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\triangle ABC中,∠ABD=30°,∠DBC=40°,∠DCB=20°,∠ACD=50°,求\angle DAC=?

解:
         如图,



\angle DAC=α,\angle DAB=β,\angle ADC=γ,则: \begin{cases} α+β=40°\ \ \ \ (1)\\ α+γ=130°\ \ \ \ (2) \end{cases} 由等式(2)-(1)的两边得 γ-β=90°\ \ \ \ (3) \triangle ABD中,由正弦定理得 \frac{BD}{sinβ}=\frac {AD}{sin30°}\ \ \ \ (4) \triangle ACD中,由正弦定理得 \frac{AD}{sin50°}=\frac{CD}{sinα}\ \ \ \ (5) (4)(5)两式联立得 \frac{BD}{sinβ}=\frac{sin50°}{sin30°sinα}CD\ \ \ \ (5) \triangle BCD中,由正弦定理得 \frac{CD}{sin40°}=\frac{BD}{sin20°}\ \ \ \ (6) (5)(6)两式联立得 \begin{align} \frac{BD}{sinβ}=&\frac{sin50°}{sin30°sinα}\times \frac{sin40°}{sin20°}BD\\ =>\ sin20°sin30°sinα=&sin40°sin50°sinβ\\ =>\ \frac{sinα}{sinβ}=&\frac{sin40°sin50°}{sin20°sin30°}\\ =&\frac{-\frac 12(cos90°-cos10°)}{sin20°sin30°}\\ =&\frac{\frac 12cos10°}{\frac 12 \times sin20°}\\ =&\frac{cos10°}{sin20°}\\ =&\frac{cos10°}{2sin10°cos10°}\\ =&\frac12 \frac 1{sin10°}\\ =&\frac{sin30°}{sin10°}\\ =>\ \frac{sinα}{sinβ}=&\frac{sin30°}{sin10°}\\ ∵\ α+β=&40°\\ ∴\ β=&40°-α\\ =>\ \frac{sinα}{sin(40°-α)}=&\frac{sin30°}{sin10°}\ \ \ \ (7)\\ \end{align} 由等式(7)通过比较,可以看出 α=30° 即所求的\angle DAC为:\angle DAC=30°



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