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

If in △ABC , m(∢BAC)>90° then: cosA + cosB cosC = (F/(aR))

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

Commented bymr W last updated on 24/Jun/22

please check! maybe you meant cosA + cosB cosC = ((aR)/F) with F=area of triangle?

$${please}\:{check}!\:{maybe}\:{you}\:{meant} \\ $$ $$\mathrm{cosA}\:+\:\mathrm{cosB}\:\mathrm{cosC}\:=\:\frac{{aR}}{{F}} \\ $$ $${with}\:{F}={area}\:{of}\:{triangle}? \\ $$

Commented byShrinava last updated on 24/Jun/22

Yes-yes dear professor

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

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

R=((abc)/(4F)) (a/(sin A))=(b/(sin B))=(c/(sin C))=2R cos A=cos (π−(B+C)) =−cos (B+C) =−cos B cos C+sin B sin C cosA + cosB cosC =sin B sin C =((2R)/b)×((2R)/c) =((4R^2 )/(bc)) =((4R)/(bc))×((abc)/(4F)) =((aR)/F) ✓

$${R}=\frac{{abc}}{\mathrm{4}{F}} \\ $$ $$\frac{{a}}{\mathrm{sin}\:{A}}=\frac{{b}}{\mathrm{sin}\:{B}}=\frac{{c}}{\mathrm{sin}\:{C}}=\mathrm{2}{R} \\ $$ $$ \\ $$ $$\mathrm{cos}\:{A}=\mathrm{cos}\:\left(\pi−\left({B}+{C}\right)\right) \\ $$ $$=−\mathrm{cos}\:\left({B}+{C}\right) \\ $$ $$=−\mathrm{cos}\:{B}\:\mathrm{cos}\:{C}+\mathrm{sin}\:{B}\:\mathrm{sin}\:{C} \\ $$ $$ \\ $$ $$\mathrm{cosA}\:+\:\mathrm{cosB}\:\mathrm{cosC} \\ $$ $$=\mathrm{sin}\:{B}\:\mathrm{sin}\:{C} \\ $$ $$=\frac{\mathrm{2}{R}}{{b}}×\frac{\mathrm{2}{R}}{{c}} \\ $$ $$=\frac{\mathrm{4}{R}^{\mathrm{2}} }{{bc}} \\ $$ $$=\frac{\mathrm{4}{R}}{{bc}}×\frac{{abc}}{\mathrm{4}{F}} \\ $$ $$=\frac{{aR}}{{F}}\:\checkmark \\ $$

Commented bySotoberry last updated on 24/Jun/22

thankyou so much sir

$${thankyou}\:{so}\:{much}\:{sir} \\ $$

Commented byShrinava last updated on 25/Jun/22

Cool dear professor thank you

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

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