In PQR, right-angled at Q (see Fig. 8.20), PQ = 3 cm and PR = 6 cm. Determine QPR and PRQ.
Step-by-Step Solution
Key Concept: In a right‑angled triangle, the trigonometric ratios (sin, cos, tan) relate the lengths of the sides to the acute angles. Since the side adjacent to ∠QPR (PQ) and the hypotenuse (PR) are known, we can use the cosine ratio to find ∠QPR, and then use the fact that the two acute angles of a right triangle are complementary.
1. Identify the known sides\
- Right angle at Q ⇒ PR is the hypotenuse.\
- Given: \(PQ = 3\text{ cm}\) (adjacent to \(\angle QPR\)), \(PR = 6\text{ cm}\) (hypotenuse).\
2. Use the cosine definition for \(\angle QPR\):\
$$\cos \angle QPR = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{PQ}{PR} = \frac{3}{6} = \frac{1}{2}.$$\
3. Find the angle whose cosine is \(\frac12\):\
$$\angle QPR = \cos^{-1}\left(\frac12\right) = 60^{\circ}.$$\
4. Determine the other acute angle \(\angle PRQ\) using the complementary property of a right triangle (the two acute angles sum to \(90^{\circ}\)):\
$$\angle PRQ = 90^{\circ} - \angle QPR = 90^{\circ} - 60^{\circ} = 30^{\circ}.$$\
5. (Optional) Find the remaining side QR for verification using the sine ratio:\
$$\sin 60^{\circ} = \frac{QR}{PR} \;\Rightarrow\; QR = PR \cdot \sin 60^{\circ} = 6 \times \frac{\sqrt3}{2} = 3\sqrt3\text{ cm}.$$\
Thus, \(\angle QPR = 60^{\circ}\) and \(\angle PRQ = 30^{\circ}\).
Correct Answer: ∠QPR = 60° , ∠PRQ = 30°