A triangle has two corners with angles of # (2 pi ) / 3 # and # ( pi )/ 6 #. If one side of the triangle has a length of #1 #, what is the largest possible area of the triangle?

Question

A triangle has two corners with angles of #  (2 pi ) /  3  # and # ( pi  )/  6   #.  If one side of the triangle has a length of #1  #, what is the largest possible area of the triangle?

Sadie Allen · Accepted Answer

Maximum area: #sqrt(3)/4 ("sq.units") ~~0.433012702("sq.units")#

All solutions rely on recognizing that the third angle must be #pi/6# (radians) and therefore the triangle is isosceles.

There are two possible configurations to consider.
For brevity, you might want to skip to version 2 which has the larger area.

In Version 1, the side with length #1# is not one of the equal sides.
In Version 2, the side with length #1# is one of the equal sides.

Version 1
Solution 1: Determine the area using side with length 1 as the base and a calculated height.

By definition of the #tan# function:
#color(white)("XXX")h/(1/2)=tan(pi/6)#
#color(white)("XXXXXX")rarr h=tan(pi/6)/2#
The angle #pi/6# is one of the standard angles with #tan(pi/6)=1/sqrt(3)#
So
#color(white)("XXX")h=1/(2sqrt(3))#
and
#"Area"_triangle = 1/2 * "base" * "height" = 1/2 * 1 * 1/(2sqrt(3)) =1/(4sqrt(3))#

This gives an approximate area of #0.1443...#

Solution 2: Determine the length of the missing sides and use Heron's Formula
#color(white)("XXX")w/(1/2)=cot(pi/6) rarr w=1/sqrt(3)#

perimeter is #1+1/sqrt(3)+1/sqrt(3) = 1+2/sqrt(3)#
and
semiperimeter, #s=1/2+1/sqrt(3)#

By Heron's Formula
#"Area"_triangle=sqrt(s * (s-w) * (s-w) * (s-1))#
#color(white)("XXX")=(s-w)sqrt(s^2-s)#
#color(white)("XXX")=1/2sqrt(1/4+1/sqrt(3)+1/3-(1/2+1/sqrt(3)))#
#color(white)("XXX")=1/2sqrt(1/12)#
#color(white)("XXX")=1/4 * 1/sqrt(3)=1/(4sqrt(3))# (as in the first solution method)

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Version 2
Solution 1: Determine the area using one of the sides with length 1 as the base and a calculated height

By definition of the #sin# function
#color(white)("XXX")h/1=sin(pi/3) rarr h=sqrt(3)/2#

#"Area"_triangle = 1/2 * "base" * "height" = 1/2 * 1 * sqrt(3)/2=sqrt(3)/4#

Note that #sqrt(3)/4~~0.433012702#
So Version 2 gives the larger area and thus is the version being asked for.

Solution 2: Determine the length of the missing side (using Law of Sines) and apply Heron's Formula.
(highlights only)

missing side: #t=(sin((2pi)/3)/(sin(pi/6) =sqrt(3)#

semiperimeter#s = 1+sqrt(3)/2#

#"Area"_triangle= sqrt(s * (s-1) * (s-t) * (s-t)) =sqrt(3)/4#

Sophia Alfred · Answer

I got an area of (approximately) #0.4330122019# sq. units

If two of the angles are #(2pi)/3# and #pi/6# then the third angle is #pi/6# and the triangle is isosceles. If one of its sides has a length of #1# unit then its maximum area will occur when this length is one of it's shorter sides. Suppose the long side has a length of #x# units (note it will be opposite the #(2pi)/3# angle) By the Law of Sines #color(white)("XXX")x/(sin((2pi)/3))=1/(sin(pi/6))# #color(white)("XXX")rarr x= sin((2pi)/3)/(sin(pi/6) )~~1.732050808color(white)("XX")#(using a calculator) We have a triangle with sides with lengths #1, 1, and 1.2247...# The semiperimeter, #s#, is #(1+1+1.7305...)/2~~1.866025404# and Using Heron's Formula for the area of a triangle: #color(white)("XXX")"Area"_triangle =sqrt(s * (s-1) * (s-2) * (s-x))# #color(white)("XXXXXX")~~0.4330122019#

Noah Atkinson · Answer

undefined To find the largest possible area of the triangle, we can use the formula for the area of a triangle, which is given by $ \frac{1}{2} \times \text{base} \times \text{height} $. Since we know one side length is 1, let's call it the base.

The height of the triangle can be found using trigonometry. Considering the angle $ \frac{\pi}{6} $, the height opposite to this angle can be calculated as $ \sin\left(\frac{\pi}{6}\right) \times \text{base} $.

Similarly, for the angle $ \frac{2\pi}{3} $, the height opposite to this angle can be calculated as $ \sin\left(\frac{\pi}{3}\right) \times \text{base} $.

Now, to find the largest possible area, we maximize the area formula $ \frac{1}{2} \times \text{base} \times \text{height} $. We know the base length is 1, so we just need to maximize the product of the base and the heights.

The product of the two heights is $ \sin\left(\frac{\pi}{6}\right) \times \sin\left(\frac{\pi}{3}\right) $.

Substituting the values of $ \sin\left(\frac{\pi}{6}\right) $ and $ \sin\left(\frac{\pi}{3}\right) $, we get $ \frac{1}{2} \times \frac{\sqrt{3}}{2} = \frac{\sqrt{3}}{4} $.

So, the largest possible area of the triangle is $ \frac{1}{2} \times 1 \times \frac{\sqrt{3}}{4} = \frac{\sqrt{3}}{4} $.