In the figure,if QT perp PR,angle TQR = 40^o | vedclass

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(C) In $\Delta TQR$,since $QT \perp PR$,$\angle QTR = 90^o$.By the angle sum property of a triangle,the sum of angles in $\Delta TQR$ is $180^o$:$\ang

Vedclass · Accepted Answer

$x=50^o$ and $y=80^o$

Vedclass · Answer

(C) In $\Delta TQR$,since $QT \perp PR$,$\angle QTR = 90^o$.By the angle sum property of a triangle,the sum of angles in $\Delta TQR$ is $180^o$:$\angle QTR + \angle TQR + \angle TRQ = 180^o$$90^o + 40^o + x = 180^o$$130^o + x = 180^o$$x = 180^o - 130^o = 50^o$Now,consider $\Delta P S R$. The exterior angle $y$ of $\Delta P S R$ is equal to the sum of the two interior opposite angles:$y = \angle SPR + \angle PRS$$y = 30^o + x$$y = 30^o + 50^o = 80^o$Thus,$x = 50^o$ and $y = 80^o$.

In the figure,if $QT \perp PR$,$\angle TQR = 40^o$ and $\angle SPR = 30^o$,find $x$ and $y$.

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