Question

In the given figure, AB || CD. If $\angle CAB={180}^{\circ }$ and $\angle EFC={25}^{\circ }$ then $\angle CEF=?$

$\left(a\right){65}^{\circ }$

$\left(b\right){55}^{\circ }$

$\left(c\right){45}^{\circ }$

$\left(d\right){75}^{\circ }$

Answer 1

(c) 45°

AB∥CD and AF is the transversal.

∴∠DCF=∠CAB=80° Corresponding Angles

Side EC of triangle EFC is produced to D.

∴∠CEF+∠EFC=∠DCF [Exterior Angle Theorem.]

⇒∠CEF+25°=80°

⇒∠CEF=55°