Question

In the given figure, $AB||CD$ and $CA$ has been produced to $E$ so that $\angle BAE={125}^{\circ }$.
If $\angle BAC=x^{\circ },\angle ABD=x^{\circ },\angle BDC=y^{\circ }$ and $\angle ACD=z^{\circ }$, find the values of $x,y,z$.

Answer 1

Given: $AB||CD$, $\angle BAE={125}^{\circ }$.

Since, $CE$ is a straight line, therefore,

$\Rightarrow \angle CAB+\angle BAE={180}^{\circ }$

$\Rightarrow x^{\circ }+{125}^{\circ }={180}^{\circ }$

$\Rightarrow x^{\circ }={180}^{\circ }-{125}^{\circ }={55}^{\circ }$

Therefore, $x=55$.

Now, since $AB||CD$, therefore,

$\Rightarrow \angle CAB+\angle ACD={180}^{\circ }$ and $\angle DBA+\angle CDB={180}^{\circ }$ [Angles on the same side of the transversal are supplementary\)

Therefore,

$\Rightarrow z^{\circ }=y^{\circ }={180}^{\circ }-{55}^{\circ }$

$\Rightarrow z^{\circ }={125}^{\circ }$ and $y^{\circ }={125}^{\circ }$

Hence, $x={55}^{\circ },y=z={125}^{\circ }$