1. **Problem Statement:** Calculate the angle $\angle LQR$ in triangle $LQR$ where sides are given as $LQ=21.22$ m, $QR=19.63$ m, and $LR=1.83$ m. 2. **Formula Used:** To find an angle in a triangle when all three sides are known, use the Law of Cosines: $$\cos(\theta) = \frac{a^2 + b^2 - c^2}{2ab}$$ where $\theta$ is the angle opposite side $c$, and $a$, $b$ are the other two sides. 3. **Identify Sides:** Here, $\angle LQR$ is at vertex $Q$, so the side opposite this angle is $LR = 1.83$ m. Sides adjacent to $\angle LQR$ are $LQ = 21.22$ m and $QR = 19.63$ m. 4. **Apply Law of Cosines:** $$\cos(\angle LQR) = \frac{LQ^2 + QR^2 - LR^2}{2 \times LQ \times QR}$$ Substitute values: $$= \frac{21.22^2 + 19.63^2 - 1.83^2}{2 \times 21.22 \times 19.63}$$ 5. **Calculate Numerator:** $$21.22^2 = 450.2484$$ $$19.63^2 = 385.6969$$ $$1.83^2 = 3.3489$$ So numerator: $$450.2484 + 385.6969 - 3.3489 = 832.5964$$ 6. **Calculate Denominator:** $$2 \times 21.22 \times 19.63 = 2 \times 416.6466 = 833.2932$$ 7. **Calculate Cosine:** $$\cos(\angle LQR) = \frac{832.5964}{833.2932}$$ Use cancellation to simplify: $$\cos(\angle LQR) = \frac{\cancel{832.5964}}{\cancel{833.2932}} \approx 0.99915$$ 8. **Find Angle:** $$\angle LQR = \cos^{-1}(0.99915)$$ Using a calculator: $$\angle LQR \approx 2.5^\circ$$ 9. **Final Answer:** Rounded to the nearest degree, $$\boxed{3^\circ}$$