Question

In the given fig. $AB\parallel CD,\angle BDC={40}^{\circ }$, and $\angle BAD={75}^{\circ }$. Find $x,y,z$.

Answer 1

In figure,

$AB\parallel CD$ and $BD$ is transversal.

$\Rightarrow x={40}^{\circ }\left(\text{Alternate angles}\right)$ ---(1)

Now in $△ABD$

${75}^{\circ }+y+x={180}^{\circ }\left(\text{Sum of interior angles of a triangle}\right)$

$\Rightarrow y={180}^{\circ }-{75}^{\circ }-{40}^{\circ }={180}^{\circ }-{115}^{\circ }={65}^{\circ }$

$\Rightarrow y={65}^{\circ }$ ---(2)

Now in $△BCD$

$\Rightarrow {40}^{\circ }+(y-{30}^{\circ })+z={180}^{\circ }\left(\text{Sum of interior angles of a triangle}\right)$

$\Rightarrow {40}^{\circ }+({65}^{\circ }-{30}^{\circ })+z={180}^{\circ }$ [$\because y={65}^{\circ }$]

$\Rightarrow {40}^{\circ }+{35}^{\circ }+z={180}^{\circ }$

$\Rightarrow {75}^{\circ }+z={180}^{\circ }$

$\Rightarrow z={180}^{\circ }-{75}^{\circ }$

$\Rightarrow z={105}^{\circ }$ ---(3)

Therefore, $x={40}^{\circ },y={65}^{\circ }$ and $z={105}^{\circ }$.