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Write the correct answer in each of the following:

$\text{ In Fig. 6.3, if }OP\parallel RS$ , $\angle OPQ = 110^{\circ}\text{and }$ $\angle QRS = 130^{\circ}\text{, then }$ $\angle PQR\text{ is equal to }$

$(A)\ 40^{\circ}$

$(B)$ $50^{\circ}$

$(C)$ $60^{\circ}$

$(D)$ $70^{\circ}$

Answers (1)

$OP \parallel RS$

$\angle RWV+\angle SRW=180^{\circ}$ $\text{ (sum of co- interior angles) }$ $130^{\circ} + \angle RWV =180^{\circ}$

$\angle RWV = 50^{\circ}$
$\angle PWQ = \angle RWV = 50^{\circ}\text{(vertically opposite angles are equal) }$

Also, for line OP

$\angle OPQ +\angle WPQ = 180^{\circ}\text{(linear pair)}$

$\angle WPQ = 180^{\circ} - \angle OPQ$

$= 180^{\circ} -110^{\circ}$

$\angle WPQ = 70^{\circ}$

Now, we know that the sum of angles of a triangle is 180⁰

$\text{in }\triangle PQW$ ,

$\angle WPQ + \angle PQW +\angle PWQ = 180^{\circ}$

$\angle PQW= 180^{\circ} - \angle WPQ -\angle PWQ = 180^{\circ} - 70^{\circ} - 50^{\circ}$

$\angle PQR = 180^{\circ} - 120^{\circ}$

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

Hence option C is the correct answer

Posted by

infoexpert26

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