Question 157260
{{{sin^2(A)-4sin(A)+1=0}}}
Looks like a quadratic equation in disguise.
Let's substitute,
{{{u=sin(A)}}}
Then
 {{{sin^2(A)-4sin(A)+1=0}}}
{{{u^2-4u+1=0}}}
Using the quadratic formula, solve for u.
{{{u = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}}
{{{u = (-(-4) +- sqrt( (-4)^2-4*1*1 ))/(2*1) }}}
{{{u = (4 +- sqrt( 16-4 ))/(2) }}}
{{{u = (4 +- sqrt( 12))/(2) }}}
Two solutions,
{{{u[1] = (4 + sqrt( 12))/(2) }}}
{{{u[1] = (4 + 3.46)/(2) }}}
{{{u[1] = (7.46)/(2) }}}
{{{u[1] = 3.73 }}}
Substitute {{{u=sin(A)}}}.
{{{u[1] = 3.73 }}}
{{{sin(A) = 3.73 }}}
We throw this solution out since {{{abs(sin(A))<=1}}}
{{{u[2]=(4 - sqrt( 12))/(2) }}}
{{{u[2]=(4 - 3.46)/(2) }}}
{{{u[2]=(0.54)/(2) }}}
{{{u[2]=0.27}}}
Substitute {{{u=sin(A)}}}.
{{{u[2]=0.27}}}
{{{sin(A)=0.27}}}
{{{A= sin^(-1)(0.27) }}}
{{{A=15.6}}}
and by identity,
{{{A=180-15.6=164.4}}}