Question

In the adjoining Figure, $\overline{XY}\parallel \overline{BC}$. Find the values of $x,y,z$.

Answer 1

Given that XY$||$BC

then,

In $\Delta ABC$

$\angle A+\angle B+\angle C={180}^o$(property of triangle)

${40}^o+{30}^o+\angle C={180}^o$

${70}^o+z={180}^o$

$z={180}^o-{70}^o$

$z={110}^o$

$XY||BC$

then

$\angle x^o=\angle B$

$\angle x^o={30}^o$.

Now, In $\Delta AXY$

$\angle x^o+\angle y^o+\angle A={180}^o$(property of triangle)

${30}^o+\angle y^o={40}^o={180}^o$

$\angle y^o={180}^o-{70}^o$

$\angle y^o={110}^o$

Hence, this is the answer.