# A triangle has two corners with angles of # pi / 6 # and # (3 pi )/ 8 #. If one side of the triangle has a length of #4 #, what is the largest possible area of the triangle?

Largest possible area of the triangle is 14.6556

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, we first need to determine the length of the third side. To do this, we can use the Law of Sines, which states that for any triangle with angles A, B, and C and sides a, b, and c opposite their respective angles:

a / sin(A) = b / sin(B) = c / sin(C)

Given that two angles of the triangle are π/6 and (3π)/8, we can find the third angle by subtracting the sum of the known angles from π:

Third angle = π - (π/6 + (3π)/8)

Next, we can use the Law of Sines to find the length of the third side. Let's denote the third side as x:

4 / sin(π/6) = x / sin(Third angle)

Now, solve for x.

Once we have the lengths of all three sides of the triangle, we can use Heron's formula to calculate its area:

Area = √(s(s - a)(s - b)(s - c))

Where s is the semi-perimeter of the triangle, calculated as:

s = (a + b + c) / 2

After finding the area, we can then maximize it by adjusting the length of the third side accordingly.

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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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