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

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

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Answer 1 · 2017-03-06T13:58:15

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

Explanation:

If two of the angles are $\frac{2 π}{3}$ $(2pi)/3$ and $\frac{π}{6}$ $pi/6$ then the third angle is $\frac{π}{6}$ $pi/6$ and the triangle is isosceles.

If one of its sides has a length of $1$ $1$ 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 $x$ $x$ units (note it will be opposite the $\frac{2 π}{3}$ $(2pi)/3$ angle)

By the Law of Sines
$XXX \frac{x}{sin (\frac{2 π}{3})} = \frac{1}{sin (\frac{π}{6})}$ $color(white)("XXX")x/(sin((2pi)/3))=1/(sin(pi/6))$

$XXX \to x = \frac{sin (\frac{2 π}{3})}{sin (\frac{π}{6})} \approx 1.732050808 XX$ $color(white)("XXX")rarr x= sin((2pi)/3)/(sin(pi/6) )~~1.732050808color(white)("XX")$ (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 · 2017-03-07T12:49:10

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 $s = 1+sqrt(3)/2$

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

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

2 Answers

Explanation:

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