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Question Number 169626 by Shrinava last updated on 04/May/22

Answered by mr W last updated on 04/May/22

Commented by mr W last updated on 04/May/22

$$a^{\prime }=B^{\prime }{C}^{\prime }=2R\sin \frac{\pi -A}{2}=2R\sin A^{\prime }$$$$\Rightarrow A^{\prime }=\frac{\pi }{2}-\frac{A}{2}\Rightarrow \sin A^{\prime }=\cos \frac{A}{2}$$$$similarly$$$$\Rightarrow B^{\prime }=\frac{\pi }{2}-\frac{B}{2}\Rightarrow \sin B^{\prime }=\cos \frac{B}{2}$$$$\Rightarrow C^{\prime }=\frac{\pi }{2}-\frac{C}{2}\Rightarrow \sin C^{\prime }=\cos \frac{C}{2}$$$$\left[I_{a^{\prime }}{I}_{b^{\prime }}{I}_{c^{\prime }}\right]=8R^2\cos \frac{A^{\prime }}{2}\cos \frac{B^{\prime }}{2}\cos \frac{C^{\prime }}{2}$$$$\left[I_{a^{\prime }}{I}_{b^{\prime }}{I}_{c^{\prime }}\right]=2R^2(\sin A^{\prime }+\sin B^{\prime }+\sin C^{\prime })$$$$\left[I_{a^{\prime }}{I}_{b^{\prime }}{I}_{c^{\prime }}\right]=2R^2(\cos \frac{A}{2}+\cos \frac{B}{2}+\cos \frac{C}{2})✓$$

Commented by MaxiMaths last updated on 05/May/22

$$\mathrm{please}\mathrm{sir},\mathrm{which}\mathrm{app}\mathrm{do}\mathrm{u}\mathrm{use}\mathrm{to}\mathrm{make}$$$$\mathrm{those}\mathrm{figures}?$$

Commented by mr W last updated on 05/May/22

$$notaspecialone.eachsmartphone$$$$hasanappforeditingimages.$$

Commented by Shrinava last updated on 05/May/22

$$\mathrm{Perfect}-\mathrm{elegant}\mathrm{solution}$$$$\mathrm{Thank}\mathrm{you}\mathrm{so}\mathrm{much}\mathrm{professor}$$

Commented by Tawa11 last updated on 05/May/22

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

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