\(\widehat{B}-\widehat{C}=10^o\\ \rightarrow B=\widehat{C}+10^o\\ Có:\widehat{B}+\widehat{C}=180^o\\ \rightarrow\widehat{C}+10^o+\widehat{C}=180^o\\ \rightarrow2\widehat{C}=170^o\\ \rightarrow\widehat{C}=85^o\\ \rightarrow\widehat{D}=85^o\left(tinhchathinhthangcan\right)\)