Question

In the given figure, if $\angle$ACB = ${30}^{\circ }$, the value of $\angle$ACD is

___${}^{\circ }$.

Answer 1

$\text{In the given figure,}AD=AB\text{(Given)}DC=BC\text{(Given)}AC\text{is common}\therefore \Delta ADC\cong \Delta ABC\text{(SSS congruence)}\text{Hence,}\angle ACD=\angle ACB={30}^{\circ }$