In Fig. 9.54, AB and CD are parallel lines and transversal EF intersects them at P and Q respectively. If APR=25°, RQC=30° and CQF= 65°, then

Given,

AB CD


And, EF cuts them


So, 30o + 65o + PQR = 180o


95o + PQR = 180o


PQR = 85o


APQ + PQC = 180o (Co. interior angle)


25o + y + 85o + 30o = 180o


y = 40o


In


PQR + PRQ + QPR = 180o


85o + x + y = 180o


x = 55o


Thus, x = 55o and y = 40o

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