Đáp án Câu 1 (1 điểm) — Đề 101+102: Ta có $\widehat{ACB}=180^{\circ }-\left(60^{\circ }+45^{\circ }\right)=75^{\circ }$
$AB=\frac{AC\cdot \sin C}{\sin B}=\frac{6\sin 75^{\circ }}{\sin 45^{\circ }}=3\left(1+\sqrt{3}\right)$
$BC=\frac{AC\cdot \sin A}{\sin B}=\frac{6\sin 60^{\circ }}{\sin 45^{\circ }}=3\sqrt{6}$
P/án 1: $2\left(3\sqrt{6}+3\left(1+\sqrt{3}\right)\right)\approx 31,09km$
P/án 2: $6+3\sqrt{6}+3\left(1+\sqrt{3}\right)\approx 21,54km$

Đề 103+104: Ta có $\widehat{ACB}=180^{\circ }-\left(60^{\circ }+45^{\circ }\right)=75^{\circ }$
$AB=\frac{AC\cdot \sin C}{\sin B}=\frac{\sin 75^{\circ }}{\sin 45^{\circ }}=4\left(1+\sqrt{3}\right)$
$BC=\frac{AC\cdot \sin A}{\sin B}=\frac{8\sin 60^{\circ }}{\sin 45^{\circ }}=4\sqrt{6}$
P/án 1: $2\left(4\sqrt{6}+4\left(1+\sqrt{3}\right)\right)\approx 41,45km$
P/án 2: $2\left(4\sqrt{6}+4\left(1+\sqrt{3}\right)\right)\approx 28,72km$

(Thang điểm: mỗi bước 0,25 điểm.)