In the given figure, if `OP|"|RS, /_OPQ = 110^(@)` and `/_QRS = 130^(@)`, then `/_PQR` is equal to

Options

A. `40^(@)`
B. `50^(@)`
C. `60^(@)`
D. `70^(@)`

Answer

C

Explanation

In the given figure producing OP, to intersect RQ at X.
Since, `OP|""|RS` and RX is a transversal.
So, `" " /_RXP = /_XRS " "` [alternate angles]

`rArr " " /_RXP = 130^(@) " " [:./_QRS = 130^(@)(given)]`..(i)
Now, RQ is a line segment.
So,` " " /_PXQ + /_RXP = 180^(@) " " `[linear pair axiom]
`rArr " "/_PXQ = 180^(@) - /_RXP = 180^(@) - 130^(@) " "` [from Eq. (i)]
`rArr " " /_PXQ =50^(@)`
In`DeltaPQX,/_OPQ` is an exterior angle , " "
`:." "/_OPQ = /_PXQ + /_PQX " "` [`:.`exterior angle = sum of two opposite interior angles ]
`rArr " " 110^(@) = 50^(@) + /_PQX`
`rArr " " /_PQX = 110^(@)-50^(@)`
`:." " /_PQR = 60^(@) " " [:. /_PQX = /_PQR]`