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

The remaining angle:

I am assuming that length AB (3) is opposite the smallest angle.

Using the ASA

To find the largest possible area of the triangle, we'll use the formula for the area of a triangle given two sides and the angle between them, which is ( A = \frac{1}{2}ab\sin(C) ), where (a) and (b) are the lengths of the two sides and (C) is the angle between them.

Given that one side of the triangle has a length of 3, we need to find the lengths of the other two sides using trigonometry.

Using the Law of Sines, we can relate the lengths of the sides to the angles:

[ \frac{a}{\sin(A)} = \frac{b}{\sin(B)} = \frac{c}{\sin(C)} ]

Let's denote the lengths of the sides as (a), (b), and (c) respectively.

We're given that one angle is ( \frac{3\pi}{4} ) and the other is ( \frac{\pi}{6} ). Let's find the third angle first:

[ \text{Third angle} = \pi - \left(\frac{3\pi}{4} + \frac{\pi}{6}\right) ]

[ = \pi - \frac{9\pi}{12} - \frac{2\pi}{12} ]

[ = \pi - \frac{11\pi}{12} ]

Now, using the Law of Sines, we can find the lengths of the other two sides:

[ \frac{a}{\sin\left(\frac{3\pi}{4}\right)} = \frac{3}{\sin\left(\frac{\pi}{6}\right)} ]

Solve for (a):

[ a = 3\sin\left(\frac{3\pi}{4}\right) ]

Similarly, we can find (b) using the same method:

[ b = 3\sin\left(\frac{\pi}{6}\right) ]

Now, we can find the area of the triangle using the formula ( A = \frac{1}{2}ab\sin(C) ), where (C) is the third angle we found earlier.

[ A = \frac{1}{2}(3\sin\left(\frac{3\pi}{4}\right))(3\sin\left(\frac{\pi}{6}\right))\sin\left(\pi - \frac{11\pi}{12}\right) ]

[ A = \frac{1}{2}(3\sin\left(\frac{3\pi}{4}\right))(3\sin\left(\frac{\pi}{6}\right))\sin\left(\frac{\pi}{12}\right) ]

Now, we can calculate the value of (A). Once we have (A), we can determine its maximum value.