Two corners of a triangle have angles of # (5 pi )/ 12 # and # ( pi ) / 12 #. If one side of the triangle has a length of # 6 #, what is the longest possible perimeter of the triangle?

Question

Two corners of a triangle have angles of #  (5  pi  )/  12  # and #  ( pi ) /  12  #. If one side of the triangle has a length of # 6  #, what is the longest possible perimeter of the triangle?

Evelyn Arboleda · Accepted Answer

#=13.35# Clearly this is a right angled triangle as #pi-(5pi)/12-pi/12=pi/2# One #side=hypoten use =6# ;So other sides #=6sin(pi/12) and 6cos( pi/12)# Therefore Perimeter of the triangle#=6+6sin(pi/12)+ 6cos( pi/12)# #=6+(6times0.2588)+(6times0.966)# #=6+1.55+5.8)# #=13.35#

Parker Allen · Answer

undefined To find the longest possible perimeter of the triangle, we need to determine the lengths of the other two sides given the angles provided and the length of one side.

Let's denote the angles of the triangle as $ \frac{5\pi}{12} $ and $ \frac{\pi}{12} $. The side with a length of 6 is opposite to the angle $ \frac{\pi}{12} $.

First, we can find the length of the side opposite to $ \frac{5\pi}{12} $ using the Law of Sines:
$$ \frac{a}{\sin A} = \frac{b}{\sin B} $$
where $ a $ and $ b $ are sides of the triangle, and $ A $ and $ B $ are the angles opposite to sides $ a $ and $ b $ respectively.

Let $ a $ be the side opposite $ \frac{5\pi}{12} $ and $ b $ be the side opposite $ \frac{\pi}{12} $. We already know $ b = 6 $ units.

$$ \frac{a}{\sin\left(\frac{5\pi}{12}\right)} = \frac{6}{\sin\left(\frac{\pi}{12}\right)} $$

Solve for $ a $:
$$ a = 6\left(\frac{\sin\left(\frac{5\pi}{12}\right)}{\sin\left(\frac{\pi}{12}\right)}\right) $$

Now, calculate $ a $:
$$ a = 6\left(\frac{\sin\left(\frac{5\pi}{12}\right)}{\sin\left(\frac{\pi}{12}\right)}\right) = 6\left(\frac{\frac{\sqrt{6}+\sqrt{2}}{4}}{\frac{\sqrt{2}-\sqrt{6}}{4}}\right) = 6\left(\frac{\sqrt{6}+\sqrt{2}}{\sqrt{2}-\sqrt{6}}\right) $$

This simplifies to:
$$ a = 6\left(\frac{(\sqrt{6}+\sqrt{2})(\sqrt{2}+\sqrt{6})}{(\sqrt{2}-\sqrt{6})(\sqrt{2}+\sqrt{6})}\right) = 6\left(\frac{8+4\sqrt{3}}{-4}\right) = -12 - 6\sqrt{3} $$

The perimeter $ P $ of the triangle is given by $ P = a + b + c $, where $ c $ is the third side. Since we want to find the longest possible perimeter, $ c $ should be the longest side possible. This occurs when $ c $ is opposite the largest angle $ \frac{5\pi}{12} $.

Now, we can calculate the perimeter $ P $:
$$ P = -12 - 6\sqrt{3} + 6 + c $$

The longest possible perimeter occurs when $ c = a $ (opposite the largest angle):
$$ P = -12 - 6\sqrt{3} + 6 - 12 - 6\sqrt{3} = -18 - 12\sqrt{3} $$

Therefore, the longest possible perimeter of the triangle is $ -18 - 12\sqrt{3} $ units.