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