MCQ
If $\angle\text{OCA}=80^\circ, \ \angle\text{COA}=40^\circ,$ and $\angle\text{BDO}=70^\circ$ then $x^\circ + y^\circ = ?$
- A$270^\circ$
- B$210^\circ$
- ✓$230^\circ$
- D$190^\circ$
✓
Answer
Correct option: C.
$230^\circ$
In the given figure, $\angle\text{BOD} = \angle\text{COA}$ (Vertically opposite angles)$\therefore\ \angle\text{BOD} = 40^\circ ... (\text{i})$
In $\triangle\text{ACO}$
$\angle\text{OAE}=\angle\text{OCA}+\angle\text{COA}$ (Exterior angle of a triangle is equal to the sum of two opposite interior angles)
$⇒ \text{x}^\circ = 80^\circ + 40^\circ = 120^\circ ... \ (\text{ii})$
In $\angle\text{BDO},$
$\angle\text{DBF}=\angle\text{BDO}+\angle\text{BOD}$ (Exterior angle of a triangle is equal to the sum of two opposite interior angles)
$⇒\text{y}^\circ=70^\circ+40^\circ=110^\circ\ ...\ (\text{iii})$
Adding $(2)$ and $(3)$ we get
$\text{x}^\circ+\text{y}^\circ=120^\circ+110^\circ=230^\circ$
In $\triangle\text{ACO}$
$\angle\text{OAE}=\angle\text{OCA}+\angle\text{COA}$ (Exterior angle of a triangle is equal to the sum of two opposite interior angles)
$⇒ \text{x}^\circ = 80^\circ + 40^\circ = 120^\circ ... \ (\text{ii})$
In $\angle\text{BDO},$
$\angle\text{DBF}=\angle\text{BDO}+\angle\text{BOD}$ (Exterior angle of a triangle is equal to the sum of two opposite interior angles)
$⇒\text{y}^\circ=70^\circ+40^\circ=110^\circ\ ...\ (\text{iii})$
Adding $(2)$ and $(3)$ we get
$\text{x}^\circ+\text{y}^\circ=120^\circ+110^\circ=230^\circ$
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