Side BC of a triangle ABC has been produced to a point D such that $\angle \mathrm{ACD}=120°$. If $\angle \mathrm{B}=\frac{1}{2}\angle \mathrm{A}$ then A is equal to

Given that,
Side BC of a triangle ABC has been extended to point D. 
$\angle \text{ACD }=\text{ 12}0°$ 
$\angle \text{B=}\frac{1}{2}\angle \text{A}$ …….(1)
By the exterior angle theorem in equation 1, 
$\begin{array}{l}\angle \text{ACD}=\angle \text{A }+\angle \text{B }\\ \text{12}0°=\angle \text{A+}\frac{1}{2}\angle \text{A}\end{array}$ 
$\begin{array}{l}\text{12}0°=\frac{(\text{2}\angle \text{A+}\angle \text{A})}{2}\\ \text{12}0°\times \text{2}=\text{3}\angle \text{A }\\ \text{24}0°=\text{3}\angle \text{A}\end{array}$  

We divide $\text{24}0°$ by 3,
Thus, the measure of $\mathbf{\angle }\mathbf{\text{A}}\mathbf{=}\mathbf{\text{8}}\mathbf{0}\mathbf{°}$.