# Two corners of a triangle have angles of # (5 pi )/ 8 # and # ( pi ) / 12 #. If one side of the triangle has a length of # 3 #, what is the longest possible perimeter of the triangle?

The maximum perimeter is 22.9

The maximum perimeter is achieved, when you associate the given side with the smallest angle.

Using the law of sines:

The length of side b:

The length of side c:

P = 3 + 9.2 + 10.7 = 22.9

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To find the longest possible perimeter of the triangle given the angles ( \frac{5\pi}{8} ) and ( \frac{\pi}{12} ), and one side with a length of 3 units, we first need to find the third angle of the triangle using the fact that the sum of the angles in a triangle is ( \pi ) radians.

Let's denote the third angle as ( \theta ). The sum of the angles in the triangle is ( \frac{5\pi}{8} + \frac{\pi}{12} + \theta = \pi ). Solving for ( \theta ), we get:

[ \theta = \pi - \left( \frac{5\pi}{8} + \frac{\pi}{12} \right) ]

Now, we can find the length of the other two sides using the Law of Sines, which states:

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

Where ( A, B, ) and ( C ) are the angles of the triangle, and ( a, b, ) and ( c ) are the lengths of the opposite sides.

Using the side ( 3 ) and the angles ( \frac{5\pi}{8} ) and ( \theta ), we can find the length of the second side. Then, we can use the side ( 3 ) and the angles ( \frac{\pi}{12} ) and ( \theta ) to find the length of the third side.

Once we have the lengths of all three sides, we can calculate the perimeter of the triangle, which is the sum of the lengths of all three sides.

Finally, we maximize this perimeter by maximizing the angle ( \theta ). Since ( \theta ) is an external angle of a triangle, it is maximized when ( \theta = 0 ). Therefore, the longest possible perimeter occurs when ( \theta = 0 ).

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