炼气士
幼苗
共回答了18个问题采纳率:100% 举报
∠ACB=∠CBD+∠ADB
∠CBD=∠ACB-∠ADB=60°-45°=15°
∴CD/sin15°=BC/sin45°
BC=CD×sin45°/sin15°
=CD×√2/2×4/(√6-√2)
=CD(√3+1)
=(√3+1)AC/2
AB²=AC²+BC²-2AC×BCcos60°
=AC²+(√3+1)²AC²/4-(√3+1)AC²/2
=(1+1+√3/2-√3/2-1/2)AC²
=3/2
∴AB=(√6/2)AC
sin∠ABC=AC×sin60°/AB
=AC×√3/2×2/(√6AC)
=1/√2
=√2/2
∴∠ABC=45°或135°
∵∠ABC=135°不合题意
∴∠ABC=45°
1年前
9