Question 1088317
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{{{A+B+C=2pi}}}
{{{pi/6+B+pi/2=pi}}}
{{{B=pi(1-1/6-1/2)}}}
{{{B=pi(6/6-1/6-3/6)}}}
{{{B=pi(2/6)}}}
{{{B=pi/3}}}
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Using the law of sines,
{{{sin(A)/a=sin(B)/b=sin(C)/c}}}
So then,
{{{sin(pi/2)/12=sin(pi/6)/a}}}
Solve for a.
{{{sin(pi/2)/12=sin(pi/3)/b}}}
Solve for b.
Or use the Pythagorean theorem,
{{{c^2=a^2+b^2}}}
{{{b^2=144-a^2}}}
{{{b=sqrt(144-a^2)}}}