Question 618255
We can use {{{sin^2(theta) = 1 - cos^2(theta)}}} and the fact that {{{theta}}} terminates in the 4th quadrant to find {{{sin(theta)}}}.<br>
{{{sin^2(theta) = 1 - cos^2(theta)}}}
Inserting the given value for {{{cos(theta)}}}:
{{{sin^2(theta) = 1 - (5/7)^2}}}
Simplifying...
{{{sin^2(theta) = 1 - 25/49}}}
{{{sin^2(theta) = 49/49 - 25/49}}}
{{{sin^2(theta) = 24/49}}}
So {{{sin(theta)}}} is one of the square roots of 24/49. Since {{{theta}}} terminates in the 4th quadrant and since sin is negative in the 4th quadrant, we know to use the negative square root:
{{{sin(theta) = -sqrt(24/49)}}}
which simplifies:
{{{sin(theta) = -sqrt(24)/sqrt(49)}}}
{{{sin(theta) = -sqrt(4*6)/7}}}
{{{sin(theta) = -(sqrt(4)*sqrt(6))/7}}}
{{{sin(theta) = -(2*sqrt(6))/7}}}<br>
I'll leave the drawing up to you.