Question

In figure given below, AC $\parallel$BT, $m$$\angle$$TOP={100}^{\circ }$, XY $\cong$ YZ, find $m$$\angle$X.

Answer 1

Since AC $\parallel$ BT so $\angle ZYO=\angle TOP$ [similar angles]

$\therefore \angle ZYO={100}^o\Rightarrow \angle XYZ=({180}^o-{100}^o)={80}^0$.

Now from $\Delta XYZ$ we get $\angle ZXY=\angle YZX$ [Since XY $\cong$ YZ].

Also, $\angle X+\angle Z={180}^o-{80}^o={100}^o\Rightarrow \angle X={50}^o$.

$\therefore m\angle X={50}^o$.