Two corners of a triangle have angles of #pi / 4 # and # pi / 3 #. 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 #pi  /  4  # and #  pi   /  3  #.  If one side of the triangle has a length of #6  #, what is the longest possible perimeter of the triangle?

Noah Atkinson · Accepted Answer

Longest possible perimeter of the triangle is 21.5447

Given #: /_ A = pi /4, /_B = (pi)/3# # /_C = (pi - pi /4 - (pi)/3 ) = (5pi)/12 # To get the longest perimeter, we should consider the side corresponding to the angle that is the smallest. #a / sin A = b / sin B = c / sin C# #6 / sin (pi/4) = b / sin ((5pi)/12) = c / sin ((pi)/3)# #:. b = (6 * sin ((5pi)/12)) / sin (pi/4) = 8.1962# #c = (6 * sin (pi/3)) / sin (pi/4) = 7.3485# Longest possible perimeter #P = 6 + 8.1962 + 7.3485 = 21.5447#

James Atkins · Answer

undefined The longest possible perimeter of the triangle can be found by considering the triangle inequality theorem and the fact that the sum of the lengths of any two sides of a triangle must be greater than the length of the third side.

Given that one side of the triangle has a length of 6, we can find the lengths of the other two sides using trigonometric relationships.

For the angle of $ \frac{\pi}{4} $, the opposite side (let's call it $ a $) can be found using the sine function:

$$ \sin\left(\frac{\pi}{4}\right) = \frac{a}{6} $$
$$ a = 6 \sin\left(\frac{\pi}{4}\right) = 6 \times \frac{\sqrt{2}}{2} = 3\sqrt{2} $$

For the angle of $ \frac{\pi}{3} $, the adjacent side (let's call it $ b $) can be found using the cosine function:

$$ \cos\left(\frac{\pi}{3}\right) = \frac{b}{6} $$
$$ b = 6 \cos\left(\frac{\pi}{3}\right) = 6 \times \frac{1}{2} = 3 $$

Now, we can find the perimeter of the triangle:

$$ \text{Perimeter} = 6 + a + b = 6 + 3\sqrt{2} + 3 $$

Therefore, the longest possible perimeter of the triangle is $ 9 + 3\sqrt{2} $.