解:1134°=3*360°+54°; 912°=2*360°+(180°°+12°).
连接BC.
利用余弦定理:BC^2=BD^2+CD^2-2*BD*DC*cos(3*360°+54°).
BC^2=2.9^2+5.2^2-2*2.9*5.2*cos54°.
=8.14+27.04-17.11
=18.08
BC=4.25 (km).
再用正弦定理:BD/sin∠BCD=BC/sin(3*360°+54°).
sin∠ BCD=BD*sin54/BC.
=2.9*0.8090/4.25.
=0.5520.
∠BCD=arcsin(0.5520)
=33.50°
∠ACB=912°-2*360°-33.5°
=192°°-33.5°.
∠ACB=158.5°
再用余弦定理:AB^2=AC^2+BC^2-2*AC*BC*cos∠ACB.
AB^2=3.7^2+18.04-2*3.7*4.25*cos158.5°.
=13.69+18.04+29.26
=60.99
∴AB=7.81 (km) -----即为所求。
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