$\begin{array}{l}\text { In } \triangle A P Q, \angle B+\angle D=\angle A Q P \ldots \ldots \text { (1) (ext. } \angle \text { of } \triangle) \\ \angle C+\angle E=\angle A P Q \ldots \ldots \text { (2) (ext. } \angle \text { of } \triangle) \\ \begin{array}{l} \angle A+\angle B+\angle C+\angle D+\angle E=\angle A+\angle A Q P+\angle A P Q \text { (by (1) and (2)) } \\ \quad=180^{\circ}(\angle \mathrm{s} \text { sum of } \triangle)\end{array} \\ \begin{aligned} \therefore a=180\end{aligned}\end{array}$
G1.1 In the given diagram, $\angle A+\angle B+\angle C+\angle D+\angle E=a^{\circ}$, find $a$.