Bài 6:
Xét ΔABC có \(cosABC=\dfrac{BA^2+BC^2-AC^2}{2\cdot BA\cdot BC}\)
=>\(\dfrac{\left(8\sqrt{2}\right)^2+23^2-AC^2}{2\cdot8\sqrt{2}\cdot23}=cos45=\dfrac{\sqrt{2}}{2}\)
=>\(128+529-AC^2=\dfrac{\sqrt{2}}{2}\cdot16\sqrt{2}\cdot23=368\)
=>\(AC^2=128+529-368=289\)
=>AC=17(cm)
Xét ΔABC có \(\dfrac{AC}{sinABC}=\dfrac{AB}{sinACB}\)
=>\(sinACB=AB\cdot\dfrac{sinABC}{AC}=\dfrac{8\sqrt{2}\cdot sin45}{17}=\dfrac{8}{17}\)
=>\(\widehat{ACB}\simeq28^04'\)
Bài 5: Xét ΔAHB vuông tại H có \(sinB=\dfrac{AH}{AB}\)
=>\(\dfrac{AH}{6\sqrt{2}}=sin45=\dfrac{\sqrt{2}}{2}\)
=>\(AH=6\sqrt{2}\cdot\dfrac{\sqrt{2}}{2}=6\left(cm\right)\)
Xét ΔAHC vuông tại H có \(sinC=\dfrac{AH}{AC}\)
=>\(AC=\dfrac{6}{sin60}=6:\dfrac{\sqrt{3}}{2}=\dfrac{12}{\sqrt{3}}=4\sqrt{3}\left(cm\right)\)
