# A triangle has two corners with angles of # (2 pi ) / 3 # and # ( pi )/ 6 #. If one side of the triangle has a length of #6 #, what is the largest possible area of the triangle?

Largest possible area of the triangle is 15.5885

The remaining angle:

I am assuming that length AB (6) is opposite the smallest angle.

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To find the largest possible area of the triangle, we can use the formula for the area of a triangle, which is 0.5 * base * height. We know the length of one side is 6 units. The height of the triangle can be found using trigonometry since we know the angles.

For the angle ( \frac{\pi}{3} ), the corresponding height forms a 30-60-90 triangle with the base, so the height is ( \frac{\sqrt{3}}{2} \times 6 = 3\sqrt{3} ) units.

For the angle ( \frac{\pi}{6} ), the height forms a 45-45-90 triangle with the base, so the height is ( \frac{\sqrt{2}}{2} \times 6 = 3\sqrt{2} ) units.

Now, we can calculate the areas of the triangles formed by these heights and the given base:

For the triangle with the height ( 3\sqrt{3} ), the area is ( 0.5 \times 6 \times 3\sqrt{3} = 9\sqrt{3} ) square units.

For the triangle with the height ( 3\sqrt{2} ), the area is ( 0.5 \times 6 \times 3\sqrt{2} = 9\sqrt{2} ) square units.

Comparing the areas, we see that ( 9\sqrt{3} ) is larger than ( 9\sqrt{2} ), so the largest possible area of the triangle is ( 9\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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