# How do you solve the triangle when angle W = 58˚, w = 12 cm, and h = 14 cm?

Mar 10, 2018

I'm going to make a table of all our information:

$\textcolor{w h i t e}{.} L e n g t h \textcolor{w h i t e}{.} | \textcolor{w h i t e}{.} A n g l e$

$w = 12 \textcolor{w h i t e}{\ldots .} | \textcolor{w h i t e}{.} W = 58$
h = 14 color(white)(. ...) | color(white)(.) H = color(green)(?)
x = color(green)(?)color(white)(4) color(white)(....) | X = color(green)(?)

We'll use $S \in \frac{A}{a} = S \in \frac{B}{b}$

$S \in \frac{58}{12} = S \in \frac{H}{14}$

$0.071 = S \in \frac{H}{14}$

$0.989 = S \in H$

$H = S {\in}^{-} 1 \left(0.989\right)$

$H = 81.65$

Now our table is

$w = 12 \textcolor{w h i t e}{\ldots .} | \textcolor{w h i t e}{.} W = 58$
$h = 14 \textcolor{w h i t e}{. \ldots} | \textcolor{w h i t e}{.} H = 81.65$
x = color(green)(?)color(white)(4) color(white)(....) | X = color(green)(?)

Since we know a triangle has $180$ total degree, we can solve for the last angle, $X$

$180 - 81.65 - 58 = 40.35$

$w = 12 \textcolor{w h i t e}{\ldots .} | \textcolor{w h i t e}{.} W = 58$
$h = 14 \textcolor{w h i t e}{. \ldots} | \textcolor{w h i t e}{.} H = 81.65$
x = color(green)(?)color(white)(4) color(white)(....) | X = 40.35

Now we can use $S \in \frac{A}{a} = S \in \frac{B}{b}$ again

$S \in \frac{58}{12} = S \in \frac{40.35}{x}$

$0.071 = S \in \frac{40.35}{x}$

$x = S \in \frac{40.35}{0.071}$

$x = 9.16$

Now our table is full

$w = 12 \textcolor{w h i t e}{\ldots .} | \textcolor{w h i t e}{.} W = 58$
$h = 14 \textcolor{w h i t e}{. \ldots} | \textcolor{w h i t e}{.} H = 81.65$
$x = 9.16 \textcolor{w h i t e}{.} | X = 40.35$