# A triangle has sides A,B, and C. If the angle between sides A and B is (pi)/3, the angle between sides B and C is pi/6, and the length of B is 15, what is the area of the triangle?

Dec 23, 2017

this is a 30-60-90 angle triangle .
length of hypoteneuse is given that is $B = 15$
using trignometry
A=B×sin30° C=B×cos30°
Area of $\triangle$ ABC =A×C×1/2
=1/2×B×sin30°×B×cos30°
=1/2×B^2×1/2×sqrt3/2
=1/2×15^2×sqrt3/4
=1/2×225×sqrt3/8
$\approx 48.7139289629$

Dec 23, 2017

$225 \frac{\sqrt{3}}{8}$

#### Explanation:

We know that $\frac{\pi}{6} \text{radians} = {30}^{o}$ and $\frac{\pi}{6} \text{radians} = {60}^{o}$, so we are dealing with a 30-60-90 triangle.

If side B touches both the 30-degree angle and the 60-degree angle, then side B is opposite the 90-degree angle and is the hypotenuse of the triangle.

Since we know that the sides of a 30-60-90 triangle follow the pattern $x$-$x \sqrt{3}$-$2 x$, we can say that:

$2 x = 15$

$x = \frac{15}{2}$

$x \sqrt{3} = \frac{15}{2} \cdot \sqrt{3}$

Now that we know the base and height of the triangle, we can find the area.

$\text{Area} = \frac{1}{2} \cdot \left(x\right) \cdot \left(x \sqrt{3}\right)$

$= \frac{1}{2} \cdot \frac{15}{2} \cdot \frac{15}{2} \cdot \sqrt{3}$

$= 225 \frac{\sqrt{3}}{8}$