Question

Question 153
In parallelogram LOST, $SN\perp OL$ and $SM\perp LT$. Find $\angle STM,\angle SON$ and $\angle NSM$.

Answer 1

Given, $\angle MST={40}^{\circ }$
In $\Delta MST$,
By the angle sum property of a triangle, $\angle TMS+\angle MST+\angle STM={180}^{\circ }$
$\Rightarrow \angle STM={180}^{\circ }-({90}^{\circ }+{40}^{\circ })$ $[\because SM\perp LT,\angle TMS={90}^{\circ }]$
$={50}^{\circ }$
$\therefore \angle SON=\angle STM={50}^{\circ }$ [$\because$ opposite angles of a parallelogram are equal]
Now, in the $\Delta ONS$,
$\angle ONS+\angle OSN+\angle SON={180}^{\circ }$ [angle sum property of triangle]
$\angle OSN={180}^{\circ }-({90}^{\circ }+{50}^{\circ })$
$={180}^{\circ }-{140}^{\circ }={40}^{\circ }$
Moreover, $\angle SON+\angle TSO={180}^{\circ }$
[$\because$ adjacent angles of a parallelogram are supplementary]
$\Rightarrow \angle SON+\angle TSM+\angle NSM+\angle OSN={180}^{\circ }\Rightarrow {50}^{\circ }+{40}^{\circ }+\angle NSM+{40}^{\circ }={180}^{\circ }\Rightarrow {130}^{\circ }+\angle NSM={180}^{\circ }\Rightarrow \angle NSM={180}^{\circ }-{130}^{\circ }={50}^{\circ }$ .