A triangle has sides A,B, and C. If the angle between sides A and B is (5pi)/12, the angle between sides B and C is pi/6, and the length of B is 15, what is the area of the triangle?

Jun 27, 2017

The area of the triangle is $= 56.25 {u}^{2}$

Explanation:

The angle between $A$ and $B$ is

$= \pi - \left(\frac{5}{12} \pi + \frac{2}{12} \pi\right)$

$= \pi - \left(\frac{7}{12} \pi\right)$

$= \frac{5}{12} \pi$

So, the triangle is isoceles

The side $B$ is $= 15$

Therefore, the side $C$ is $= 15$

$\sin \left(\frac{1}{6} \pi\right) = \frac{1}{2}$

The area of the triangle is

$A = \frac{1}{2} \cdot 15 \cdot 15 \sin \left(\frac{1}{6} \pi\right)$

$= \frac{225}{4} = 56.25$

Jun 27, 2017

$a r e a = 56.25$ units squared

Explanation:

Here's all the info we know (I found that last angle by subtracting all the other angles from 2pi):

So, here's what we need to know for the area:

Once we find these values, we'll be able to use the formula $a r e a = \frac{1}{2} \left(b \times h\right)$

Let's work on finding the height, $h$.

To do that, we just need to use $\sin \left(\frac{\pi}{6}\right) = \frac{h}{15}$, or $7.5 = h$

Now we know the height, all that's left is to find the base, or $c$

First, let's put all our info in a table:

length $\textcolor{w h i t e}{0000}$ angle
A = ? color(white)(0000) A= pi/6
$B = 15 \textcolor{w h i t e}{0000} B = \frac{17 \pi}{12}$
C = ? color(white)(0000) A= (5pi)/12

Let's use law of sines

$\frac{\sin \left(\frac{5 \pi}{12}\right)}{C} = \frac{\sin \left(\frac{17 \pi}{12}\right)}{15}$

$c = - 15$, but because this problem is dealing in distances, we cannot have a negative length, so $c = 15$

Now let's use our formula: $a r e a = \frac{1}{2} \left(b \times h\right)$

$a r e a = \frac{15 \times 7.5}{2}$

$a r e a = 56.25$ units squared