Answer:
[tex]\large\boxed{Q1.\ \sin\theta=\dfrac{40}{41}}\\\boxed{Q2.\ 30^o}[/tex]
Step-by-step explanation:
Q1.
[tex]\sin\theta=\dfrac{y}{r}[/tex]
[tex]\text{We have}\\\\r=1\to(\text{radius of the unit circle}),\\\\y=\dfrac{40}{41}\to\text{from the given point}\ \left(\dfrac{9}{41},\ \dfrac{40}{41}\right)\\\\\text{Substitute:}\\\\\sin\theta=\dfrac{\frac{40}{41}}{1}=\dfrac{40}{41}[/tex]
Q2.
look at the picture
210° is in III Quadrant. Therefore the reference angle for θ is:
210° - 180° = 30°
