Question

c) in \\( \triangle p q r, p q=15.0 \mathrm{~cm}, p r=11.0 \mathrm{~cm} \\), and \\( q r=10.5 \mathrm{~cm} \\)

Explanation:

Step1: Identify sides and angles

In $\triangle PQR$, $PQ = 15.0$ cm (opposite $\angle R$), $PR = 11.0$ cm (opposite $\angle Q$), $QR = 10.5$ cm (opposite $\angle P$). Use Law of Cosines for $\angle P$.
$$\cos P = \frac{PQ^2 + PR^2 - QR^2}{2 \cdot PQ \cdot PR}$$
$$\cos P = \frac{15^2 + 11^2 - 10.5^2}{2 \cdot 15 \cdot 11} = \frac{225 + 121 - 110.25}{330} = \frac{235.75}{330} \approx 0.7144$$
$$\angle P = \cos^{-1}(0.7144) \approx 44^\circ$$

Step2: Calculate $\angle Q$ with Law of Sines

$$\frac{\sin Q}{PR} = \frac{\sin P}{QR}$$
$$\sin Q = \frac{PR \cdot \sin P}{QR} = \frac{11 \cdot \sin 44^\circ}{10.5} \approx \frac{11 \cdot 0.6947}{10.5} \approx 0.720$$
$$\angle Q = \sin^{-1}(0.720) \approx 47^\circ$$

Step3: Find $\angle R$ using angle sum

$$\angle R = 180^\circ - \angle P - \angle Q = 180^\circ - 44^\circ - 47^\circ = 89^\circ$$

Answer:

$\angle P \approx 44^\circ$, $\angle Q \approx 47^\circ$, $\angle R \approx 89^\circ$