Question

数学求角，解答

Answer 1

∠ABD=180°-∠BAD-∠BDA=180°-48°-16°-58°=58°，AB=AD
∠ACB=180°-∠BAC-∠ABC=180°-48°-58°-30°=44°
设AB=AD=1，
正弦定理：
AB/sinACB=AC/sinABC，
AC=ABsinABC/sinACB=1×sin（58°+30°）/sin44°=sin88°/sin44°=2cos44°
余弦定理：
CD2=AC2+AD2-2AC.ADcosDAC
=4cos244°+1-2×2cos44°cos16°
=4cos44°（cos44°-cos16°）+1
=-8cos44°sin30°sin14°+1
=-4cos44°sin14°+1
=-2[sin58°-sin30°]+1
=2-2sin58°
2（1-sin58°）=2（sin229°-2sin29°cos29°+cos229°）=2（cos29°-sin29°）2
CD=√2（cos29°-sin29°）=2（sin45°cos29°-cos45°sin29°）=2sin16°
正弦定理：
CD/sin16°=AD/sinACD
sinACD=ADsin16°/CD=sin16°/2sin16°=1/2，
∠ACD=30°