Question 1094706
<pre>{{{drawing(110,40,0,3,-1.5,1.4,locate(0,.9,"tan[sin"^(-1)),
locate(2.17,1.2,(1/7)),locate(2.9,.35,"]")  )}}}

First we look at the inner part: {{{drawing(70,40,0,2,-1,1.7,locate(0,.9,"sin"^(-1)),locate(1.4,1.3,(1/7)) )}}}

That asks for the angle between {{{-pi/2}}} and {{{"" + pi/2}}} whose sine
is {{{1/7}}}.  

Since we know that {{{matrix(1,5, sine,""="",opposite/hypotenuse,""="",y/r)}}},
we draw such an angle by placing a right triangle
in the first quadrant whose angle at the origin is such that its
opposite side is y=1 and its hypotenuse is r=7, like this:

{{{drawing(400,199600/999,-2,7.99,-2,2.99,
locate(3.4,.5,x="???"),
graph(400,199600/999,-2,7.99,-2,2.99), locate(3.1,.9,r=7),
red(arc(0,0,5,-5,0,8)),
line(0,0,sqrt(48),1),line(sqrt(48),0,sqrt(48),1),
locate(7,.7,y=1) )}}}

The angle indicated by the red arc is the angle: {{{drawing(70,40,0,2,-1,1.7,locate(0,.9,"sin"^(-1)),locate(1.4,1.3,(1/7)) )}}}.

To find the tangent, we must calculate the adjacent side x,
using the Pythagorean theorem:

{{{r^2}}}{{{""=""}}}{{{x^2+y^2}}}
{{{7^2}}}{{{""=""}}}{{{x^2+1^2}}}
{{{49}}}{{{""=""}}}{{{x^2+1}}}
{{{48}}}{{{""=""}}}{{{x^2}}}
{{{sqrt(48)}}}{{{""=""}}}{{{x}}}
{{{sqrt(16*3)}}}{{{""=""}}}{{{x}}}
{{{4sqrt(3)}}}{{{""=""}}}{{{x}}}

So we now have all three sides of the right triangle:

{{{drawing(400,199600/999,-2,7.99,-2,2.99,

graph(400,199600/999,-2,7.99,-2,2.99), locate(3.1,.9,r=7),
red(arc(0,0,5,-5,0,8)),locate(3.4,.5,x=4sqrt(3)),
line(0,0,sqrt(48),1),line(sqrt(48),0,sqrt(48),1),
locate(7,.7,y=1) )}}}

The problem is 

{{{drawing(110,40,0,3,-1.5,1.4,locate(0,.9,"tan[sin"^(-1)),
locate(2.17,1.2,(1/7)),locate(2.9,.35,"]")  )}}}

and since we know that {{{matrix(1,5, tangent,""="",opposite/adjacent,""="",y/x)}}}

{{{drawing(110,40,0,3,-1.5,1.4,locate(0,.9,"tan[sin"^(-1)),
locate(2.17,1.2,(1/7)),locate(2.9,.35,"]")  )}}}{{{""=""}}}{{{1/(4sqrt(3))}}}{{{""=""}}}{{{sqrt(3)/12}}}.

Edwin</pre>