Question

Trong tam giác abc, có ac = 10cm, acb = 45° , abc =30° .dường cao ah. Hãy tính độ dài Ah, Ab

Answer 1

\(\Delta AHC\) vuông tại \(H\)

\(\Rightarrow sin45^0=\dfrac{AH}{AC}=\dfrac{AH}{10}\)

\(\Rightarrow AH=10.sin45^0=5\sqrt{2}\left(cm\right)\)

\(\Delta AHB\) vuông tại \(H\)

\(\Rightarrow sin30^0=\dfrac{AH}{AB}=\dfrac{5\sqrt{2}}{AB}\)

\(\Rightarrow AB=\dfrac{5\sqrt{2}}{sin30^0}=10\sqrt{2}\left(cm\right)\)

Answer 2

Ta có: \(\sin ACB=\dfrac{AH}{AC}\)\(\Rightarrow\sin45^o=\dfrac{AH}{10}\)\(\Rightarrow AH=5\sqrt{2}\)(cm)

\(\sin ABH=\dfrac{AH}{AB}\)\(\Rightarrow\sin30^o=\dfrac{5\sqrt{2}}{AB}\)\(\Rightarrow AB=10\sqrt{2}\)