Question

If the straight lines AB and XY intersect at the point $O$ and $\angle AOX=3\angle XOB$, then the four angles formed at $O$ are

• A. ${30}^0,{30}^0,{90}^0,{210}^0$
• B. ${30}^0,{30}^0,{150}^0,{150}^0$
• C. ${45}^0,{135}^0,{90}^0,{90}^0$
• D. ${45}^0,{45}^0,{135}^0,{135}^0$

Answer 1

The correct option is D ${45}^0,{45}^0,{135}^0,{135}^0$

Given lines AB and XY intersect at point O and $\angle AOX=3\angle XOB$

AB and XY inter sect at point O then opposite angle are equal each others

Then$\angle XOB=\angle AOY$

And $\angle AOX=\angle BOY=3\angle XOB$

We know that sum of all angle at a point total is 360

Then $\angle AOX+\angle BOX+\angle BOY+\angle AOY={360}^0$

$\Rightarrow 3\angle XOB+\angle XOB+3\angle XOB+\angle XOB={360}^0$

$\Rightarrow 8\angle XOB={360}^0\Rightarrow \angle XOB={45}^0$

$\angle XOB=\angle AOY={45}^0$

$\angle AOX=\angle BOY=3\angle XOB=3\times 45={135}^0$