In the right triangle $ABC$, we need to find the sine of angle $A$.

In a right triangle, the sine of an angle is defined as:

$\sin(\theta) = \frac{\text{opposite side}}{\text{hypotenuse}}$

Here, for $\angle A$:

• The side opposite to $\angle A$ is $6$ (side $AC$).
• The hypotenuse is $7$ (side $AB$).

So, the sine of $\angle A$ is:

$\sin(A) = \frac{6}{7}$

Thus, the correct answer is:

$\boxed{\frac{6}{7}}$

Would you like more details or have any questions?

Here are 5 related questions to expand on this concept:

1. How do you find the cosine of $\angle A$ in a right triangle?
2. What is the tangent of $\angle A$ in this triangle?
3. How can you determine $\sin(B)$ from the given information?
4. How does the Pythagorean theorem apply to this triangle?
5. What is the relationship between sine and cosine for complementary angles?

Tip: Remember that in a right triangle, the sine and cosine of complementary angles (like $\angle A$ and $\angle B$) are related: $\sin(A) = \cos(B)$ and vice versa.