Look at the picture.
Definitions:
[tex]\sin\theta=\dfrac{y}{r}\\\\\cos\theta=\dfrac{x}{r}\\\\\tan\theta=\dfrac{y}{x}\\\\\cot\theta=\dfrac{x}{y}[/tex]
We have:
[tex]\sin\theta=\dfrac{5}{6}[/tex] and Quadrant II (x < 0, y > 0)
Therefore y = 5 and r = 6. Calculate x:
[tex]r=\sqrt{x^2+y^2}\to\sqrt{x^2+5^2}=6\ \ \ |^2\\\\x^2+25=36\ \ \ |-25\\\\x^2=11\to x=-\sqrt{11}[/tex]
Sunstitute to the formula of tan:
[tex]\tan\theta=\dfrac{5}{-\sqrt{11}}=-\dfrac{5}{\sqrt{11}}\cdot\dfrac{\sqrt{11}}{\sqrt{11}}=-\dfrac{5\sqrt{11}}{11}[/tex]
Answer: d.
