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

$A r e a = \frac{25 \sqrt{3}}{12} = 3.60844$ square units

#### Explanation:

The triangle is isosceles. The base angles $A = \frac{\pi}{6} = {30}^{\circ}$ and $B = \frac{\pi}{6} = {30}^{\circ}$. We can readily see that $\frac{1}{2}$ of the base $b$ equal $\frac{5}{2}$. Use $\frac{5}{2}$ to compute side $c$.

$C o s \left(\frac{\pi}{6}\right) = \frac{\frac{5}{2}}{c}$

$c = \frac{\frac{5}{2}}{\cos} \left(\frac{\pi}{6}\right) = \frac{\frac{5}{2}}{\frac{\sqrt{3}}{2}} = \frac{5}{\sqrt{3}} = \frac{5 \sqrt{3}}{3}$

Compute the Area now using the formula for two sides and an included angle

$A r e a = \frac{1}{2} \cdot b \cdot c \cdot \sin A$

$A r e a = \frac{1}{2} \cdot 5 \cdot \frac{5 \sqrt{3}}{3} \cdot \sin \left(\frac{\pi}{6}\right)$

$A r e a = \frac{1}{2} \cdot 5 \cdot \frac{5 \sqrt{3}}{3} \cdot \frac{1}{2}$

$A r e a = \frac{25 \sqrt{3}}{12} = 3.60844$ square units

Have a nice day!!! from the Philippines