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Question Number 172214 by Shrinava last updated on 24/Jun/22

$$\mathrm{If}\mathrm{in}△\mathrm{ABC},\mathrm{m}\left(\sphericalangle \mathrm{BAC}\right)>90°\mathrm{then}:$$ $$\mathrm{cosA}+\mathrm{cosB}\mathrm{cosC}=\frac{\mathrm{F}}{\mathrm{aR}}$$

Commented bymr W last updated on 24/Jun/22

$$pleasecheck!maybeyoumeant$$ $$\mathrm{cosA}+\mathrm{cosB}\mathrm{cosC}=\frac{aR}{F}$$ $$withF=areaoftriangle?$$

Commented byShrinava last updated on 24/Jun/22

$$\mathrm{Yes}-\mathrm{yes}\mathrm{dear}\mathrm{professor}$$

Answered by mr W last updated on 24/Jun/22

$$R=\frac{abc}{4F}$$ $$\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=2R$$ $$$$ $$\cos A=\cos (\pi -(B+C))$$ $$=-\cos (B+C)$$ $$=-\cos B\cos C+\sin B\sin C$$ $$$$ $$\mathrm{cosA}+\mathrm{cosB}\mathrm{cosC}$$ $$=\sin B\sin C$$ $$=\frac{2R}{b}\times \frac{2R}{c}$$ $$=\frac{4R^2}{bc}$$ $$=\frac{4R}{bc}\times \frac{abc}{4F}$$ $$=\frac{aR}{F}✓$$

Commented bySotoberry last updated on 24/Jun/22

$$thankyousomuchsir$$

Commented byShrinava last updated on 25/Jun/22

$$\mathrm{Cool}\mathrm{dear}\mathrm{professor}\mathrm{thank}\mathrm{you}$$

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