Given that $\theta_1$ is located in Quadrant II and $\sin(\theta_1) = \dfrac{9}{41}$, we need to find $\cos(\theta_1)$.

We can use the Pythagorean identity for sine and cosine, which is:

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

Substitute $\sin(\theta_1) = \dfrac{9}{41}$ into the equation:

$\left(\dfrac{9}{41}\right)^2 + \cos^2(\theta_1) = 1$

First, square $\dfrac{9}{41}$:

$\left(\dfrac{9}{41}\right)^2 = \dfrac{81}{1681}$

Now substitute this into the identity:

$\dfrac{81}{1681} + \cos^2(\theta_1) = 1$

Subtract $\dfrac{81}{1681}$ from both sides:

$\cos^2(\theta_1) = 1 - \dfrac{81}{1681}$

To subtract, express $1$ as $\dfrac{1681}{1681}$:

$\cos^2(\theta_1) = \dfrac{1681}{1681} - \dfrac{81}{1681} = \dfrac{1600}{1681}$

Now, take the square root of both sides:

$\cos(\theta_1) = \pm \dfrac{\sqrt{1600}}{41} = \pm \dfrac{40}{41}$

Since $\theta_1$ is in Quadrant II, where cosine is negative, we choose the negative value:

$\cos(\theta_1) = -\dfrac{40}{41}$

Thus, the exact value of $\cos(\theta_1)$ is:

$\cos(\theta_1) = -\dfrac{40}{41}$

Would you like further details or have any questions?

Here are 5 related questions you can explore:

1. How do you determine the sign of trigonometric functions in each quadrant?
2. What is the value of $\tan(\theta_1)$ based on $\sin(\theta_1)$ and $\cos(\theta_1)$?
3. Can you find $\sec(\theta_1)$ using the value of $\cos(\theta_1)$?
4. How does the Pythagorean identity help in finding trigonometric values?
5. What are the values of $\sin(\theta)$, $\cos(\theta)$, and $\tan(\theta)$ in other quadrants?

Tip: Remember, in Quadrant II, sine is positive and cosine is negative!