ΔABC vuông tại A có AH là đường cao
nên AH^2=HB*HC
=>AH=6cm
ΔABH vuông tại H
=>AB^2=AH^2+HB^2=6^2+4^2=52
=>AB=2*căn 13(cm)
\(\dfrac{sinHAB+tanHAC}{tanB+tanC}\)
\(=\dfrac{\dfrac{HB}{AB}+\dfrac{HC}{AH}}{\dfrac{AH}{HB}+\dfrac{AH}{HC}}=\dfrac{\dfrac{4}{2\sqrt{13}}+\dfrac{9}{6}}{\dfrac{6}{4}+\dfrac{9}{4}}=\dfrac{12+9\sqrt{13}}{6\sqrt{13}}:\dfrac{15}{4}\)
\(=\dfrac{4\left(12+9\sqrt{13}\right)}{90\sqrt{13}}=\dfrac{12\left(4+\sqrt{13}\right)}{90\sqrt{13}}=\dfrac{2}{15}\cdot\dfrac{4+\sqrt{13}}{\sqrt{13}}\)
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