Xét ΔIGH có \(\widehat{I}+\widehat{G}+\widehat{H}=180^0\)
=>\(\widehat{H}=180^0-60^0-50^0=70^0\)
Xét ΔIHG có \(\dfrac{IH}{sinG}=\dfrac{IG}{sinH}=\dfrac{HG}{sinI}\)
=>\(\dfrac{IH}{sin50}=\dfrac{4}{sin70}=\dfrac{HG}{sin60}\)
=>\(IH=4\cdot\dfrac{sin50}{sin70}\simeq3,26\left(cm\right)\); \(HG=4\cdot\dfrac{sin60}{sin70}\simeq3,69\left(cm\right)\)