Two corners of a triangle have angles of # (5 pi )/ 8 # and # ( pi ) / 4 #. If one side of the triangle has a length of # 7 #, what is the longest possible perimeter of the triangle?

Question

Two corners of a triangle have angles of #  (5  pi  )/  8  # and #  ( pi ) /  4  #. If one side of the triangle has a length of # 7  #, what is the longest possible perimeter of the triangle?

Aurora Ayala · Accepted Answer

Perimeter #color(blue)( P = a + b + c = 36.83#

#hat A = (5pi)/8, hat B = pi/4, hat C = pi - (5pi)/8 - pi/4 = pi/8#

Least angle #hat C = pi/8# should correspond to the side 7 to get the longest perimeter.

Applying Law of Sines,

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

#a = (c sin A) / sin C = (7 * sin((5pi)/8)) / sin (pi/8) = 16.9#

#b = (7 * sin (pi/4)) / sin (pi/8) = 12.93#

Perimeter #color(blue)( P = a + b + c = 16.9 + 12.93 + 7 = 36.83#

Sadie Allen · Answer

undefined To find the longest possible perimeter of the triangle, we need to maximize the length of the third side, given that one side has a length of 7 and two angles are given.

1. Use the Triangle Sum Theorem to find the third angle.
2. Apply the Law of Sines to find the possible lengths of the third side.
3. Choose the maximum value for the perimeter.

The third angle can be found by subtracting the sum of the given angles from 180 degrees (or $ \pi $ radians):

$$ \text{Third angle} = \pi - \left(\frac{5\pi}{8} + \frac{\pi}{4}\right) = \pi - \frac{7\pi}{8} = \frac{\pi}{8} $$

Now, using the Law of Sines:

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

Substituting the known values:

$$ \frac{\sin\left(\frac{5\pi}{8}\right)}{7} = \frac{\sin\left(\frac{\pi}{4}\right)}{b} = \frac{\sin\left(\frac{\pi}{8}\right)}{c} $$

Solving for $ b $ and $ c $:

$$ b = \frac{7\sin\left(\frac{\pi}{4}\right)}{\sin\left(\frac{5\pi}{8}\right)} $$
$$ c = \frac{7\sin\left(\frac{\pi}{8}\right)}{\sin\left(\frac{\pi}{4}\right)} $$

Using the trigonometric identities $ \sin\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2} $ and $ \sin\left(\frac{\pi}{8}\right) = \frac{\sqrt{2-\sqrt{2}}}{2} $, we can calculate $ b $ and $ c $.

$$ b \approx 5.33 $$
$$ c \approx 9.9 $$

The longest possible perimeter occurs when $ b $ and $ c $ are maximized:

$$ \text{Perimeter} = 7 + b + c $$
$$ \text{Perimeter} \approx 7 + 5.33 + 9.9 $$
$$ \text{Perimeter} \approx 22.23 $$

So, the longest possible perimeter of the triangle is approximately 22.23 units.