Question

I want to solve cos B but I dont get the steps! click attached.!

Answer 1

HELP

Answer 2

you have it

Answer 3

if you want to solve for B itself (which is probably what you want) you have to take the inverse cosine off that mess on the right

Answer 4

i.e.
\[B=\cos^{-1}\left(\frac{a^2+c^2-b^2}{2ac}\right)\]

Answer 5

this what happened.

you have:

b^2 = a^2 + c^2 - 2ac(cosB)

transfaer -2ac(cosB) to the left side.

b^2 + 2ac(cosB) = a^2 + c^2

transfer b^2 to the right side..

2ac(cosB) = a^2 + c^2 - b^2

now divide both sides by 2ac

cosB = (a^2 + c^2 - b^2)/2ac

Answer 6

@satellite73 but how did 2ac at the the denominators?

Answer 7

oh i am sorry, you wanted the algebra to solve, i didn't understand that

Answer 8

\[b^2=a^2+c^2-2ac\cos(B)\]
\[b^2-a^2-c^2=-2ac\cos(B)\]
\[\cos(B)=\frac{b^2-a^2-c^2}{-2ac}\] but it is easier to compute
\[\cos(B)=\frac{a^2+c^2-b^2}{2ac}\] i.e. multiply top and bottom by \(-1\)

Answer 9

@hbaldon I got it! thank you!!! @satellite73 @Callisto , i got it now! thank you so much!!

Answer 10

\[b^2=a^2+c^2-2accosB\]\[b^2+2accosB=a^2+c^2-2accosB+2accosB\]\[b^2+2accosB=a^2+c^2\]\[b^2+2accosB-b^2=a^2+c^2-b^2\]\[2accosB=a^2+c^2-b^2\]\[\frac{2accosB}{2ac}=\frac{a^2+c^2-b^2}{2ac}\]\[cosB=\frac{a^2+c^2-b^2}{2ac}\]
Hmm... I'm always late...

Answer 11

@Callisto thank you so much for taking your time to do this all for me. really helped!!

Answer 12

Welcome :)

Answer 13

you're welcome! :)