Let's break down the questions and solve them step by step:

Problem 3:

For each angle, identify the quadrant it terminates in and the reference angle.

a) 200°

• Quadrant: 200° lies between 180° and 270°, so it is in Quadrant III.
• Reference Angle: The reference angle is $200° - 180° = 20°$.

b) 130°

• Quadrant: 130° lies between 90° and 180°, so it is in Quadrant II.
• Reference Angle: The reference angle is $180° - 130° = 50°$.

c) 90°

• Quadrant: 90° lies directly on the positive $y$-axis. It is not in a specific quadrant.
• Reference Angle: Since it lies on the axis, the reference angle is 0°.

d) 330°

• Quadrant: 330° lies between 270° and 360°, so it is in Quadrant IV.
• Reference Angle: The reference angle is $360° - 330° = 30°$.

Problem 4:

The terminal arm passes through the point $(-3, 6)$.

a) Draw this angle in standard position:

• The terminal arm passes through $(-3, 6)$, so it is in Quadrant II.

b) Determine the exact distance from the origin to the point Q:

• The distance is calculated using the distance formula: $r = \sqrt{x^2 + y^2} = \sqrt{(-3)^2 + 6^2} = \sqrt{9 + 36} = \sqrt{45} = 3\sqrt{5}.$

c) Determine the exact values for $\sin \theta$, $\cos \theta$, and $\tan \theta$:

• Using $r = 3\sqrt{5}$:
  • $\sin \theta = \frac{y}{r} = \frac{6}{3\sqrt{5}} = \frac{2}{\sqrt{5}} = \frac{2\sqrt{5}}{5}$,
  • $\cos \theta = \frac{x}{r} = \frac{-3}{3\sqrt{5}} = \frac{-1}{\sqrt{5}} = \frac{-\sqrt{5}}{5}$,
  • $\tan \theta = \frac{y}{x} = \frac{6}{-3} = -2$.

d) Determine the value of $\theta$:

• The angle is in Quadrant II, and the reference angle $\theta_{\text{ref}}$ is calculated as: $\theta_{\text{ref}} = \arctan\left|\frac{y}{x}\right| = \arctan\left|\frac{6}{-3}\right| = \arctan(2).$ So, $\theta = 180° - \theta_{\text{ref}}$.

Problem 5:

Find points corresponding to the reference angle of $P(2, -5)$.

• The reference angle is the angle formed in Quadrant IV where $\tan \theta = \frac{-5}{2}$.
• Symmetrical points:
  • Quadrant I: $(2, 5)$,
  • Quadrant II: $(-2, 5)$,
  • Quadrant III: $(-2, -5)$.

Would you like a detailed sketch for these solutions or further explanation on any part?

Follow-up Questions:

1. How is the reference angle related to the terminal angle?
2. Can you explain why trigonometric ratios change signs in different quadrants?
3. What is the geometric meaning of a reference angle?
4. How would you compute $\theta$ if given only a tangent ratio and its quadrant?
5. Why does the distance formula use squares of coordinates?

Tip:

Always associate the reference angle with the closest axis—it simplifies solving trigonometric problems.