Question

In the figure given below, if OP || RS, ∠OPQ=110^o and ∠QRS=130^o, then ∠PQR is equal to:

• A.
• B.
• C.
• D.

Answer 1

The correct option is A

In the given figure, if we extend OP then a triangle PQT will be formed, where T is the point where OP cuts QR after extending

Now from the above figure:

$\angle$OPQ + $\angle$QPT = ${180}^{\circ }$

$\Rightarrow$ $\angle$QPT = ${180}^{\circ }$ - ${110}^{\circ }$ = ${70}^{\circ }$

Now, since if OP || RS, $\angle$SRQ and $\angle$UTQ are corresponding angles hence, $\angle$UTQ = $\angle$SRQ = ${130}^{\circ }$.

Therefore we have,

$\Rightarrow$​ $\angle$UTQ + $\angle$PTQ = ${180}^{\circ }$

$\Rightarrow$​ $\angle$PTQ = ${180}^{\circ }$ - ${130}^{\circ }$ = ${50}^{\circ }$

In triangle PTQ

$\angle$PTQ + $\angle$TQP + $\angle$QPT = ${180}^{\circ }$

$\Rightarrow$​ ${50}^{\circ }$ + $\angle$TQP + ${70}^{\circ }$ = ${180}^{\circ }$

$\Rightarrow$$\angle$TQP = ${180}^{\circ }$ - ${120}^{\circ }$ = ${60}^{\circ }$

Since, $\angle$TQP = $\angle$PQR = ${60}^{\circ }$.