Question

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

• A. 60${}^{\circ }$
• B. 65${}^{\circ }$
• C. 40${}^{\circ }$
• D. 45${}^{\circ }$

Answer 1

The correct option is A

60${}^{\circ }$

Extend OP. Then, a triangle PQT will be formed; where T is the point at which OP cuts QR.

Now,

$\angle$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,

$\angle$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${}^{\circ }$

$\Rightarrow \angle$TQP = $\angle$PQR = 60${}^{\circ }$