Question 1113581
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Let  t = sin(A).  Then


t + 1 = {{{2*sqrt(1-t^2)}}}.


Square both sides:


t^2 + 2t + 1 = 4*(1-t^2)


t^2 + 2t + 1 = 4 - 4t^2


5t^2 + 2t - 3 = 0


{{{t[1,2]}}} = {{{(-2 +- sqrt(2^2 -4*5*(-3)))/(2*5)}}} = {{{(-2 +- sqrt(64))/10}}} = {{{(-2 +- 8)/10}}}


{{{t[1]}}} = {{{(-2 + 8)/10}}} = {{{6/10}}} = {{{3/5}}} = 0.6.


{{{t[2]}}} = {{{(-2 - 8)/10}}} = -1.


Since the angle A is in QI, the only value for sin(A) is 0.6.


<U>Answer</U>.  At given conditions, sin(A) = 0.6.
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Solved.