MCQ
In the given figure, find the value of x
- A$75^{\circ}$
- B$90^{\circ}$
- ✓$120^{\circ}$
- D$60^{\circ}$
✓
Answer
Correct option: C.
$120^{\circ}$
(C) $120^{\circ}$
Hint : In $\triangle A B C$
$\angle A+\angle B+\angle A C B=180^{\circ}$ $\quad$[angle sum property of a triangle]
$\therefore \angle A C B=180^{\circ}-\left(25^{\circ}+35^{\circ}\right)=120^{\circ}$
Now, $\angle A C D+\angle A C B=180^{\circ}$ $\quad$[linear pair]
$\therefore \quad \angle A C D=180^{\circ}-120^{\circ}=60^{\circ}$
So, $x=\angle O C D+\angle C D O$$\quad$[exterior angle of a triangle]
$x=60^{\circ}+60^{\circ}$
Hence, $x=120^{\circ}$
Hint : In $\triangle A B C$
$\angle A+\angle B+\angle A C B=180^{\circ}$ $\quad$[angle sum property of a triangle]
$\therefore \angle A C B=180^{\circ}-\left(25^{\circ}+35^{\circ}\right)=120^{\circ}$
Now, $\angle A C D+\angle A C B=180^{\circ}$ $\quad$[linear pair]
$\therefore \quad \angle A C D=180^{\circ}-120^{\circ}=60^{\circ}$
So, $x=\angle O C D+\angle C D O$$\quad$[exterior angle of a triangle]
$x=60^{\circ}+60^{\circ}$
Hence, $x=120^{\circ}$
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