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

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To find the longest possible perimeter of the triangle, we can use the Law of Sines.

Let (a) be the side opposite the angle (\frac{\pi}{3}), and (b) be the side opposite the angle (\frac{\pi}{2}). Since we know the length of one side ((b)), we need to find (a) and (c), the side opposite the third angle.

Using the Law of Sines, we have:

[\frac{a}{\sin\left(\frac{\pi}{3}\right)} = \frac{b}{\sin\left(\frac{\pi}{2}\right)}]

Given that (b = 2), and (\sin\left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{2}) and (\sin\left(\frac{\pi}{2}\right) = 1), we can solve for (a):

[\frac{a}{\frac{\sqrt{3}}{2}} = \frac{2}{1}]

[a = \frac{2\sqrt{3}}{3}]

Now, to find (c), the side opposite the remaining angle, we can use the fact that the angles in a triangle sum up to (\pi):

[\frac{\pi}{3} + \frac{\pi}{2} + \theta = \pi]

[\theta = \pi - \frac{\pi}{3} - \frac{\pi}{2}]

[\theta = \frac{\pi}{6}]

Now, we can use the Law of Sines again:

[\frac{c}{\sin\left(\frac{\pi}{6}\right)} = \frac{2}{\sin\left(\frac{\pi}{2}\right)}]

[\frac{c}{\frac{1}{2}} = \frac{2}{1}]

[c = 4]

The perimeter of the triangle is (a + b + c), so substituting the values we found:

[P = \frac{2\sqrt{3}}{3} + 2 + 4]

[P = \frac{2\sqrt{3}}{3} + 6]

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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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