15 In \( \triangle P Q R, P Q Q R=84^{\circ}, Q \widehat{P} R=43^{\circ} \) and \( I P Q /=5 \mathrm{~cm} \). Find \( / Q R / \) in cm, correct to 1 decimal place. (WASSCE)

To solve for \( QR \) in triangle \( \triangle PQR \), we'll use the **Law of Sines**. Here's a step-by-step breakdown: 1. **Given:** - Angles: - \( \angle PQR = 84^\circ \) - \( \angle QPR = 43^\circ \) - Side: - \( PQ = 5 \) cm 2. **Find the third angle (\( \angle PRQ \)):** \[ \angle PRQ = 180^\circ - \angle PQR - \angle QPR = 180^\circ - 84^\circ - 43^\circ = 53^\circ \] 3. **Apply the Law of Sines:** \[ \frac{QR}{\sin(\angle QPR)} = \frac{PQ}{\sin(\angle PRQ)} \] Plugging in the known values: \[ \frac{QR}{\sin(43^\circ)} = \frac{5}{\sin(53^\circ)} \] 4. **Solve for \( QR \):** \[ QR = \frac{\sin(43^\circ)}{\sin(53^\circ)} \times 5 \] Calculate the sine values: \[ \sin(43^\circ) \approx 0.681998, \quad \sin(53^\circ) \approx 0.798636 \] \[ QR \approx \frac{0.681998}{0.798636} \times 5 \approx 4.265 \text{ cm} \] 5. **Round to one decimal place:** \[ QR \approx 4.3 \text{ cm} \] **Answer:** \( QR \approx 4.3 \) cm