- Hạ đường cao AH của \(\Delta ABC\).
- \(\Delta ACH\) vuông tại H có:.
\(\left\{{}\begin{matrix}\sin\widehat{ACH}=\dfrac{AH}{AC}\\\cos\widehat{ACH}=\dfrac{CH}{AC}\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{AH}{AC}=\sin50^0\approx0,8\\\dfrac{CH}{AC}=\cos50^0\approx0,6\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}AH\approx0,8.AC=0,8.35=28\left(cm\right)\\CH\approx0,6.AC=0,6.35=21\left(cm\right)\end{matrix}\right.\)
- \(\Delta ABH\) vuông tại H có:
\(\tan\widehat{ABH}=\dfrac{AH}{BH}\)
\(\Rightarrow\dfrac{AH}{BH}=\tan60^0=\sqrt{3}\)
\(\Rightarrow BH=\dfrac{AH}{\sqrt{3}}\approx\dfrac{28}{\sqrt{3}}=\dfrac{28\sqrt{3}}{3}\left(cm\right)\)
- Vậy \(S_{ABC}=\dfrac{1}{2}AH.BC\approx\dfrac{1}{2}28.\left(\dfrac{28\sqrt{3}}{3}+21\right)=\dfrac{392\sqrt{3}+882}{3}\left(cm^2\right)\)
