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?

2 Answers
Mar 6, 2017

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

Explanation:

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#

Mar 7, 2017

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

Explanation:

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.

enter image source here

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#