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 #(5pi)/12#, and the length of B is 2, what is the area of the triangle?

Question

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 #(5pi)/12#, and the length of B is 2, what is the area of the triangle?

Daniel Acosta · Accepted Answer

Area of triangle #A_t = (1/2) * a * b * sin C = color(red)(2.5758#

#hatC = pi/3, hat A = (5pi)/12, b = 2#

#hatB = pi - (5pi)/12 - (pi/3) = pi/4#

#a / sin A = b / sin B = c / sin C#

#a = (2 * sin ((5pi)/12)) / sin (pi/4) = 2.2307#

Area of triangle #A_t = (1/2) * a * b * sin C #

#A_t = (1/2) * 2.2307 * 2 sin (pi/3) = color(red)(2.5758#

Dylan Arnold · Answer

undefined The area of the triangle can be calculated using Heron's formula, which states that the area $A$ of a triangle with side lengths $a$, $b$, and $c$ is given by:

$$A = \sqrt{s(s - a)(s - b)(s - c)}$$

Where $s$ is the semi-perimeter of the triangle, calculated as:

$$s = \frac{a + b + c}{2}$$

Given that side $b$ is 2 units, and the angles between sides $a$ and $b$ and between sides $b$ and $c$ are $ \frac{\pi}{3} $ and $ \frac{5\pi}{12} $ respectively, we can use the law of cosines to find the lengths of sides $a$ and $c$. Then, we can calculate the semi-perimeter and finally the area of the triangle using Heron's formula.

Using the law of cosines, we have:

$$c^2 = a^2 + b^2 - 2ab \cdot \cos(\frac{5\pi}{12})$$

$$a^2 = b^2 + c^2 - 2bc \cdot \cos(\frac{\pi}{3})$$

After calculating $a$ and $c$, we can find the semi-perimeter $s$ and then use Heron's formula to find the area of the triangle.