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

Largest possible area of the triangle is 103.4256

The remaining angle:

I am assuming that length AB (1) is opposite the smallest angle.

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To find the longest possible perimeter of the triangle, we need to consider the maximum lengths for the other two sides given the angles provided.

Let's label the triangle ABC, where angle A is π/3 and angle B is π/12.

- Determine the length of the side opposite angle A (side BC) using the law of sines:

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

Given that angle A = π/3, we have:

[ \frac{\sin(\frac{\pi}{3})}{8} = \frac{\sin(\frac{\pi}{12})}{b} ]

[ b = \frac{8 \sin(\frac{\pi}{12})}{\sin(\frac{\pi}{3})} ]

- Find the length of the side opposite angle B (side AC) using the law of sines:

[ \frac{\sin B}{b} = \frac{\sin A}{a} ]

[ \frac{\sin(\frac{\pi}{12})}{8} = \frac{\sin(\frac{\pi}{3})}{c} ]

[ c = \frac{8 \sin(\frac{\pi}{3})}{\sin(\frac{\pi}{12})} ]

- Find the length of the side opposite angle C (side AB) using the law of cosines:

[ c^2 = a^2 + b^2 - 2ab\cos C ]

Given that angle C = π - (π/3 + π/12) = π - π/4 = 3π/4:

[ (8)^2 = a^2 + b^2 - 2ab\cos(\frac{3\pi}{4}) ]

[ 64 = a^2 + b^2 + \sqrt{2}ab ]

- The perimeter of the triangle is given by ( P = a + b + c ).

Substitute the values of a, b, and c into the equation for the perimeter:

[ P = 8 + \frac{8 \sin(\frac{\pi}{12})}{\sin(\frac{\pi}{3})} + \frac{8 \sin(\frac{\pi}{3})}{\sin(\frac{\pi}{12})} ]

Simplify this expression to find the longest possible perimeter of the triangle.

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