MCQ
In the given figure, $AOB$ is a straight line. If $\angle\text{AOC}=(3\text{x}-10)^\circ,\angle\text{COD}=50^\circ$ and $\angle\text{BOD}=(\text{x}+20)^\circ$ then $\angle\text{AOC}=?$
- A$50^\circ$
- B$60^\circ$
- ✓$80^\circ$
- D$40^\circ$
✓
Answer
Correct option: C.
$80^\circ$
We have,$\angle\text{AOC}+\angle\text{COD}+\angle\text{BOD}=180^\circ$ [Since $AOB$ is a straight line]
$\Rightarrow 3x - 10 + 50 + x + 20 = 180$
$\Rightarrow 4x = 120$
$\Rightarrow x = 30$
$\therefore\angle\text{AOC}=[3\times30−10]^\circ$
$\Rightarrow\angle\text{AOC}=80^\circ.$
$\Rightarrow 3x - 10 + 50 + x + 20 = 180$
$\Rightarrow 4x = 120$
$\Rightarrow x = 30$
$\therefore\angle\text{AOC}=[3\times30−10]^\circ$
$\Rightarrow\angle\text{AOC}=80^\circ.$
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