Two corners of a triangle have angles of # (5 pi )/ 8 # and # ( pi ) / 6 #. 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 ) /  6  #. If one side of the triangle has a length of # 7  #, what is the longest possible perimeter of the triangle?

Ella Armstrong · Accepted Answer

Largest possible area of the triangle is 27.5587

Given are the two angles #(5pi)/8# and #pi/6# and the length 7 The remaining angle: #= pi - (((5pi)/8) + pi/6) = (5pi)/24# I am assuming that length AB (1) is opposite the smallest angle. Using the ASA Area#=(c^2*sin(A)*sin(B))/(2*sin(C)# Area#=( 7^2*sin((5pi)/24)*sin((5pi)/8))/(2*sin(pi/6))# Area#=27.5587#

Parker Allen · Answer

undefined To find the longest possible perimeter of the triangle given that two corners have angles of $ \frac{5\pi}{8} $ and $ \frac{\pi}{6} $, and one side has a length of 7, we can use the law of sines.

First, let's denote the angle between the known side (of length 7) and the side we're looking for as $ \theta $. Then, we can use the law of sines to find the relationship between the sides and angles:

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

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

Given that one side has a length of 7, we have:

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

Solving for $ b $:

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

Now, we want to maximize the perimeter, which is $ P = 7 + b + c $. Since $ b $ is already expressed in terms of $ \theta $, we need to find $ c $ in terms of $ \theta $. We know that the sum of angles in a triangle is $ \pi $, so $ C = \pi - \theta - \frac{5\pi}{8} $.

Now, using the law of sines again:

$$ \frac{c}{\sin(C)} = \frac{7}{\sin\left(\frac{\pi}{6}\right)} $$

Solving for $ c $:

$$ c = \frac{7 \cdot \sin\left(\frac{\pi}{6}\right)}{\sin\left(\pi - \theta - \frac{5\pi}{8}\right)} $$

Now, substitute $ b $ and $ c $ back into the perimeter formula $ P = 7 + b + c $ to get the perimeter in terms of $ \theta $. Then, find the value of $ \theta $ that maximizes this expression to get the longest possible perimeter of the triangle.