Q. 73.6( 11 Votes )

# In Fig. 6.3, if O

Answer :

In the figure, it is given to us that –

OP || RS

OPQ = 110°

QRS = 130°

We have to find PQR.

Let us extend OP, so as to intersect QR at point D. The figure is as -

Now, we have OP || RS and, DR is a transversal.

We know, if a transversal intersects two parallel lines, then each pair of alternate angles are equal.

DRS = RDP

It is given to us that QRS = 130°

DRS = 130°

RDP = 130° - - - - (i)

Now, we have QR as a line segment. By linear pair axiom,

PDQ + PDR = 180°

QDP + RDP = 180°

QDP = 180° - RDP

QDP = 180° - 130°

QDP = 50°

PDQ = 50° - - - - (ii)

In ΔPQD,

OPQ is an exterior angle.

Also, we know that exterior angle is equal to the sum of the two interior opposite angles.

OPQ = PQD + PDQ - - - - (iii)

We have OPQ = 110° and from equation (ii) we have PDQ = 50°.

So, equation (iii) can be written as –

110° = PQD + 50°

PQD = 60°

So, PQR = 60°

Thus, option (C) is correct.

Option (A) is not correct. If PQR is equal to 40°, then it won’t satisfy the linear axiom. So, PQR is not 40°.

Option (B) is not correct. We have got PDQ = 50°. If PQR = 50°, then the exterior angle, i.e., OPQ = 110° is not equal to the sum of the opposite interior angles, PDQ and PQR which are now 50° each. So, PQR is not equal to 50°.

Option (D) is not correct. We have, OPQ = 110°. By linear pair axiom, QPD = 70°. So, PQR is not equal to 700

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