# A triangle has two corners with angles of # ( pi ) / 3 # and # ( pi )/ 6 #. If one side of the triangle has a length of #8 #, what is the largest possible area of the triangle?

Largest possible area of the triangle is 55.4256

The remaining angle:

I am assuming that length AB (8) is opposite the smallest angle.

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To find the largest possible area of the triangle given two angles ( \frac{\pi}{3} ) and ( \frac{\pi}{6} ), and one side with a length of 8, we can use the formula for the area of a triangle:

[ \text{Area} = \frac{1}{2} \times \text{base} \times \text{height} ]

Given that one side of the triangle is 8 units, let's call it ( b ). We need to find the height corresponding to this side. Since the height is perpendicular to the base, we can use trigonometric ratios to find it.

Using the angle ( \frac{\pi}{3} ), which is 60 degrees, we find that the height (opposite side) is ( h = 8 \times \sin(\frac{\pi}{3}) ).

[ \sin(\frac{\pi}{3}) = \frac{\sqrt{3}}{2} ]

[ h = 8 \times \frac{\sqrt{3}}{2} = 4\sqrt{3} ]

Now, we can calculate the area using the formula:

[ \text{Area} = \frac{1}{2} \times 8 \times 4\sqrt{3} = 16\sqrt{3} ]

Therefore, the largest possible area of the triangle is ( 16\sqrt{3} ) square units.

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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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