Two corners of a triangle have angles of # (2 pi )/ 3 # and # ( pi ) / 6 #. If one side of the triangle has a length of # 16 #, what is the longest possible perimeter of the triangle?

Question

Two corners of a triangle have angles of #  (2  pi  )/  3  # and #  ( pi ) /  6  #. If one side of the triangle has a length of # 16  #, what is the longest possible perimeter of the triangle?

Aaliyah Anderson · Accepted Answer

Longest possible perimeter of the triangle is #color(purple)(P_t = 71.4256)#

Given angles #A = (2pi)/3, B = pi /6#

#C = pi - (2pi)/3 - pi/6 = pi/6#

It’s an isosceles triangle with sides b & c equal.

Sadie Adams · Answer

undefined To find the longest possible perimeter of the triangle, you need to maximize the length of the third side.

Using the Law of Sines:

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

Where $ a $, $ b $, and $ c $ are the lengths of the sides opposite to angles $ A $, $ B $, and $ C $ respectively.

Given:
$ A = \frac{2\pi}{3} $ and $ B = \frac{\pi}{6} $

Let's find the third angle $ C $:

$ C = \pi - A - B $
$ C = \pi - \frac{2\pi}{3} - \frac{\pi}{6} $
$ C = \frac{\pi}{6} $

Now, let's find the length of the third side $ c $:

$$ \frac{16}{\sin(\frac{\pi}{6})} = \frac{c}{\sin(\frac{2\pi}{3})} $$

Solving for $ c $:

$$ c = 16\frac{\sin(\frac{2\pi}{3})}{\sin(\frac{\pi}{6})} $$

$$ c = 16\frac{\sqrt{3}}{\frac{1}{2}} = 32\sqrt{3} $$

So, the longest possible perimeter of the triangle is:

$$ P = 16 + 16 + 32\sqrt{3} = 32 + 32\sqrt{3} $$