Question

cho tam giác ABC vuông ở A, đường cao AH. CH=6cm, sin B = \(\dfrac{\sqrt{3}}{2}\) . Tính AH.

Answer 1

Ta co ^B=^HAC nen \(SinB=SInHAC=\dfrac{\sqrt{3}}{2}\Rightarrow AC=\dfrac{HC}{\dfrac{\sqrt{3}}{2}}=4\sqrt{3}\Rightarrow AH=\sqrt{16.3-6^2}=2\sqrt{3}\)

Answer 2

Xét ΔABC vuông tại A có:

\(\sin B=\dfrac{\sqrt{3}}{2}\\ \Rightarrow\widehat{B}=60^0\\ \widehat{B}+\widehat{BAH}=90^0\\ \Rightarrow\widehat{BAH}=90^0-60^0=40^0\\ \widehat{BAH}+\widehat{HAC}=90^0\\ \Rightarrow\widehat{HAC}=90^0-40^0=50^0\\ \tan HAC=\dfrac{HC}{AH}\\ \Leftrightarrow\tan50=\dfrac{6}{AH}\\ \Rightarrow AH\approx5,03cm\)