Question

In $\Delta XYZ,\angle Y={90}^o$ and $\angle Z={40}^o$. Arranging sides (according to length) in ascending order, we get:

• A. $XY<YZ<YZ$
• B. $XY<YZ<ZX$
• C. $YZ<ZX<XY$
• D. $YZ<XY<ZX$

Answer 1

The correct option is B $XY<YZ<ZX$

$\Delta XYZ,\angle Y={90}^o$ and $\angle Z={40}^o$
$\therefore \angle X={50}^o(\because \angle X+\angle Y+\angle Z={180}^o)$
Now, arranging the angles in ascending order, we get $\angle Z,\angle X$ and $\angle Y$

Sides opposite to greater angle are greater in length and the angles and the opposite sides of a triangle have same relation.
Thus, arranging the sides in ascending order $XY<YZ<ZX.$