Two corners of a triangle have angles of #(7 pi ) / 12 # and # (3 pi ) / 8 #. If one side of the triangle has a length of #8 #, what is the longest possible perimeter of the triangle?

Question

Two corners of a triangle have angles of #(7 pi ) /  12  # and # (3 pi  ) /  8  #.  If one side of the triangle has a length of #8  #, what is the longest possible perimeter of the triangle?

Claire Akin · Accepted Answer

Largest possible area of the triangle is 218.7819

Given are the two angles #(7pi)/12# and #(3pi)/8# and the length 8 The remaining angle: #= pi - (((7pi)/12) + (3pi)/8) = pi/24# I am assuming that length AB (8) is opposite the smallest angle. Using the ASA Area#=(c^2*sin(A)*sin(B))/(2*sin(C)# Area#=( 8^2*sin((3pi)/8)*sin((7pi)/12))/(2*sin(pi/24))# Area#=218.7819#

Sadie Allen · Answer

undefined To find the longest possible perimeter of the triangle, we need to determine the length of the remaining side and then calculate the perimeter.

Given that one side of the triangle has a length of 8, we can use the law of sines to find the lengths of the other sides.

The law of sines states:
$$ \frac{\sin A}{a} = \frac{\sin B}{b} = \frac{\sin C}{c} $$

Let's denote the angles of the triangle as A, B, and C, and the corresponding sides as a, b, and c.

We know:
Angle A = $ \frac{7\pi}{12} $
Angle B = $ \frac{3\pi}{8} $
Angle C = $ \pi - \left(\frac{7\pi}{12} + \frac{3\pi}{8}\right) $ (since the sum of angles in a triangle is $ \pi $)

First, we find the angle C:
$$ \text{Angle C} = \pi - \left(\frac{7\pi}{12} + \frac{3\pi}{8}\right) = \pi - \left(\frac{14\pi + 9\pi}{24}\right) = \frac{24\pi - 23\pi}{24} = \frac{\pi}{24} $$

Now, using the law of sines:
$$ \frac{\sin A}{8} = \frac{\sin B}{b} = \frac{\sin C}{c} $$

We can solve for the lengths of sides b and c using the known angles and the side length 8.

$$ b = \frac{8 \cdot \sin B}{\sin A} $$
$$ c = \frac{8 \cdot \sin C}{\sin A} $$

Plug in the values and calculate.

Then, the perimeter of the triangle is:
$$ \text{Perimeter} = 8 + b + c $$