Two corners of a triangle have angles of # (2 pi )/ 3 # and # ( pi ) / 6 #. If one side of the triangle has a length of # 1 #, what is the longest possible perimeter of the triangle?

Question

Two corners of a triangle have angles of #  (2  pi  )/  3  # and #  ( pi ) /  6  #. If one side of the triangle has a length of # 1  #, what is the longest possible perimeter of the triangle?

Savannah Aguilar · Accepted Answer

Perimeter of isosceles triangle #color(green)(P = a + 2b = 4.464#

#hatA = (2pi)/3, hatB = pi/6, side = 1#

To find the longest possible perimeter of the triangle.

Third angle #hatC = pi - (2pi)/3 - pi/6 = pi/6#

It’s an isosceles triangle with
#hat B = hat C = pi/6#

Least angle #pi/6# should correspond to the side 1 to get the longest perimeter.

Applying sine law, #a / sin A = c / sin C#

#a = (1 * sin ((2pi)/3)) / sin (pi/6) = sqrt3 = 1.732#

Perimeter of isosceles triangle #color(green)(P = a + 2b = 1 + (2 * 1.732) = 4.464#

Jayden Jackson · Answer

undefined To find the longest possible perimeter of the triangle, we need to maximize the length of the third side. Since the sum of the angles in a triangle is always $ \pi $ radians (or 180 degrees), we can find the measure of the third angle by subtracting the measures of the given angles from $ \pi $.

$ \text{Third angle} = \pi - \left(\frac{2\pi}{3} + \frac{\pi}{6}\right) = \pi - \frac{5\pi}{6} = \frac{\pi}{6} $

Now, we have two angles measuring $ \frac{\pi}{6} $ radians each and one side of length 1. To maximize the perimeter, we need to make the third side as long as possible. Since the third side connects the two angles of $ \frac{\pi}{6} $ radians each, we can maximize its length by forming an equilateral triangle.

In an equilateral triangle, all sides are equal. Therefore, the third side, which is common to both angles of $ \frac{\pi}{6} $ radians each, will have the same length as the given side, which is 1.

So, the perimeter of the triangle is $ 1 + 1 + 1 = 3 $. Therefore, the longest possible perimeter of the triangle is 3 units.