Question
✓
Answer
- 230º
Solution:
In $\triangle\text{ACO}$
$\angle\text{ACO} + \angle\text{COA} + \angle\text{OAC} = 180^\circ$
Now, $\angle\text{OAC} = 180^\circ$
$⇒\ 80^\circ + 40^\circ + 180^\circ - \text{x}^\circ= 180^\circ$
$⇒\ \text{x}^\circ = 120^\circ$
$ \angle\text{BOD} = \angle\text{COA} = 40^\circ$ (Opposite angles)
$\angle\text{BDO} = 70^\circ$
In $\triangle\text{OBD}$
$\angle\text{OBD} = 180^\circ - 40^\circ - 70^\circ = 70^\circ$
Also, $\text{y}^\circ = 180^\circ - \angle\text{OBD} = 180^\circ - 700^\circ = 110^\circ$
$⇒\ \text{x}^\circ + \text{y}^\circ = 120^\circ + 110^\circ = 230^\circ$
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