Question

In fig 3.7 and fig. 3.8, seg AB $\cong$ seg AC, seg BP $\cong$ seg PC, $\angle$A = 50$°$, $\angle$P = 110$°$. Then find $\angle$ABP and $\angle$ACP and hence find the relation between them.

Answer 1

$$\begin{gathered} \mathrm{In}\mathrm{fig}\left(1\right), \\ \angle \mathrm{A}=50°,\angle \mathrm{P}=110°, \\ \mathrm{AB}=\mathrm{AC},\mathrm{BP}=\mathrm{PC} \\ \mathrm{Let}\angle \mathrm{ABC}=\angle \mathrm{ACB}=x\mathit{}\mathrm{and} \\ \angle \mathrm{CBP}=\angle \mathrm{BCP}=y \\ \mathrm{Then},\mathrm{in}\mathrm{ACPB}, \\ \angle \mathrm{A}+\angle \mathrm{B}+\angle \mathrm{C}+\angle \mathrm{D}=360°\left(\text{a}\mathrm{ngle}\mathrm{addition}\mathrm{property}\mathrm{of}\text{a}\mathrm{quadrilateral}\right) \\ \Rightarrow 50°+x+y+x+y+110°=360° \\ \Rightarrow 2(x+y)=200° \\ \Rightarrow (x+y)=100° \\ \mathrm{Thus},\angle \mathrm{ABP}=\angle \mathrm{ACP}=100°. \end{gathered}$$

$$\begin{gathered} \mathrm{In}\mathrm{fig}\left(2\right), \\ \mathrm{AB}=\mathrm{AC},\mathrm{BP}=\mathrm{PC},\angle \mathrm{A}=50°,\angle \mathrm{P}=110° \\ \mathrm{Let}\angle \mathrm{B}=\angle \mathrm{C}=\mathit{}y \\ \mathrm{Let}\angle \mathrm{PBC}=\angle \mathrm{PCB}=\mathit{}x \\ \mathrm{In}△\mathrm{ABC}, \\ 2\mathrm{y}+50°=180°\left(\text{a}\mathrm{ngle}\mathrm{sum}\mathrm{property}\right) \\ \Rightarrow 2y=130° \\ \Rightarrow y=65°...\left(\mathrm{i}\right) \\ \text{In}△\mathrm{PBC}, \\ 2x+110°=180° \\ \Rightarrow 2x=70° \\ \Rightarrow x=35°...\left(\mathrm{ii}\right) \\ \mathrm{Thus},\mathrm{from}\left(\mathrm{i}\right)\mathrm{and}\left(\mathrm{ii}\right), \\ \angle \mathrm{ABP}=\angle \mathrm{ACP}=\left(y-x\right)=30° \end{gathered}$$