Question

$\text{In the given figure, lines AB and CD intersect at a point O . The sides CA and OB have been produced to E and F respectively such that}$

$\left(a\right){190}^{\circ }\left(b\right){230}^{\circ }\left(c\right){210}^{\circ }\left(d\right){270}^{\circ }$

Answer 1

ANSWER:
In the given figure, $\angle BOD=\angle COA$
∴ $\angle BOD={40}^{\circ }$
.....(1)
(Vertically opposite angles)In $△ACO$,
$\angle OAE=\angle OCA+\angle COA$ (Exterior angle of a triangle is equal to the sum of two opposite interior angles)
⇒ $x={80}^{\circ }+{40}^{\circ }={120}^{\circ }$ .....(2)
In $△BDO$,
$\angle DBF=\angle BDO+\angle BOD$ (Exterior angle of a triangle is equal to the sum of two opposite interior angles)
⇒ $y={70}^{\circ }+{40}^{\circ }={110}^{\circ }$ [Using (1)]
.....(3)
Adding (2) and (3), we get
x° + y° = 120° + 110° = 230°
Hence, the correct answer is option (b).