# A triangle has two corners with angles of # pi / 12 # and # (7 pi )/ 8 #. 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 = 8.0031

The remaining angle:

I am assuming that length AB (6) 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, which is ( \frac{1}{2} \times \text{base} \times \text{height} ).

First, find the height of the triangle using trigonometry. Let the angle ( \theta ) be the angle opposite the side of length 6.

Using the sine function: [ \sin(\theta) = \frac{\text{opposite side}}{\text{hypotenuse}} = \frac{h}{6} ] [ h = 6 \sin(\theta) ]

Given ( \theta = \frac{\pi}{12} ) or ( \theta = \frac{7\pi}{8} ), calculate ( h ) for each case.

Once you have ( h ), use the area formula ( A = \frac{1}{2} \times \text{base} \times \text{height} ) with the known base length of 6 and the calculated height to find the areas for both cases.

The largest area will be the maximum value obtained from these calculations.

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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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