Question

In the fig seg NR $\cong$ seg NT, seg MR $\cong$ seg ST and $\angle N={70}^{\circ }$ then find $\angle S$ and $\angle NRT$

Answer 1

In $△NRT$,
$NR=NT$
$\angle NRT=\angle NTR=x$ (isosceles triangle property)
Sum of angles = 180
$\angle NRT+\angle NTR+\angle RNT=180$
$x+x+70=180$
$2x=110$
$x={55}^{\circ }$
Hence, $\angle NRT={55}^{\circ }$
In $△MNS$,
$MN=NS$ (Given, NR = NT and MR = ST)
$\angle S=\angle M=y$ (isosceles triangle property)
Sum of angles = 180
$\angle S+\angle M+\angle N=180$
$y+y+70=180$
$2y=110$
$y={55}^{\circ }$
Hence, $\angle S={55}^{\circ }$