Question

Please help me right away.

Answer 1

Look at the picture.

Definitions:

$\sin\theta=\dfrac{y}{r}\\\\\cos\theta=\dfrac{x}{r}\\\\\tan\theta=\dfrac{y}{x}\\\\\cot\theta=\dfrac{x}{y}$

We have:

$\sin\theta=\dfrac{5}{6}$ and Quadrant II (x < 0, y > 0)

Therefore y = 5 and r = 6. Calculate x:

$r=\sqrt{x^2+y^2}\to\sqrt{x^2+5^2}=6\ \ \ |^2\\\\x^2+25=36\ \ \ |-25\\\\x^2=11\to x=-\sqrt{11}$

Sunstitute to the formula of tan:

$\tan\theta=\dfrac{5}{-\sqrt{11}}=-\dfrac{5}{\sqrt{11}}\cdot\dfrac{\sqrt{11}}{\sqrt{11}}=-\dfrac{5\sqrt{11}}{11}$

Answer: d.