Two corners of a triangle have angles of #(3 pi ) / 8 # and # pi / 3 #. If one side of the triangle has a length of #2 #, what is the longest possible perimeter of the triangle?

Question

Two corners of a triangle have angles of #(3 pi ) /  8  # and #  pi   /  3  #.  If one side of the triangle has a length of #2  #, what is the longest possible perimeter of the triangle?

Autumn Armstrong · Accepted Answer

Largest possible area of the triangle is 2.017

Given are the two angles #(3pi)/8# and #pi/3# and the length 2 The remaining angle: #= pi - (((3pi)/8) + pi/3) = (7pi)/24# I am assuming that length AB (2) is opposite the smallest angle. Using the ASA Area#=(c^2*sin(A)*sin(B))/(2*sin(C)# Area#=( 2^2*sin(pi/3)*sin((3pi)/8))/(2*sin((7pi)/24))# Area#=2.017#

Violet Aldrich · Answer

undefined To find the longest possible perimeter of the triangle, we need to maximize the length of the third side. We can use the Law of Sines to find the length of this side.

Given the two angles:
- Angle 1: $ \frac{3\pi}{8} $
- Angle 2: $ \frac{\pi}{3} $

We can find the third angle by subtracting the sum of the given angles from $ \pi $:
$ \pi - \left(\frac{3\pi}{8} + \frac{\pi}{3}\right) = \frac{8\pi}{24} - \frac{9\pi}{24} - \frac{8\pi}{24} = \frac{11\pi}{24} $

Now, we use the Law of Sines:

$$ \frac{a}{\sin(A)} = \frac{b}{\sin(B)} = \frac{c}{\sin(C)} $$

Let's choose $ a = 2 $ (the given side), $ A = \frac{3\pi}{8} $, and $ C = \frac{11\pi}{24} $:

$$ \frac{2}{\sin\left(\frac{3\pi}{8}\right)} = \frac{c}{\sin\left(\frac{11\pi}{24}\right)} $$

Solve for $ c $:
$$ c = 2 \times \frac{\sin\left(\frac{11\pi}{24}\right)}{\sin\left(\frac{3\pi}{8}\right)} $$

Now, find the perimeter:
$$ P = 2 + 2 + c $$

Plug in the values:
$$ P = 4 + 2 \times \frac{\sin\left(\frac{11\pi}{24}\right)}{\sin\left(\frac{3\pi}{8}\right)} $$

Calculate $ \sin\left(\frac{11\pi}{24}\right) $ and $ \sin\left(\frac{3\pi}{8}\right) $, then substitute them into the equation to find the longest possible perimeter of the triangle.