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Question Number 74994 by aliesam last updated on 05/Dec/19

Commented by mind is power last updated on 05/Dec/19

Commented by mind is power last updated on 05/Dec/19

$$\mathrm{\angle }\mathrm{ACB}=\alpha ,\mathrm{\angle }\mathrm{ABC}=\beta ,\mathrm{\angle }\mathrm{BAC}=\phi$$$$\mathrm{AB}=\mathrm{c},\mathrm{BC}=\mathrm{a},\mathrm{AC}=\mathrm{b}$$$$\mathrm{OL}=\mathrm{OT}={\mathrm{r}}_3\Rightarrow \mathrm{MLOT}\mathrm{Squar}$$$$\Rightarrow \mathrm{MT}={\mathrm{r}}_3=\mathrm{r}$$$$\mathrm{OBS}\equiv \mathrm{OBT}$$$$\Rightarrow \mathrm{\angle }\mathrm{OBC}=\frac{1}{2}\mathrm{\angle }\left(\mathrm{MBC}\right)$$$$\mathrm{\angle }\mathrm{MBC}=180-(90+\alpha )=90-\alpha$$$$\Rightarrow \mathrm{\angle }\mathrm{OBS}=\mathrm{\angle }\mathrm{OBC}=\frac{90-\alpha }{2}=\frac{\pi }{4}-\frac{\alpha }{2}\left(\mathrm{i}\mathrm{prefer}\mathrm{radian}\right)$$$$\mathrm{BM}=\mathrm{asin}\left(\alpha \right)$$$$\mathrm{BT}=\mathrm{asin}\left(\alpha \right)+\mathrm{r}$$$$\mathrm{BO}=\frac{\mathrm{r}}{\sin (\frac{\pi }{4}-\frac{\alpha }{2})},{\mathrm{BO}}^2=\frac{{\mathrm{r}}^2}{{\sin }^2(\frac{\pi }{4}-\frac{\alpha }{2})}=\frac{{\mathrm{r}}^2}{{(\frac{\sqrt{2}}{2}\cos \left(\frac{\alpha }{2}\right)-\frac{\sqrt{2}}{2}\sin \left(\frac{\alpha }{2}\right))}^2}$$$$=\frac{{2\mathrm{r}}^2}{1-\sin \left(\alpha \right)}$$$${\mathrm{BO}}^2={\mathrm{BT}}^2+{\mathrm{r}}^2={(\mathrm{asin}\left(\alpha \right)+\mathrm{r})}^2+{\mathrm{r}}^2=\frac{{2\mathrm{r}}^2}{1-\sin \left(\alpha \right)}$$$$\iff {2\mathrm{r}}^2+2\mathrm{arsin}\left(\alpha \right)+{\mathrm{a}}^2{\sin }^2\left(\alpha \right)=\frac{{2\mathrm{r}}^2}{1-\sin \left(\alpha \right)}$$$$\iff \frac{-{2\mathrm{r}}^2\sin \left(\alpha \right)}{1-\sin \left(\alpha \right)}+2\mathrm{arsin}\left(\alpha \right)+{\mathrm{a}}^2{\sin }^2\left(\alpha \right)=0$$$$\iff -{2\mathrm{r}}^2+2\mathrm{a}(1-\sin \left(\alpha \right)\mathrm{r}+{\mathrm{a}}^2\sin \left(\alpha \right)(1-\sin \alpha )=0$$$$\mathrm{\Delta }={4\mathrm{a}}^2{(1-\sin \alpha )}^2+{8\mathrm{a}}^2\sin \left(\alpha \right)(1-\sin \left(\alpha \right))$$$$={4\mathrm{a}}^2(1+{\sin }^2\left(\alpha \right)-2\sin \left(\alpha \right)+2\sin \left(\alpha \right)-{2\sin }^2\left(\alpha \right))$$$$={4\mathrm{a}}^2(1-{\sin }^2\left(\alpha \right))={4\mathrm{a}}^2{\cos }^2\left(\alpha \right)$$$$\mathrm{r}=\frac{-2\mathrm{a}(1-\sin \left(\alpha \right))-2\mathrm{acos}\left(\alpha \right)}{-4}$$$$=\frac{\mathrm{a}(\cos \left(\alpha \right)-\sin \left(\alpha \right)+1)}{2}={\mathrm{r}}_3$$$$\mathrm{sam}\mathrm{idea}\mathrm{Symetric}\mathrm{approche}$$$${\mathrm{r}}_4=\frac{\mathrm{c}(\cos \left(\phi \right)-\sin \left(\phi \right)+1)}{2},{\mathrm{r}}_5=\frac{\mathrm{b}(\cos \left(\phi \right)-\sin \left(\phi \right)+1)}{2}$$$${\mathrm{r}}_6=\frac{\mathrm{a}(\cos \left(\beta \right)-\sin \left(\beta \right)+1)}{2},{\mathrm{r}}_2=\frac{\mathrm{b}(\cos \left(\alpha \right)-\sin \left(\alpha \right)+1)}{2}$$$${\mathrm{r}}_1=\frac{\mathrm{c}(\cos \left(\beta \right)-\sin \left(\beta \right)+1)}{2}$$$${\mathrm{r}}_1.{\mathrm{r}}_3{\mathrm{r}}_5=\frac{\mathrm{abc}(\cos \left(\beta \right)-\sin \left(\beta \right)+1)(\cos \left(\alpha \right)-\sin \left(\alpha \right)+1)(\cos \left(\phi \right)-\sin \left(\phi \right)+1)}{8}$$$${\mathrm{r}}_2{\mathrm{r}}_4{\mathrm{r}}_6=\frac{\mathrm{abc}(\cos \left(\beta \right)-\sin \left(\beta \right)+1)(\cos \left(\alpha \right)-\sin \left(\alpha \right)+1)(\cos \left(\phi \right)-\sin \left(\phi \right)+1)}{8}$$$$\iff r_1.r_3.r_5=r_2.r_4.r_6$$$$$$$$$$

Commented by aliesam last updated on 05/Dec/19

$$thankyousomuchsir.godblessyou$$$$$$$$butwhatis\mathrm{\Delta }$$

Commented by mind is power last updated on 05/Dec/19

$$\mathrm{in}\mathrm{Franc}\mathrm{we}\mathrm{use}\mathrm{\Delta }\mathrm{discriminant}\mathrm{of}{\mathrm{aX}}^2+\mathrm{bX}+\mathrm{c}$$$$:\mathrm{\Delta }={\mathrm{b}}^2-4\mathrm{ac}$$

Commented by aliesam last updated on 05/Dec/19

$$thankyousir$$

Commented by mind is power last updated on 05/Dec/19

$${\mathrm{y}}^{\prime }\mathrm{re}\mathrm{Welcom}$$

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