# Two corners of a triangle have angles of # (2 pi )/ 3 # and # ( pi ) / 6 #. If one side of the triangle has a length of # 16 #, what is the longest possible perimeter of the triangle?

Longest possible perimeter of the triangle is

Given angles

It’s an isosceles triangle with sides b & c equal.

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To find the longest possible perimeter of the triangle, you need to maximize the length of the third side.

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 to angles ( A ), ( B ), and ( C ) respectively.

Given: ( A = \frac{2\pi}{3} ) and ( B = \frac{\pi}{6} )

Let's find the third angle ( C ):

( C = \pi - A - B ) ( C = \pi - \frac{2\pi}{3} - \frac{\pi}{6} ) ( C = \frac{\pi}{6} )

Now, let's find the length of the third side ( c ):

[ \frac{16}{\sin(\frac{\pi}{6})} = \frac{c}{\sin(\frac{2\pi}{3})} ]

Solving for ( c ):

[ c = 16\frac{\sin(\frac{2\pi}{3})}{\sin(\frac{\pi}{6})} ]

[ c = 16\frac{\sqrt{3}}{\frac{1}{2}} = 32\sqrt{3} ]

So, the longest possible perimeter of the triangle is:

[ P = 16 + 16 + 32\sqrt{3} = 32 + 32\sqrt{3} ]

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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