A triangle has sides A, B, and C. The angle between sides A and B is (5pi)/6 and the angle between sides B and C is pi/12. If side B has a length of 1, what is the area of the triangle?

Jan 3, 2016

Sum of angles gives an isosceles triangle. Half of the enter side is calculated from $\cos$ and the height from $\sin$. Area is found like that of a square (two triangles).

$A r e a = \frac{1}{4}$

Explanation:

The sum of all triangles in degrees is ${180}^{o}$ in degrees or π in radians. Therefore:

a+b+c=π

π/12+x+(5π)/6=π

x=π-π/12-(5π)/6

x=(12π)/12-π/12-(10π)/12

x=π/12

We notice that the angles $a = b$. This means that the triangle is isosceles, which leads to $B = A = 1$. The following image shows how the height opposite of $c$ can be calculated:

For the $b$ angle:

$\sin {15}^{o} = \frac{h}{A}$

$h = A \cdot \sin 15$

$h = \sin 15$

To calculate half of the $C$:

$\cos {15}^{o} = \frac{\frac{C}{2}}{A}$

$\left(\frac{C}{2}\right) = A \cdot \cos {15}^{o}$

$\left(\frac{C}{2}\right) = \cos {15}^{o}$

Therefore, the area can be calculated via the area of the square formed, as shown in the following image:

$A r e a = h \cdot \left(\frac{C}{2}\right)$

$A r e a = \sin 15 \cdot \cos 15$

Since we know that:

$\sin \left(2 a\right) = 2 \sin a \cos a$

$\sin a \cos a = \sin \frac{2 a}{2}$

So, finally:

$A r e a = \sin 15 \cdot \cos 15$

$A r e a = \sin \frac{2 \cdot 15}{2}$

$A r e a = \sin \frac{30}{2}$

$A r e a = \frac{\frac{1}{2}}{2}$

$A r e a = \frac{1}{4}$