Question

In the given figure AS || BT; $\angle 4=\angle 5\overrightarrow{SB}$ bisects $\angle AST$. Find the measure of $\angle 1$

Answer 1

$AS\parallel BT$ [ Given ]

$\therefore$ $\angle 2=\angle 5$ [ Alternate angles are equal ]

$\Rightarrow$ $\angle 2=\angle 3$

$\Rightarrow$ Let, $\angle 2=\angle 3=\angle 5=x$

$\Rightarrow$ In $△BST,$ $\angle RSB$ is an exterior angle.

$\therefore$ $\angle RSB=\angle SBT+\angle BTS$ [ By exterior angle property ]

$\therefore$ $\angle 1+\angle 2=\angle 5+\angle 4$

$\therefore$ $\angle 1+x=x+x$

$\therefore$ $\angle 1+x=2x$

$\therefore$ $\angle 1=x$

$\Rightarrow$ $\angle 1+\angle 2+\angle 3={180}^o$ [ Since, $\angle 1,\angle 2$ and $\angle 3$ forms a linear pair ]

$\Rightarrow$ $x+x+x={180}^o$

$\Rightarrow$ $3x={180}^o$

$\therefore$ $x=\frac{{180}^o}{3}={60}^o$

$\therefore$ Measure of $\angle 1$ is ${60}^o$.