Given that $\cos(\theta) = -\frac{7}{25}$ and $\theta$ is in Quadrant II, we need to find the exact values of $\sin(\theta)$ and $\tan(\theta)$.

Part 1: Finding $\sin(\theta)$

We can use the Pythagorean identity:

$\sin^2(\theta) + \cos^2(\theta) = 1$

Substituting the value of $\cos(\theta)$:

$\sin^2(\theta) + \left(-\frac{7}{25}\right)^2 = 1$

$\sin^2(\theta) + \frac{49}{625} = 1$

Subtract $\frac{49}{625}$ from both sides:

$\sin^2(\theta) = 1 - \frac{49}{625}$

To subtract the fractions, write $1$ as $\frac{625}{625}$:

$\sin^2(\theta) = \frac{625}{625} - \frac{49}{625} = \frac{576}{625}$

Taking the square root of both sides:

$\sin(\theta) = \pm \sqrt{\frac{576}{625}} = \pm \frac{24}{25}$

Since $\theta$ is in Quadrant II, where sine is positive:

$\sin(\theta) = \frac{24}{25}$

Part 2: Finding $\tan(\theta)$

The tangent function is defined as:

$\tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)}$

Substitute the values we have:

$\tan(\theta) = \frac{\frac{24}{25}}{-\frac{7}{25}} = \frac{24}{25} \times \frac{25}{-7} = \frac{24}{-7} = -\frac{24}{7}$

Summary of Results:

• $\sin(\theta) = \frac{24}{25}$
• $\tan(\theta) = -\frac{24}{7}$

Would you like more details on any part of this solution?

Here are some related questions to consider:

1. How do you determine the sign of the sine function in different quadrants?
2. What are the key identities used in solving trigonometric equations?
3. How do you convert between radians and degrees?
4. What is the significance of the Pythagorean identity?
5. How can the tangent function be interpreted geometrically?

Tip: Always check the quadrant of the angle to determine the correct sign of trigonometric functions.