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

Dec 8, 2017

$A \approx 2.29 {\text{ units}}^{2}$

#### Explanation:

Given: A triangle with angles $\frac{\pi}{12} \left({15}^{o}\right) , \frac{\pi}{6} \left({30}^{o}\right) , \frac{3 \pi}{4} \left({135}^{o}\right)$ and $B = 5$ There are multiple ways to solve this problem. You can use the formula: $A r e a = \frac{1}{2} \text{Base" * "height}$ or you can use Heron's formula which requires you to find the length of all sides.

Find the area using $A r e a = \frac{1}{2} \text{Base" * "height}$:

Find the height:

$\text{ "sin (pi/12) = h/5 " so } h = 5 \sin \left(\frac{\pi}{12}\right) \approx 1.294$

Find the base A:

You can use the Pythagorean Theorem or trigonometry to find $A + h$:

$\sin \left({30}^{o} + {45}^{o}\right) = \sin \left({75}^{o}\right) = \frac{A + h}{5}$

$A + h = 5 \sin {75}^{o}$

$A = 5 \sin {75}^{o} - h = 5 \sin {75}^{o} - 5 \sin \left(\frac{\pi}{12}\right) = 5 \left(\sin {75}^{o} - \sin \left(\frac{\pi}{12}\right)\right) \approx 3.536$

$A r e a = \frac{1}{2} \cdot 3.536 \cdot 1.294 \approx 2.29 {\text{units}}^{2}$

Find the Area using Heron's Formula:

s = (A + B + C)/2; Area = sqrt(s(s-A)(s-B)(s-C))

Since we have a ${45}^{o} - {45}^{o} - {90}^{o}$ triangle formed with the height, we have the ratio: $h : h : h \sqrt{2}$ for side lengths.

This means $C = h \sqrt{2} = 5 \sqrt{2} \sin \left(\frac{\pi}{12}\right) \approx 1.83$

From above, $A = 5 \left(\sin {75}^{o} - \sin \left(\frac{\pi}{12}\right)\right) \approx 3.536$

$s \approx 5.183$

$A r e a = \sqrt{5.183 \left(1.647\right) \left(.183\right) \left(3.353\right)} \approx 2.29 {\text{units}}^{2}$