Question

If AB || CD, then find $\angle$PCD and $\angle$CPD from the adjoining figure 1.88.

Answer 1

Given: $\mathrm{AB}\parallel \mathrm{CD}$

We have,
$$\begin{gathered} \angle \mathrm{APC}+\angle \mathrm{CPD}=180°(\mathrm{Linear}\mathrm{pair}) \\ \Rightarrow 100°+\angle \mathrm{CPD}=180° \\ \Rightarrow \angle \mathrm{CPD}=80° \\ \mathrm{Also},\angle \mathrm{BAP}=\angle \mathrm{PDC}=56°(\mathrm{Alternate}\mathrm{angles}) \\ \mathrm{In}△\mathrm{CPD},\mathrm{we}\mathrm{have}, \\ \angle \mathrm{CPD}+\angle \mathrm{PDC}+\angle \mathrm{PCD}=180°[\mathrm{Angle}\mathrm{sum}\mathrm{property}] \\ \Rightarrow 80°+56°+\angle \mathrm{PCD}=180° \\ \Rightarrow 136°+\angle \mathrm{PCD}=180° \\ \Rightarrow \angle \mathrm{PCD}=44° \\ \therefore \angle \mathrm{CPD}=80°\mathrm{and}\angle \mathrm{PCD}=44° \end{gathered}$$