Question

In a $△$PQR, PS is bisector of $\angle P$ and $\angle Q={70}^o\angle R={30}^o$, then

• A. $QS>PQ>PR$
• B. $QS<PQ<PR$
• C. $QS<PQ>PR$
• D. None of these

Answer 1

Answer: (B)

$QS<PQ<PR$

Given: $\angle Q={70}^o,\angle R={30}^o$

$\angle QPR={180}^o-({70}^o+{30}^o)={80}^o$ {angle sum property of a triangle}
$\therefore \angle QPS=\angle SPR={40}^o$
$\angle PRQ<\angle PQR$
$\therefore PQ<PR$ ....... (i) {Side opposite to the larger angle is larger}
$\angle QPS<PQS$
$\therefore QS<PR$ ...... (ii) {Side opposite to the larger angle is larger}

In $\Delta PQS$,
$\angle PSQ={180}^o-({70}^o+{40}^o)={70}^o$ {angle sum property of a triangle}
$\angle QPS<\angle PSQ$
$\therefore QS<PQ$ ........ (iii) {Side opposite to the larger angle is larger}
From (i), (ii) & (iii)
$QS<PQ<PR$.