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Question Number 188515 by mnjuly1970 last updated on 02/Mar/23

in AB^Δ C : a=3 , b=6 , c=7 find the value of : E = (a+b )cos(C) + (b+c)cos(A)+ (a+c )cos(B)=?

$$ \\ $$$$\:\:\:\:\:{in}\:\:{A}\overset{\Delta} {{B}C}\:\::\:\:\:{a}=\mathrm{3}\:\:,\:\:{b}=\mathrm{6}\:\:,\:\:{c}=\mathrm{7} \\ $$$$\:\:\: \\ $$$$\: \\ $$$$\:\:\:\:{find}\:\:{the}\:{value}\:\:{of}\:: \\ $$$$\:\:\: \\ $$$$\:\:\:\:\:\:\:{E}\:=\:\left({a}+{b}\:\right){cos}\left({C}\right)\:+\:\left({b}+{c}\right){cos}\left({A}\right)+\:\left({a}+{c}\:\right){cos}\left({B}\right)=?\:\:\:\:\:\:\:\:\:\:\:\:\: \\ $$$$\:\:\:\:\:\:\: \\ $$

Answered by mr W last updated on 02/Mar/23

E=(a+b+c)(cos A+cos B+cos C)−(a cos A+b cos B+c cos C) =(a+b+c)(1+4 sin (A/2) sin (B/2) sin (C/2))−R(sin 2A+sin 2B+sin 2C) =(a+b+c)+4(a+b+c) sin (A/2) sin (B/2) sin (C/2)−4R sin A sinB sinC =(a+b+c)+8R(sin A+sin B+sin C) sin (A/2) sin (B/2) sin (C/2)−4R sin A sinB sinC =(a+b+c)+8R×4 cos (A/2) cos (B/2) cos (C/2) sin (A/2) sin (B/2) sin (C/2)−4R sin A sinB sinC =(a+b+c)+4R sin A sin B sin C−4R sin A sinB sinC =a+b+c =3+6+7=16 ✓

$${E}=\left({a}+{b}+{c}\right)\left(\mathrm{cos}\:{A}+\mathrm{cos}\:{B}+\mathrm{cos}\:{C}\right)−\left({a}\:\mathrm{cos}\:{A}+{b}\:\mathrm{cos}\:{B}+{c}\:\mathrm{cos}\:{C}\right) \\ $$$$\:=\left({a}+{b}+{c}\right)\left(\mathrm{1}+\mathrm{4}\:\mathrm{sin}\:\frac{{A}}{\mathrm{2}}\:\mathrm{sin}\:\frac{{B}}{\mathrm{2}}\:\mathrm{sin}\:\frac{{C}}{\mathrm{2}}\right)−{R}\left(\mathrm{sin}\:\mathrm{2}{A}+\mathrm{sin}\:\mathrm{2}{B}+\mathrm{sin}\:\mathrm{2}{C}\right) \\ $$$$\:=\left({a}+{b}+{c}\right)+\mathrm{4}\left({a}+{b}+{c}\right)\:\mathrm{sin}\:\frac{{A}}{\mathrm{2}}\:\mathrm{sin}\:\frac{{B}}{\mathrm{2}}\:\mathrm{sin}\:\frac{{C}}{\mathrm{2}}−\mathrm{4}{R}\:\mathrm{sin}\:{A}\:\mathrm{sin}{B}\:\mathrm{sin}{C} \\ $$$$\:=\left({a}+{b}+{c}\right)+\mathrm{8}{R}\left(\mathrm{sin}\:{A}+\mathrm{sin}\:{B}+\mathrm{sin}\:{C}\right)\:\mathrm{sin}\:\frac{{A}}{\mathrm{2}}\:\mathrm{sin}\:\frac{{B}}{\mathrm{2}}\:\mathrm{sin}\:\frac{{C}}{\mathrm{2}}−\mathrm{4}{R}\:\mathrm{sin}\:{A}\:\mathrm{sin}{B}\:\mathrm{sin}{C} \\ $$$$\:=\left({a}+{b}+{c}\right)+\mathrm{8}{R}×\mathrm{4}\:\mathrm{cos}\:\frac{{A}}{\mathrm{2}}\:\mathrm{cos}\:\frac{{B}}{\mathrm{2}}\:\mathrm{cos}\:\frac{{C}}{\mathrm{2}}\:\mathrm{sin}\:\frac{{A}}{\mathrm{2}}\:\mathrm{sin}\:\frac{{B}}{\mathrm{2}}\:\mathrm{sin}\:\frac{{C}}{\mathrm{2}}−\mathrm{4}{R}\:\mathrm{sin}\:{A}\:\mathrm{sin}{B}\:\mathrm{sin}{C} \\ $$$$\:=\left({a}+{b}+{c}\right)+\mathrm{4}{R}\:\mathrm{sin}\:{A}\:\mathrm{sin}\:{B}\:\mathrm{sin}\:{C}−\mathrm{4}{R}\:\mathrm{sin}\:{A}\:\mathrm{sin}{B}\:\mathrm{sin}{C} \\ $$$$\:={a}+{b}+{c} \\ $$$$\:=\mathrm{3}+\mathrm{6}+\mathrm{7}=\mathrm{16}\:\checkmark \\ $$

Commented by manxsol last updated on 03/Mar/23

perpendicular projection

$${perpendicular}\:{projection} \\ $$

Commented by manxsol last updated on 03/Mar/23

Commented by mnjuly1970 last updated on 03/Mar/23

thanks alot sir W

$${thanks}\:{alot}\:{sir}\:{W} \\ $$

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