Question

In the fig. 3.9 seg NR $\cong$ seg NT, seg MR $\cong$ seg ST and $\angle$N = 70$°$ then find $\angle$S and $\angle$NRT

Answer 1

$$\begin{gathered} \mathrm{Given}:\mathrm{NR}=\mathrm{NT}\Rightarrow \angle \mathrm{NRT}=\angle \mathrm{NTR}, \\ \mathrm{MR}=\mathrm{ST},\angle \mathrm{N}=70° \\ \mathrm{In}△\mathrm{NRT}, \\ \angle \mathrm{RNT}+\angle \mathrm{NRT}+\angle \mathrm{NTR}=180° \\ \Rightarrow 70°+2\angle \mathrm{NRT}=180°(\because \angle \mathrm{NRT}=\angle \mathrm{NTR}) \\ \Rightarrow 2\angle \mathrm{NRT}=110° \\ \Rightarrow \angle \mathrm{NRT}=55° \\ \mathrm{In}△\mathrm{MNS}, \\ \mathrm{MR}=\mathrm{TS} \\ \Rightarrow \mathrm{MR}+\mathrm{RN}=\mathrm{TS}+\mathrm{TN} \\ \Rightarrow \mathrm{MN}=\mathrm{SN} \\ \Rightarrow \angle \mathrm{M}=\angle \mathrm{S} \\ \mathrm{Thus},\angle \mathrm{N}+\angle \mathrm{M}+\angle \mathrm{S}=180°\left(\text{a}\mathrm{ngle}\mathrm{sum}\mathrm{property}\right) \\ \Rightarrow 70°+2\angle \mathrm{S}=180° \\ \Rightarrow \angle \mathrm{S}=55° \end{gathered}$$