20/10/01 08:54:00.32 n2o6aWK1.net
>>240
AC = sin(B)/sin(A),
CD = sinβ/sin(β+γ),
ここに α = ∠CAD,
β = ∠DBC = 30°, γ = ∠DCB = 10°,
正弦定理
sin(150-α)：sinα = AC：CD,
より
1/tanα = AC/(CD･sin(30)) - 1/tan(30),
α = 40°
>>241
BC = sin(A)/sin(C),
BD = sinα/sin(α+β),
ここに α = ∠DAB = 3°, β = ∠DBA = 51°,
正弦定理
sin(156-y)：sin(y) = BC：BD,
より
1/tan(y) = BC/(BD･sin(156)) + 1/tan(156),
y = 15°
x = 96°- y = 81°