In the given figure, BE is parallel to CF. Evaluate the value of x.
13∘
Construction: Draw a line XY such that XY∥BE∥CF and is passing through O.
So, BE || XY and AO is the transversal.
⇒∠BOY=∠ABE
(Since corresponding angles are equal)
⇒∠BOY=4x
Similarly, CF || XY and CO is the transversal.
⇒∠COY=∠DCF
(Since corresponding angles are equal)
⇒∠COY=6x
Now, ∠BOC=∠BOY+∠COY (from the figure)
⇒130∘=4x+6x
⇒10x=130∘
⇒x=13∘