6. Find the distance from A to B for the hexagonal nut shown below: А 1.50 in BYo I've asked tutors and they have been unable to answer, after all it's only given one side and I need some help.

Let
x ------> the length side of the regular polygon
we have a regular hexagon
that means
the interior angle of this polygon is
180(6-2)/6=120 degrees
A regular hexagon can be divided into 6 congruent equilateral triangles
see the attached figure to better understand the problem
in the right triangle of the figure
we have that
sin(60)=0.75/x
solve for x
x=0.75/sin(60)
Remember that
[tex]\sin (60^o)=\frac{\sqrt[]{3}}{2}[/tex]substitute
[tex]\begin{gathered} x=0.75\colon\frac{\sqrt[]{3}}{2} \\ \\ x=\frac{1.50}{\sqrt[]{3}}\cdot\frac{\sqrt[]{3}}{\sqrt[]{3}}=\frac{1.50\sqrt[]{3}}{3}=\frac{\sqrt[]{3}}{2} \end{gathered}[/tex]Part 2
Find the distance AB
Applying the Pythagorean Theorem
AB^2=1.5^2+x^2
substitute the value of x
AB^2=2.25+(3/4)
AB^2=3
[tex]AB=\sqrt[]{3}\text{ in}[/tex]