# Two corners of a triangle have angles of # ( pi )/ 3 # and # ( pi ) / 4 #. If one side of the triangle has a length of # 8 #, what is the longest possible perimeter of the triangle?

Longest possible perimeter = 28.726

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

Let's denote the third angle of the triangle as (\theta). The sum of angles in a triangle is (\pi) radians, so we have:

[ \frac{\pi}{3} + \frac{\pi}{4} + \theta = \pi ]

Solving for (\theta), we find:

[ \theta = \pi - \left(\frac{\pi}{3} + \frac{\pi}{4}\right) = \pi - \frac{7\pi}{12} = \frac{5\pi}{12} ]

Now, using the Law of Cosines, we can find the length of the third side of the triangle:

[ c^2 = a^2 + b^2 - 2ab \cos(\theta) ]

Substituting (a = 8), (b = 8), and (\theta = \frac{5\pi}{12}), we find (c \approx 9.89).

Therefore, the longest possible perimeter of the triangle is (8 + 8 + 9.89 = 25.89).

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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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