# A triangle has sides A, B, and C. The angle between sides A and B is (7pi)/12 and the angle between sides B and C is pi/12. If side B has a length of 12, what is the area of the triangle?

May 19, 2017

$A = 24.053$ $u n i t {s}^{2}$

#### Explanation:

$A \textcolor{w h i t e}{00} \textcolor{b l a c k}{\frac{\pi}{12}} \textcolor{w h i t e}{0000000} a \textcolor{w h i t e}{00} \textcolor{w h i t e}{0}$

$B \textcolor{w h i t e}{00} \textcolor{w h i t e}{\frac{\pi}{1} 3} \textcolor{w h i t e}{0000000} b \textcolor{w h i t e}{00} \textcolor{b l a c k}{12}$

$C \textcolor{w h i t e}{00} \textcolor{b l a c k}{\frac{7 \pi}{12}} \textcolor{w h i t e}{0000000} c \textcolor{w h i t e}{00} \textcolor{w h i t e}{0}$

Let's find the remaining angle, $B$:

$\pi - \frac{\pi}{12} - \frac{7 \pi}{12}$ leaves us with $\frac{\pi}{3}$

$A \textcolor{w h i t e}{00} \textcolor{b l a c k}{\frac{\pi}{12}} \textcolor{w h i t e}{0000000} a \textcolor{w h i t e}{00} \textcolor{w h i t e}{0}$

$B \textcolor{w h i t e}{00} \textcolor{b l a c k}{\frac{\pi}{3}} \textcolor{w h i t e}{.} \textcolor{w h i t e}{0000000} b \textcolor{w h i t e}{00} \textcolor{b l a c k}{12}$

$C \textcolor{w h i t e}{00} \textcolor{b l a c k}{\frac{7 \pi}{12}} \textcolor{w h i t e}{0000000} c \textcolor{w h i t e}{00} \textcolor{w h i t e}{0}$

Now we should use law of sines

$\frac{\sin \left(\frac{\pi}{3}\right)}{12} = \frac{\sin \left(\frac{\pi}{12}\right)}{a}$

$a \approx 3.59$

$\frac{\sin \left(\frac{\pi}{3}\right)}{12} = \frac{\sin \left(\frac{7 \pi}{12}\right)}{c}$

$c \approx 13.4$

Now, to find the area, we use the equation $A = \frac{h \times b}{2}$, where $h$ is the height.

$\textcolor{w h i t e}{a} \textcolor{w h i t e}{- - - - - - - - - -} \textcolor{b l a c k}{/} \textcolor{b l a c k}{|}$
$\textcolor{w h i t e}{a} \textcolor{w h i t e}{- - - - - - - - -} \textcolor{b l a c k}{/} \textcolor{w h i t e}{-} \textcolor{b l a c k}{|}$
$\textcolor{w h i t e}{a} \textcolor{w h i t e}{- - - - - - - -} \textcolor{b l a c k}{/} \textcolor{w h i t e}{- - .} \textcolor{b l a c k}{|}$
$\textcolor{w h i t e}{a} \textcolor{w h i t e}{- - - - - - -} \textcolor{b l a c k}{/} \textcolor{w h i t e}{- - - 0} \textcolor{b l a c k}{|}$
$\textcolor{w h i t e}{- - - -} \textcolor{b l a c k}{a} \textcolor{w h i t e}{00000} \textcolor{b l a c k}{/} \textcolor{w h i t e}{- - - - 0} \textcolor{b l a c k}{|} \textcolor{b l a c k}{h}$
$\textcolor{w h i t e}{a} \textcolor{w h i t e}{- - - - -} \textcolor{b l a c k}{/} \textcolor{w h i t e}{- - - - - 0} \textcolor{b l a c k}{|}$
$\textcolor{w h i t e}{a} \textcolor{w h i t e}{- - - -} \textcolor{b l a c k}{/} \textcolor{w h i t e}{- - - - - - 0} \textcolor{b l a c k}{|}$
$\textcolor{w h i t e}{a} \textcolor{w h i t e}{- - -} \textcolor{b l a c k}{/} \textcolor{w h i t e}{- - - - - - - 0} \textcolor{b l a c k}{|}$
$\textcolor{w h i t e}{a} \textcolor{w h i t e}{- -} \textcolor{b l a c k}{/} \textcolor{w h i t e}{- - - - - - - - 0} \textcolor{b l a c k}{|}$
$\textcolor{w h i t e}{-} \textcolor{b l a c k}{/} \textcolor{b l a c k}{. .} \textcolor{b l a c k}{B} \textcolor{b l a c k}{\ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots .}$

$a = 3.59$, and the angle $B$ is $\frac{\pi}{3}$. We need to find the remaining length, $h$.

$\sin \left(\frac{\pi}{3}\right) = \frac{h}{3.59}$

$\sin \left(\frac{\pi}{3}\right) \times 3.59 = h$

$h = 3.11$

Now we can find the area:

$A = \frac{h \times b}{2}$

$A = \frac{3.59 \times 13.4}{2}$

$A = 24.053$ $u n i t {s}^{2}$