Two corners of a triangle have angles of # (2 pi )/ 3 # and # ( pi ) / 6 #. 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 #  (2  pi  )/  3  # and #  ( pi ) /  6  #. If one side of the triangle has a length of # 8  #, what is the longest possible perimeter of the triangle?

Violet Aldrich · Accepted Answer

The longest perimeter is #P ~~ 29.856#

Let #angle A = pi/6# Let #angle B = (2pi)/3#

Then #angle C = pi - A - B#

#C = pi - pi/6 - (2pi)/3#

#C = pi/6#

Because the triangle has two equal angles, it is isosceles. Associate the given length, 8, with the smallest angle. By coincidence, this is both side "a" and side "c". because this will give us the longest perimeter.

#a = c = 8#

Use the Law of Cosines to find the length of side "b":

#b = sqrt(a^2 + c^2 - 2(a)(c)cos(B))#

#b = 8sqrt(2(1 - cos(B)))#

#b = 8sqrt(2(1 - cos((2pi)/3)))#

#b = 8sqrt(3)#

The perimeter is:

#P = a + b + c#

#P = 8 + 8sqrt(3) + 8#

#P ~~ 29.856#

Sophia Alfred · Answer

undefined The longest possible perimeter of the triangle can be found using the Law of Sines. Given the angles $ \frac{2\pi}{3} $ and $ \frac{\pi}{6} $, the remaining angle is $ \pi - \left(\frac{2\pi}{3} + \frac{\pi}{6}\right) = \frac{\pi}{2} $.

Using the Law of Sines:

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

where $a$, $b$, and $c$ are the lengths of the sides opposite angles $A$, $B$, and $C$ respectively.

Let $x$ be the length of the side opposite the angle $ \frac{2\pi}{3} $. Then:

$$
\frac{x}{\sin\left(\frac{2\pi}{3}\right)} = \frac{8}{\sin\left(\frac{\pi}{6}\right)}
$$

Solving for $x$:

$$
x = 8 \times \frac{\sin\left(\frac{2\pi}{3}\right)}{\sin\left(\frac{\pi}{6}\right)} = 8 \times \frac{\sqrt{3}}{\frac{1}{2}} = 16\sqrt{3}
$$

Thus, the longest side has a length of $16\sqrt{3}$. The perimeter of the triangle is then $8 + 8 + 16\sqrt{3} = 16 + 16\sqrt{3}$.