In the figure,lines XY and MN intersect at O. | vedclass

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(A) $XOY$ is a straight line.Therefore,$\angle b + \angle a + \angle POY = 180^o$ (Linear pair axiom).Given that $\angle POY = 90^o$,so $\angle b + \a

Vedclass · Accepted Answer

$126$

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(A) $XOY$ is a straight line.Therefore,$\angle b + \angle a + \angle POY = 180^o$ (Linear pair axiom).Given that $\angle POY = 90^o$,so $\angle b + \angle a + 90^o = 180^o$,which implies $\angle b + \angle a = 90^o$.Given $a: b = 2: 3$,let $a = 2x$ and $b = 3x$.Then $2x + 3x = 90^o$,so $5x = 90^o$,which gives $x = 18^o$.Thus,$\angle a = 2 	imes 18^o = 36^o$ and $\angle b = 3 	imes 18^o = 54^o$.Since lines $XY$ and $MN$ intersect at $O$,$\angle XON$ and $\angle MOY$ are vertically opposite angles,and $\angle XOM$ and $\angle NOY$ are vertically opposite angles.Here,$\angle c = \angle XOM = \angle a + \angle POY$ is not correct based on the figure. Looking at the figure,$\angle c = \angle XON$.Since $XOY$ is a straight line,$\angle XON + \angle NOY = 180^o$.Also,$\angle NOY = \angle b = 54^o$ (vertically opposite angles).Therefore,$\angle c = 180^o - 54^o = 126^o$.

In the figure,lines $XY$ and $MN$ intersect at $O$. If $\angle POY = 90^o$ and $a: b = 2: 3$,find $c$. (in $^o$)

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