How do you solve the triangle given A=40, b=7, c=6?

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Answer 1 · 2017-01-13T19:09:03

$angle B= 81°56'14'', a = 4.544, angle C=58°03'46''$

Cosine rule:

$a^2=b^2+c^2-2bc CosA$

$a^2=(7^2)+(6^2)-2 xx 7 xx 6 xx Cos40°$

$a^2=49+36-84 xx 0.766044443$

$a^2=85-64.34773322$

$a^2=20.65226678$

$a=sqrt20.65226678$

$a=4.544=BC$

$b^2=a^2+c^2-2ac CosB$

$b^2=(4.544)^2+6^2-2 xx 4.544 xx CosB$

$7^2=(4.544)^2+36-54.528 xx CosB$

$7^2=20.648+36-54.528 xx CosB$

$7^2=56.648-54.528 xx CosB$

$-54.528cosB=49-56.648$

$54.528CosB=56.648-49$

$CosB=7.648/54.528$

$CosB=0.140258216$

$angleB=81° 56’ 14’’$

$180°-(40°+81° 56' 14'')=58° 03' 46'' =angle C$

Check:

$(BC)/(Sin40)=6/(Sin58° 03’ 46’’)$

$BC=(6 xx Sin40°)/(Sin58° 03’ 46’’)$

$BC=3.856725658/0.848628208$

$BC=4.545=a$