The problem asks to find $\cos \theta$ given that $\sin \theta = \frac{7}{10}$, using the Pythagorean identity:

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

Step-by-Step Solution:

1. First, use the Pythagorean identity: $\sin^2 \theta + \cos^2 \theta = 1$ Plug in the value of $\sin \theta = \frac{7}{10}$: $\left(\frac{7}{10}\right)^2 + \cos^2 \theta = 1$

2. Square $\frac{7}{10}$: $\frac{49}{100} + \cos^2 \theta = 1$

3. Subtract $\frac{49}{100}$ from both sides: $\cos^2 \theta = 1 - \frac{49}{100} = \frac{100}{100} - \frac{49}{100} = \frac{51}{100}$

4. Take the square root of both sides to find $\cos \theta$: $\cos \theta = \pm \sqrt{\frac{51}{100}} = \pm \frac{\sqrt{51}}{10}$

Therefore, the possible values for $\cos \theta$ are:

$\cos \theta = \pm \frac{\sqrt{51}}{10}$

The final step is to choose the correct sign depending on the quadrant of $\theta$, which is not specified in this case.

Would you like further details on how to determine the sign or anything else?

Here are 5 related questions you might find helpful:

1. How do you determine the correct sign for $\cos \theta$?
2. What is the geometric interpretation of the Pythagorean identity?
3. Can you apply the same identity to find $\tan \theta$?
4. How can you simplify $\sqrt{51}$ further?
5. What happens if $\sin \theta$ is negative?

Tip: Always check the quadrant of the angle when determining the sign of trigonometric functions.