\alpha\:\mathrm{sin}\:\mathrm{15}°=\frac{\left(\mathrm{7}+\mathrm{4}\sqrt{\mathrm{3}}\right){r}+\mathrm{2}{R}}{\mathrm{2}{R}\left(\sqrt{\mathrm{6}}+\sqrt{\mathrm{2}}\right)} \\ $$$$\frac{\mathrm{3}}{\:\sqrt{\mathrm{61}−\mathrm{30}\sqrt{\mathrm{3}}}}×\frac{\sqrt{\mathrm{6}}+\sqrt{\mathrm{2}}}{\mathrm{4}}−\frac{\sqrt{\mathrm{52}−\mathrm{30}\sqrt{\mathrm{3}}}}{\:\sqrt{\mathrm{61}−\mathrm{30}\sqrt{\mathrm{3}}}}×\frac{\sqrt{\mathrm{6}}−\sqrt{\mathrm{2}}}{\mathrm{4}}=\frac{\left(\mathrm{7}+\mathrm{4}\sqrt{\mathrm{3}}\right){r}+\mathrm{2}{R}}{\mathrm{2}{R}\left(\sqrt{\mathrm{6}}+\sqrt{\mathrm{2}}\right)} \\ $$$$\mathrm{3}\left(\mathrm{2}+\sqrt{\mathrm{3}}\right)−\sqrt{\mathrm{52}−\mathrm{30}\sqrt{\mathrm{3}}}=\frac{\left(\mathrm{7}+\mathrm{4}\sqrt{\mathrm{3}}\right){r}}{\mathrm{2}}+{R} \\ $$$$\Rightarrow{r}=\mathrm{2}\left(\mathrm{7}−\mathrm{4}\sqrt{\mathrm{3}}\right)\left(\mathrm{6}+\mathrm{3}\sqrt{\mathrm{3}}−\sqrt{\mathrm{52}−\mathrm{30}\sqrt{\mathrm{3}}}−\sqrt{\mathrm{61}−\mathrm{30}\sqrt{\mathrm{3}}}\right)\approx\mathrm{1}.\mathrm{1478} \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com