Question 824914
All angles in quadrant 1 are acute angles, and all acute angles are in quadrant 1.
In the right triangle below, the acute angle {{{6theta}}} has
{{{cos(6theta)}}}= (adjacent side)/(hypotenuse) ={{{4/5}}}
 
{{{drawing(300,200,-1,5,-0.5,3.5,rectangle(3.8,0,4,0.2),
triangle(0,0,4,0,4,3),locate(0.5,0.35,6theta),
arc(0,0,1,1,-37,0),
locate(1.8,0,4),locate(4.1,1.8,x),locate(1.8,1.8,5)
)}}}
 
According to the Pythagorean theorem,
{{{4^2+x^2=5^2}}}
{{{16+x^2=25}}}
{{{x^2=25-16}}}
{{{x^2=9}}}
{{{x=3}}}
So {{{tan(6theta)}}}= (opposite side)(adjacent side) ={{{highlight(3/4)}}}