Two corners of a triangle have angles of #pi / 3 # and # pi / 6 #. If one side of the triangle has a length of #4 #, what is the longest possible perimeter of the triangle?

Question

Two corners of a triangle have angles of #pi  /  3  # and #  pi   /  6  #.  If one side of the triangle has a length of #4  #, what is the longest possible perimeter of the triangle?

Violet Aldrich · Accepted Answer

The maximum perimeter is #P=12+4sqrt(3)#

As the sum of the internal angles of a triangle is always #pi#, if two angles are #pi/3# and #pi/6# the third angle equals:

#pi-pi/6-pi/3 = pi/2#

So this is a right triangle and if #H# is the length of the hypotenuse, the two legs are:

#A=Hsin(pi/6)=H/2# #B = Hsin(pi/3)=Hsqrt(3)/2#

The perimeter is maximum if the side length we have is the shortest of the three, and as evidenty #A < B < H# then:

#A=4# #H=8# #B=4sqrt(3)#

And the maximum perimeter is:

#P=A+B+H=12+4sqrt(3)#

Caroline Aucoin · Answer

undefined To find the longest possible perimeter of the triangle, we need to maximize the length of the remaining side, which is opposite the angle of π/3.

Using the Law of Sines, we can find the lengths of the other sides:

Let x be the length of the side opposite the angle of π/3.

Using the Law of Sines:
sin(π/6) / 4 = sin(π/3) / x

Solving for x:
x = (4 * sin(π/3)) / sin(π/6)
x = (4 * (√3/2)) / (1/2)
x = 4 * √3

Now, we can find the perimeter:
Perimeter = 4 + 4 + 4√3
Perimeter = 8 + 4√3

So, the longest possible perimeter of the triangle is 8 + 4√3.