Question Number 223954 by MASANJAJJ last updated on 10/Aug/25
Answered by som(math1967) last updated on 11/Aug/25
$$\:{x}=\angle{BDC} \\ $$$$\angle{DBA}=\angle{BCD} \\ $$$$\:\therefore{x}=\mathrm{10}+\mathrm{90}−{x}\Rightarrow{x}=\mathrm{50} \\ $$
Answered by som(math1967) last updated on 11/Aug/25
$${A}\cup{D}={D} \\ $$$${A}\cap{B}=\phi,\:\:\left({i}\right)\:{true}\:{A}\:{is}\:{proper} \\ $$$${subset}\:{of}\:{D},\:{it}\:{should}\:{be} \\ $$$$\:{A}\:\subseteq{D} \\ $$$$ \\ $$