In OPQ, right-angled at P, OP = 7 cm and OQ – PQ = 1 cm (see Fig. 8.12). Determine the values of sin Q and cos Q.
Step-by-Step Solution
Key Concept: Use Pythagoras theorem to find the unknown sides of the right‑angled triangle and then apply the definitions of sine and cosine for the acute angle Q: \(\sin Q = \frac{\text{opposite side}}{\text{hypotenuse}}\) and \(\cos Q = \frac{\text{adjacent side}}{\text{hypotenuse}}\).
1. Let \(PQ = x\) cm. Then \(OQ = x + 1\) cm (given).\
2. Since the triangle is right‑angled at P, apply Pythagoras theorem:
$$OP^{2} + PQ^{2} = OQ^{2}$$
Substituting the known values:
$$7^{2} + x^{2} = (x+1)^{2}$$
3. Expand and simplify:
$$49 + x^{2} = x^{2} + 2x + 1$$
$$49 = 2x + 1$$
$$2x = 48$$
$$x = 24$$
Hence, \(PQ = 24\) cm and \(OQ = 24 + 1 = 25\) cm.
4. Identify the sides with respect to angle \(Q\):
- Opposite side to \(Q\) = \(OP = 7\) cm.
- Adjacent side to \(Q\) = \(PQ = 24\) cm.
- Hypotenuse = \(OQ = 25\) cm.
5. Use the definitions:
$$\sin Q = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{7}{25}$$
$$\cos Q = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{24}{25}$$
Correct Answer: \(\sin Q = \frac{7}{25},\; \cos Q = \frac{24}{25}\)