# A triangle has two corners with angles of # pi / 6 # and # (3 pi )/ 8 #. If one side of the triangle has a length of #12 #, what is the largest possible area of the triangle?

Largest possible area of the triangle is 131.9005

The remaining angle:

I am assuming that length AB (12) is opposite the smallest angle.

Using the ASA

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To find the largest possible area of the triangle, use the formula for the area of a triangle:

[A = \frac{1}{2} \times b \times h]

where (b) is the base and (h) is the height. The base of the triangle is the side with length 12.

To find the height, use the formula for the height of a triangle:

[h = b \times \sin(\theta)]

where (\theta) is the angle opposite the base.

Given the angles (\pi/6) and (3\pi/8), the angle opposite the base is (\pi/6) or (3\pi/8).

So, calculate the height for both cases:

For (\theta = \pi/6): [h_1 = 12 \times \sin(\pi/6) = 12 \times \frac{1}{2} = 6]

For (\theta = 3\pi/8): [h_2 = 12 \times \sin(3\pi/8)]

Now, find the larger area between the two cases.

[A_1 = \frac{1}{2} \times 12 \times 6]

[A_2 = \frac{1}{2} \times 12 \times h_2]

Compare (A_1) and (A_2) to find the larger area.

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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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