How do you use the law of cosines to find BC given a triangle ABC with Angle A=93, AC=17 and AB=28?

Let's change the notation to be where angles are in uppercase and the side opposite that is the same letter but in lowercase:

`angle A` is unchanged `= 93^@`

Side AC becomes side `b = 17`

Side AB becomes side `c = 28`

Side BC becomes side `a = unknown`

Also unknown are `angle B and angle C` but they are not need for this problem.

The one of the three forms of the Law of Cosines that we need is:

`a^2 = b^2 + c^2 - 2(b)(c)cos(A)`

Substitute in the known values

`a^2 = 17^2 + 28^2 - 2(17)(28)cos(93^@)`

solve for a:

`a = sqrt(17^2 + 28^2 - 2(17)(28)cos(93^@))`

`a ~~ 33.5`

Given
`color(white)("XXX")/_A=93^@`

(I've assumed the "degree" part)

and using the standard convention lettering
`color(white)("XXX")a= "length of side opposite "/_ A=abs(BC)`
`color(white)("XXX")b= "length of side opposite "/_B = abs(AC)`
`color(white)("XXX")c= "length of side opposite "/_C = abs(AB)`

The Law of Cosines tells us that
`color(white)("XXX")a^2=b^2+c^2-2bc * cos(A)`

In this case
`color(white)("XXX")a^2=17^2+28^2-2 * 17 * 28 *cos(93^@)`

using a calculator
`color(white)("XXX")a^2~~1122.82383`
and
`color(white)("XXX")a~~33.51`