Question Number 177975 by aurpeyz last updated on 11/Oct/22
Answered by mr W last updated on 11/Oct/22
$$\angle{ACB}={x}/\mathrm{2} \\ $$$$\angle{BAC}={y} \\ $$$$\angle{B}=\mathrm{180}−{x}/\mathrm{2}−{y} \\ $$$$\angle{ADC}=\mathrm{180}−\angle{B}={x}/\mathrm{2}+{y} \\ $$
Commented by Tawa11 last updated on 11/Oct/22
$$\mathrm{Great}\:\mathrm{sir} \\ $$
Commented by aurpeyz last updated on 11/Oct/22
$${pls}\:{how}\:{is}\:\angle{ACB}={x}/\mathrm{2}.\:{thats}\:{the}\:{only}\:{part} \\ $$$${that}\:{i}\:{didnt}\:{get}.\:{thanks} \\ $$
Commented by mr W last updated on 12/Oct/22
Commented by aurpeyz last updated on 12/Oct/22
$${thanks}\:{sir} \\ $$