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?

Answer 1

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`

Answer 2

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.

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`