Question 885621
{{{cot(90^o-A)=tan(A)}}} so you have to find {{{tan(A)}}} , and {{{tan(A)=sin(A)/cos(A)}}} .
You know that {{{cos^2(A)+sin^2(A)=1}}} so you can find {{{sin(A)}}} .
{{{(15/17)^2+sin^2(A)=1}}}
{{{sin^2(A)=1-(15/17)^2}}}
{{{sin^2(A)=1-15^2/17^2}}}
{{{sin^2(A)=17^2/17^2-15^2/17^2}}}
{{{sin^2(A)=(17^2-15^2)/17^2}}}
{{{sin^2(A)=(289-225)/17^2}}}
{{{sin^2(A)=64/17^2}}}
{{{sin(A)=sqrt(64/17^2)}}}
{{{sin(A)=sqrt(64)/17)}}}
{{{sin(A)=8/17}}}
{{{tan(A)=sin(A)/cos(A)=((8/17))/((15/17))=(8/17)*(17/15)=highlight(8/15)}}}