# How do you solve the triangle given trianglePQR, p=12, mangleQ=80, mangleR=30?

May 12, 2017

$P \textcolor{w h i t e}{.} {70}^{o} \textcolor{w h i t e}{\ldots \ldots \ldots \ldots .} p \textcolor{w h i t e}{.} 12$
$Q \textcolor{w h i t e}{.} {80}^{o} \textcolor{w h i t e}{\ldots \ldots \ldots \ldots .} q \textcolor{w h i t e}{.} 12.576$
$R \textcolor{w h i t e}{.} {30}^{o} \textcolor{w h i t e}{\ldots \ldots \ldots \ldots .} r \textcolor{w h i t e}{.} 6.385$

#### Explanation:

This si a table of what we know:

I like to put all the information in one place so I can keep track of everything
Pcolor(white)(.) ?color(white)(.............)pcolor(white)(.)12
Qcolor(white)(.)80 color(white)(.............)qcolor(white)(.)?
Rcolor(white)(.)30 color(white)(.............)rcolor(white)(.)?

Capital letters are angles, lowercase are lengths

We know that all angles in a triangle must add to ${180}^{o}$. So, if we have ${80}^{o}$ and ${30}^{o}$ (${110}^{o}$), then the last remaining angle must be ${70}^{o}$ $\left(180 - 110 = 70\right)$.

$P \textcolor{w h i t e}{.} {70}^{o} \textcolor{w h i t e}{\ldots \ldots \ldots \ldots .} p \textcolor{w h i t e}{.} 12$
Qcolor(white)(.)80^o color(white)(.............)qcolor(white)(.)?
Rcolor(white)(.)30^o color(white)(.............)rcolor(white)(.)?

Now we know one angle-length pair (Q), so we can find the remaining lengths.

$\textcolor{w h i t e}{0}$
Solving for $\textcolor{red}{q}$
$\frac{\sin \left(70\right)}{12} = \frac{\sin \left(80\right)}{q}$

$\sin \left(80\right) \cdot \frac{12}{\sin \left(70\right)} = q$

$q \approx 12.576$

$\textcolor{w h i t e}{0}$

Solving for $\textcolor{red}{r}$

$\frac{\sin \left(70\right)}{12} = \frac{\sin \left(30\right)}{r}$

$\sin \left(30\right) \cdot \frac{12}{\sin \left(70\right)} = r$

$r \approx 6.385$

Now we have everything:

$P \textcolor{w h i t e}{.} 70 \textcolor{w h i t e}{\ldots \ldots \ldots \ldots .} p \textcolor{w h i t e}{.} 12$
$Q \textcolor{w h i t e}{.} 80 \textcolor{w h i t e}{\ldots \ldots \ldots \ldots .} q \textcolor{w h i t e}{.} 12.576$
$R \textcolor{w h i t e}{.} 30 \textcolor{w h i t e}{\ldots \ldots \ldots \ldots .} r \textcolor{w h i t e}{.} 6.385$