Question: top-left: Solve for side $q$. Round to the nearest tenth.
center: A triangle with vertices $P$ at the bottom-left, $Q$ at the top, and $R$ at the bottom-right; side $PQ$ is labeled $10$ cm, side $QR$ is labeled $14$ cm, and the angle at $Q$ is labeled $98^\circ$.
bottom-left: $335.0$ cm
bottom-left: $257.0$ cm
bottom-left: $16.0$ cm
bottom-left: $18.3$ cm
1. **State the problem:** We need to find the length of side $q$ in triangle $PQR$ where $PQ=10$ cm, $QR=14$ cm, and the angle at $Q$ is $98^\circ$.
2. **Identify the sides and angle:** Let side $q$ be the side opposite angle $Q$. We know two sides and the included angle, so we can use the Law of Cosines.
3. **Law of Cosines formula:** $$q^2 = PQ^2 + QR^2 - 2 \times PQ \times QR \times \cos(\angle Q)$$
4. **Substitute known values:** $$q^2 = 10^2 + 14^2 - 2 \times 10 \times 14 \times \cos(98^\circ)$$
5. **Calculate each term:**
$$10^2 = 100$$
$$14^2 = 196$$
$$2 \times 10 \times 14 = 280$$
$$\cos(98^\circ) \approx -0.1392$$
6. **Plug in cosine value:**
$$q^2 = 100 + 196 - 280 \times (-0.1392)$$
7. **Simplify:**
$$q^2 = 296 + 38.976 = 334.976$$
8. **Take the square root:**
$$q = \sqrt{334.976} \approx 18.3$$
9. **Round to the nearest tenth:**
$$q \approx 18.3 \text{ cm}$$
**Final answer:** The length of side $q$ is approximately $18.3$ cm.