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

Dec 31, 2016

There are 3 forms of The Law of Cosines so we pick the appropriate one and substitute in the values. Please see below.

#### Explanation:

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 $= {93}^{\circ}$

Side AC becomes side $b = 17$

Side AB becomes side $c = 28$

Side BC becomes side $a = u n k n o w n$

Also unknown are $\angle B \mathmr{and} \angle C$ but they are not need for this problem.

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

${a}^{2} = {b}^{2} + {c}^{2} - 2 \left(b\right) \left(c\right) \cos \left(A\right)$

Substitute in the known values

${a}^{2} = {17}^{2} + {28}^{2} - 2 \left(17\right) \left(28\right) \cos \left({93}^{\circ}\right)$

solve for a:

$a = \sqrt{{17}^{2} + {28}^{2} - 2 \left(17\right) \left(28\right) \cos \left({93}^{\circ}\right)}$

$a \approx 33.5$

Dec 31, 2016

$\left\mid B C \right\mid \approx 33.51$

#### Explanation:

Given
$\textcolor{w h i t e}{\text{XXX}} \angle A = {93}^{\circ}$

(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
$\textcolor{w h i t e}{\text{XXX}} {a}^{2} = {b}^{2} + {c}^{2} - 2 b c \cdot \cos \left(A\right)$

In this case
$\textcolor{w h i t e}{\text{XXX}} {a}^{2} = {17}^{2} + {28}^{2} - 2 \cdot 17 \cdot 28 \cdot \cos \left({93}^{\circ}\right)$

using a calculator
$\textcolor{w h i t e}{\text{XXX}} {a}^{2} \approx 1122.82383$
and
$\textcolor{w h i t e}{\text{XXX}} a \approx 33.51$