# A triangle has sides A, B, and C. The angle between sides A and B is (pi)/2 and the angle between sides B and C is pi/12. If side B has a length of 24, what is the area of the triangle?

May 22, 2017

$A = 77.172$ $u n i t {s}^{2}$

#### Explanation:

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

$B \textcolor{w h i t e}{00} \textcolor{w h i t e}{\frac{\pi}{1} 3} \textcolor{w h i t e}{0000000} b \textcolor{w h i t e}{00} \textcolor{b l a c k}{24}$

$C \textcolor{w h i t e}{00} \textcolor{b l a c k}{\frac{\pi}{2}} \textcolor{w h i t e}{0000000} c \textcolor{w h i t e}{00} \textcolor{w h i t e}{0}$

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

$\pi - \frac{\pi}{12} - \frac{\pi}{2}$ leaves us with $\frac{5 \pi}{12}$

$A \textcolor{w h i t e}{00} \textcolor{b l a c k}{\frac{\pi}{12}} \textcolor{w h i t e}{00000000} a \textcolor{w h i t e}{00} \textcolor{w h i t e}{0}$

$B \textcolor{w h i t e}{00} \textcolor{b l a c k}{\frac{5 \pi}{12}} \textcolor{w h i t e}{.} \textcolor{w h i t e}{0000000} b \textcolor{w h i t e}{00} \textcolor{b l a c k}{24}$

$C \textcolor{w h i t e}{00} \textcolor{b l a c k}{\frac{\pi}{2}} \textcolor{w h i t e}{000000000} c \textcolor{w h i t e}{00} \textcolor{w h i t e}{0}$

Now we could use law of sines, but I noticed something... $\frac{\pi}{2}$ is a right angle! Here's what our triangle looks like:

$\textcolor{w h i t e}{a} \textcolor{w h i t e}{- - - - - - - - - -} \textcolor{b l a c k}{/} \textcolor{b l a c k}{|}$
$\textcolor{w h i t e}{a} \textcolor{w h i t e}{- - - - - - - - -} \textcolor{b l a c k}{/} \textcolor{w h i t e}{-} \textcolor{b l a c k}{|}$
$\textcolor{w h i t e}{a} \textcolor{w h i t e}{- - - - - - - -} \textcolor{b l a c k}{/} \textcolor{w h i t e}{- - .} \textcolor{b l a c k}{|}$
$\textcolor{w h i t e}{a} \textcolor{w h i t e}{- - - - - - -} \textcolor{b l a c k}{/} \textcolor{w h i t e}{- - - 0} \textcolor{b l a c k}{|}$
$\textcolor{w h i t e}{- - - -} \textcolor{b l a c k}{c} \textcolor{w h i t e}{00000} \textcolor{b l a c k}{/} \textcolor{w h i t e}{- - - -} \textcolor{w h i t e}{/} \textcolor{b l a c k}{|} \textcolor{b l a c k}{a}$
$\textcolor{w h i t e}{a} \textcolor{w h i t e}{- - - - -} \textcolor{b l a c k}{/} \textcolor{w h i t e}{- - - - - 0} \textcolor{b l a c k}{|}$
$\textcolor{w h i t e}{a} \textcolor{w h i t e}{- - - -} \textcolor{b l a c k}{/} \textcolor{w h i t e}{- - - - - - 0} \textcolor{b l a c k}{|}$
$\textcolor{w h i t e}{a} \textcolor{w h i t e}{- - -} \textcolor{b l a c k}{/} \textcolor{w h i t e}{- - - - - - - 0} \textcolor{b l a c k}{|}$
$\textcolor{w h i t e}{a} \textcolor{w h i t e}{- -} \textcolor{b l a c k}{/} \textcolor{w h i t e}{- - - - - - - - 0} \textcolor{b l a c k}{|}$
$\textcolor{w h i t e}{-} \textcolor{b l a c k}{/} \textcolor{b l a c k}{} \textcolor{b l a c k}{A} \textcolor{b l a c k}{\ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots .}$
$\textcolor{w h i t e}{0000000000000000} b$

What we need to find is the height, or $\textcolor{red}{a}$ and the base, or $\textcolor{red}{b}$

$A \textcolor{w h i t e}{00} \textcolor{b l a c k}{\frac{\pi}{12}} \textcolor{w h i t e}{00000000} a \textcolor{w h i t e}{00} \textcolor{w h i t e}{0}$

$B \textcolor{w h i t e}{00} \textcolor{b l a c k}{\frac{5 \pi}{12}} \textcolor{w h i t e}{.} \textcolor{w h i t e}{0000000} b \textcolor{w h i t e}{00} \textcolor{red}{24}$

$C \textcolor{w h i t e}{00} \textcolor{b l a c k}{\frac{\pi}{2}} \textcolor{w h i t e}{000000000} c \textcolor{w h i t e}{00} \textcolor{w h i t e}{0}$

To find length $\textcolor{red}{a}$, we should use $\tan$:

$\tan \left(A\right) = \frac{a}{b}$

$\tan \left(\frac{\pi}{12}\right) = \frac{a}{24}$

$\tan \left(\frac{\pi}{12}\right) \times 24 = a$

$a \approx 6.431$

Now we can find the area:

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

$A = \frac{6.431 \times 24}{2}$

$A = 77.172$ $u n i t {s}^{2}$