# Q). In the adjoining figure, ABCD is a rectangle. If $\angle CEB:\angle ECB=3:2$, find (i) $\angle CEB$ (ii) $\angle DCF$.

in triangle BEC
angle B = 90   (angle of rectangle)
angle E = 3x
Angle C = 2x
by angle sum property of triangle
90 + 3x + 2x = 180
90 + 5x = 180
5x = 180 - 90 = 90
x = 90/5
x =  18
it gives us,  angle CEB = 3x  = 3 X 18 = 54
angle DCE =  angle CEB (alternate angles)
angle DCE + angle DCF = 180  (angles on straight line / linear pair)
it gives us,  54 + angle DCF = 180
angle DCF = 180 - 54 = 126
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