Two corners of a triangle have angles of #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 #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?

Aurora Ayala · Accepted Answer

Longest possible perimeter of the triangle is 31.0412

Given are the two angles #(pi)/6# and #(pi)/8# and the length 1 The remaining angle: #= pi - (((pi)/6) + (p)/8) = (17pi)/24# I am assuming that length AB (7) is opposite the smallest angle #a / sin A = b / sin B = c / sin C# #7 / sin ((pi)/6) = b / sin ((pi) /8) = c / ((17pi) / 24)# #b = (7*sin((3pi)/8)) / sin ((pi) /6) = 12.9343# #c = (7*sin ((17pi)/24)) / sin ((pi)/6) = 11.1069# Longest possible perimeter of the triangle is =# (a+b+c) = (7+12.9343+11.1069) = 31.0412#

Maverick Smith · Answer

undefined To find the longest possible perimeter of the triangle given that two angles are $ \frac{\pi}{8} $ and $ \frac{\pi}{6} $, we first need to find the third angle.

Using the fact that the sum of angles in a triangle is $ \pi $ radians (180 degrees), we can find the third angle:

$$ \text{Third angle} = \pi - \left( \frac{\pi}{8} + \frac{\pi}{6} \right) = \pi - \frac{7\pi}{24} = \frac{17\pi}{24} $$

Now, we have all three angles of the triangle. To maximize the perimeter, we need to maximize the lengths of the sides opposite to the given angles.

Let's call the side opposite the angle $ \frac{\pi}{8} $ as $ a $, and the side opposite the angle $ \frac{\pi}{6} $ as $ b $. Then, the side opposite the angle $ \frac{17\pi}{24} $ would be $ c $.

Using the law of sines:

$$ \frac{a}{\sin(\frac{\pi}{8})} = \frac{7}{\sin(\frac{\pi}{6})} $$

$$ a = \frac{7\sin(\frac{\pi}{8})}{\sin(\frac{\pi}{6})} $$

Similarly,

$$ b = \frac{7\sin(\frac{\pi}{6})}{\sin(\frac{\pi}{8})} $$

Now, we can find the length of side $ c $ using the Law of Cosines:

$$ c^2 = a^2 + b^2 - 2ab\cos\left(\frac{17\pi}{24}\right) $$

$$ c = \sqrt{a^2 + b^2 - 2ab\cos\left(\frac{17\pi}{24}\right)} $$

After finding the lengths of all sides, we can calculate the perimeter $ P $ of the triangle:

$$ P = a + b + c $$

This will give us the longest possible perimeter of the triangle.