[tex]\it m(\widehat{C})=180^o-(75^O+30^o)=180^o-105^o=75^o\\ \\ m(\widehat{A})=m(\widehat{C})=75^o \Rightarrow \Delta ABC-isoscel,\ BC=BA=10\ cm\\ \\ \\ \mathcal{A}=\dfrac{BC\cdot BA\cdot sin30^o}{2}=\dfrac{10\cdot10\cdot\dfrac{1}{2}}{2}=25\ cm^2[/tex]
