The question you shared involves solving for the angle $\theta$ when given:

$\cos \theta = -\frac{1}{2}, \quad \tan \theta > 0$

Step 1: Recognize the conditions

• The cosine value is negative, meaning $\theta$ lies either in the second or third quadrant (because cosine is negative there).
• The tangent value is positive, which occurs in the second quadrant (since both sine and cosine are negative, but sine is more negative than cosine in the third quadrant).

Step 2: Reference Angle

To find the reference angle, first consider the angle where $\cos \theta = \frac{1}{2}$, which is $60^\circ$ or $\frac{\pi}{3}$ radians. Since we are in the second quadrant (based on the conditions above), the reference angle is:

$180^\circ - 60^\circ = 120^\circ$

Step 3: Least positive value of $\theta$

The least positive value of $\theta$ in the second quadrant where $\cos \theta = -\frac{1}{2}$ and $\tan \theta > 0$ is $120^\circ$.

Conclusion:

• The reference angle is $60^\circ$.
• The least positive value of $\theta$ is $120^\circ$.

Would you like further details or help with other parts of the question?

Here are some related questions for practice:

1. How do you find the reference angle for cosine in different quadrants?
2. What are the key differences between cosine and sine in terms of their signs in various quadrants?
3. How do you use the unit circle to determine angle values for trigonometric functions?
4. What does it mean when a trigonometric function is positive or negative in different quadrants?
5. Can you solve for angles when given other trigonometric functions like sine or tangent?

Tip: Remember, the reference angle always gives you the acute angle in the right triangle that corresponds to your given angle. Use it to determine other function values!