Question

Determine the value of x, y, and z in the following figure :

Answer 1

Now,

$\begin{array}{c}\Rightarrow {40}^0+x^0+{30}^0={180}^0\\ \Rightarrow {70}^0+x^0={180}^0\\ \therefore x^0={110}^0\\ \therefore {40}^0=z^0\left(oppositeangle\right)\\ Now,\\ \Rightarrow {40}^0+y^0={180}^0\\ \therefore y^0={140}^0\\ Hence,\\ weget\\ x={110}^0\\ y={140}^0\\ z={40}^0\end{array}$