Question

In the fig. 3.23 seg AB$\cong$seg AC, seg PB $\cong$ seg PC, $\angle$A = 40$°$, $\angle$P = 140$°$, Find x and y.

Answer 1

$$\begin{gathered} \mathrm{Given}:\mathrm{seg}\mathrm{AB}\cong \hspace{0.17em}\mathrm{seg}\mathrm{AC}\mathrm{or}\mathrm{AB}=\mathrm{AC} \\ \mathrm{seg}\mathrm{PB}\cong \mathrm{seg}\mathrm{PC}\mathrm{or}\mathrm{PB}=\mathrm{PC} \\ \angle \mathrm{A}=40°,\angle \mathrm{P}=140° \\ \mathrm{In}△\mathrm{PBC}, \\ \angle \mathrm{PBC}=\angle \mathrm{PCB}=x\mathit{}(\because \mathrm{PB}=\mathrm{PC}) \\ \mathrm{Also},\angle \mathrm{BPC}+\angle \mathrm{PBC}+\angle \mathrm{PCB}=180°\left(\mathrm{Angle}\mathrm{sum}\mathrm{property}\right) \\ \Rightarrow 140°+2x\mathit{}=180° \\ \Rightarrow 2x=40° \\ \Rightarrow x\mathit{}=20°.....\left(1\right) \\ \mathrm{In}△\mathrm{ABC}, \\ \angle \mathrm{ABC}=\angle \mathrm{ACB}=x+y\mathit{}(\because \mathrm{AB}=\hspace{0.17em}\mathrm{AC}) \\ \mathrm{Also}, \\ \angle \mathrm{BAC}+\angle \mathrm{ABC}+\angle \mathrm{ACB}=180° \\ \Rightarrow 40°+2(x+y)=180° \\ \Rightarrow 2(x+y)=140° \\ \Rightarrow (x+y)=70° \\ \Rightarrow y\mathit{}=70°-20°=50°(\mathrm{from}(1\left)\right) \\ \mathrm{Thus},x=20°\mathrm{and}y=50° \end{gathered}$$