Answer:
[tex]\huge\boxed{\sin\theta=\dfrac{\sqrt{35}}{6},\ \tan\theta=\sqrt{35}}\\\\or\\\\\large\boxed{\sin\theta=-\dfrac{\sqrt{35}}{6},\ \tan\theta=-\sqrt{35}}[/tex]
Step-by-step explanation:
[tex]\text{Use}\ \sin^2\theta+\cos^2\theta=1\\\\\cos\theta=\dfrac{1}{6},\ \text{substitute:}\\\\\sin^2\theta+\left(\dfrac{1}{6}\right)^2=1\\\\\sin^2\theta+\dfrac{1}{36}=1\qquad\text{subtract}\ \dfrac{1}{36}\ \text{from both sides}\\\\\sin^2\theta+\dfrac{1}{36}-\dfrac{1}{36}=\dfrac{36}{36}-\dfrac{1}{36}\\\\\sin^2\theta=\dfrac{36-1}{36}\\\\\sin^2\theta=\dfrac{35}{36}\Rightarrow\sin\theta=\pm\sqrt{\dfrac{35}{36}}\\\\\sin\theta=\pm\dfrac{\sqrt{35}}{\sqrt{36}}\\\\\sin\theta=\pm\dfrac{\sqrt{35}}{6}[/tex]
[tex]\text{If}\ \theta\ \text{is in Quadrant I}\ (0^o,\ 90^o),\ \text{then the value of sine is positive}.\\\\\bold{\sin\theta=\dfrac{\sqrt{35}}{6}}[/tex]
[tex]\text{For}\ \tan\theta\ \text{use}\ \tan\theta=\dfrac{\sin\theta}{\cos\theta}.\\\\\text{Substitute:}\\\\\tan\theta=\dfrac{\frac{\sqrt{35}}{6}}{\frac{1}{6}}=\dfrac{\sqrt{35}}{6}\cdot\dfrac{6}{1}=\bold{\sqrt{35}}[/tex]