91Ó°ÊÓ

Line segments $\stackrel{¯}{SB}\text{and}\stackrel{¯}{RT}$are transversal to parallel lines

$∠1$and $∠4$forms a pair of corresponding angles

Using Postulate 10: If two parallel lines are cut by transversal, then corresponding angles are congruent.

$∠1â‰\dots ∠4$

$m∠1=m∠4\left(1\right)$

$∠2$and $∠5$forms a pair of alternate interior angles

Using theorem 3-2: If two parallel lines are cut by transversal, then alternate interior angles are congruent.

$∠2â‰\dots ∠5$

$m∠2=m∠5\left(2\right)$

From statement $\stackrel{→}{SB}$bisects$∠AST$

$m∠2=m∠3$

Substitute $m∠5$for $m∠2$in above equation

$m∠5=m∠3\left(3\right)$

Since, sum of angles of a triangle is 180

Use equation (3) and given equation

$\begin{array}{c}m∠3+m∠4+m∠5=180\\ m∠5+m∠5+m∠5=180\end{array}$

$3m∠5=180$

$m∠5=60$

Substitute 60 for $m∠5$and role="math" localid="1637818811573" $m∠1$for $m∠4$(from equation (1)) in given equation

$m∠1=60$

Hence, measure of $∠1$is 60.