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S-Area-of-AB-C-in-AB-C-a-2-4S-1-2-cot-B-cot-C-a-2-4S-a-2-4-1-2-bc-sin-A-4R-2-sin-2-A-8R-2-sin-B-sin-




Question Number 201134 by mnjuly1970 last updated on 30/Nov/23
         S :  Area  of   AB^Δ C       in    AB^Δ C  :   (a^( 2) /(4S)) =^?  (1/2) (cot(B)+cot(C))               (a^( 2) /(4S)) = (( a^( 2) )/(4 ((1/2) bc sin(A))))=((4R^2 sin^( 2) (A))/(8R^( 2) sin(B)sin(C)sin(A)))                    = ((sin (A ))/(2sin(B)sin(C)))             =^(A+B+C=π)   ((sin(B)cos(C)+cosBsin(C))/(2sin(B)sin(C)))                    = (1/2) (cot(B)+cot(C))   ■
$$\: \\ $$$$\:\:\:\:\:\:{S}\::\:\:{Area}\:\:{of}\:\:\:{A}\overset{\Delta} {{B}C} \\ $$$$\:\:\:\:\:{in}\:\:\:\:{A}\overset{\Delta} {{B}C}\:\::\:\:\:\frac{{a}^{\:\mathrm{2}} }{\mathrm{4}{S}}\:\overset{?} {=}\:\frac{\mathrm{1}}{\mathrm{2}}\:\left({cot}\left({B}\right)+{cot}\left({C}\right)\right) \\ $$$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\frac{{a}^{\:\mathrm{2}} }{\mathrm{4}{S}}\:=\:\frac{\:{a}^{\:\mathrm{2}} }{\mathrm{4}\:\left(\frac{\mathrm{1}}{\mathrm{2}}\:{bc}\:{sin}\left({A}\right)\right)}=\frac{\mathrm{4}{R}^{\mathrm{2}} {sin}^{\:\mathrm{2}} \left({A}\right)}{\mathrm{8}{R}^{\:\mathrm{2}} {sin}\left({B}\right){sin}\left({C}\right){sin}\left({A}\right)} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\frac{{sin}\:\left({A}\:\right)}{\mathrm{2}{sin}\left({B}\right){sin}\left({C}\right)} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\overset{{A}+{B}+{C}=\pi} {=}\:\:\frac{{sin}\left({B}\right){cos}\left({C}\right)+{cosBsin}\left({C}\right)}{\mathrm{2}{sin}\left({B}\right){sin}\left({C}\right)} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\frac{\mathrm{1}}{\mathrm{2}}\:\left({cot}\left({B}\right)+{cot}\left({C}\right)\right)\:\:\:\blacksquare \\ $$$$ \\ $$

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