Question

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

Answer 1

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

Explanation:

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`