Question 1042536
Use,
{{{cos^2(theta)+sin^2(theta)=1}}}
{{{cos^2(theta)+(2/7)^2=1}}}
{{{cos^2(theta)=1-4/49}}}
{{{cos^2(theta)=45/49}}}
{{{cos(theta)=0 +- (3/7)sqrt(5)}}}
Since you know that {{{tan(theta)=sin(theta)/cos(theta)>0}}} and {{{sin(theta)>0}}}, then {{{cos(theta)>0}}}
So,
{{{cos(theta)=(3/7)sqrt(5)}}}