計算2
由\(\displaystyle sinB=\frac{AF}{AH} \Rightarrow \overline{AH}=\frac{bcosA}{sinB}\)
且\(\displaystyle tanB=\frac{CD}{DH}=\frac{bcosC}{tanB} \Rightarrow \overline{DH}=\frac{bcosBcosC}{sinB}\)
所求\(\displaystyle \frac{AH}{DH}=\frac{cosA}{cosBcosC}=\frac{-cos(B+C)}{cosBcosC}=tanBtanC-1\)