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We are given a right-angled triangle \(\triangle OPQ\), where:

- \( \angle P = 90^\circ \)
- \( OP = 7 \, \text{cm} \)
- \( OQ - PQ = 1 \, \text{cm} \)

We are asked to determine the values of \( \sin Q \) and \( \cos Q \).

### Step 1: Define the relationship
We know that:
\[
OQ = PQ + 1 \, \text{cm}
\]
Let \( PQ = x \). Therefore, \( OQ = x + 1 \).

### Step 2: Apply Pythagoras' Theorem
For the right-angled triangle \( \triangle OPQ \), Pythagoras' theorem states:
\[
OQ^2 = OP^2 + PQ^2
\]
Substitute the known values:
\[
(x + 1)^2 = 7^2 + x^2
\]
Expanding both sides:
\[
x^2 + 2x + 1 = 49 + x^2
\]
Simplify the equation:
\[
2x + 1 = 49
\]
\[
2x = 48
\]
\[
x = 24
\]
So, \( PQ = 24 \, \text{cm} \) and \( OQ = 24 + 1 = 25 \, \text{cm} \).

### Step 3: Calculate \( \sin Q \) and \( \cos Q \)
- \( \sin Q = \frac{OP}{OQ} = \frac{7}{25} \)
- \( \cos Q = \frac{PQ}{OQ} = \frac{24}{25} \)

### Final Answer:
- \( \sin Q = \frac{7}{25} \)
- \( \cos Q = \frac{24}{25} \)

Would you like further details or have any questions?

Here are 5 related questions to expand your understanding:
1. How can we find \( \tan Q \) in this triangle?
2. What is the length of the hypotenuse in any right-angled triangle given two sides?
3. How would you find the angles if only the side lengths are known?
4. What are the other trigonometric identities for this triangle?
5. How can the Pythagorean theorem be applied in three dimensions?

**Tip:** When working with trigonometry, always check which angle you are focusing on to avoid mixing up opposite and adjacent sides.

In triangle OPQ, right-angled at P, OP = 7 cm and OQ - PQ = 1 cm. Determine the values of sin Q and cos Q.

In this problem, we calculate the values of sin Q and cos Q in a right-angled triangle using Pythagoras' theorem. Given OP = 7 cm and OQ - PQ = 1 cm, we find PQ = 24 cm and OQ = 25 cm, leading to sin Q = 7/25 and cos Q = 24/25. Suitable for students in Grades 9-10.

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