In The given figure determine <P+<Q+<R+<S+<T

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$$\begin{gathered} \mathrm{By} \mathrm{linear} \mathrm{pair} \mathrm{axiom} \mathrm{we} \mathrm{have}, \\ \angle \mathrm{DEP}=180-\angle 5 \\ \angle \mathrm{EDP}=180-\angle 4 \\ \mathrm{In} △\mathrm{PDE}, \mathrm{sum} \mathrm{of} \mathrm{all} \mathrm{angles} \mathrm{will} \mathrm{be} 180. \\ \Rightarrow \angle \mathrm{P}+180-\angle 5+180-\angle 4=180 \\ \Rightarrow \angle \mathrm{P}=\angle 5+\angle 4-180 \\ \mathrm{Similarly} \\ \angle \mathrm{Q}=\angle 5+\angle 1-180 \\ \angle \mathrm{R}=\angle 1+\angle 2-180 \\ \angle \mathrm{S}=\angle 2+\angle 3-180 \\ \angle \mathrm{T}=\angle 3+\angle 4-180 \\ \mathrm{Adding} \mathrm{above} 5 \mathrm{equations}, \mathrm{we} \mathrm{get}, \\ \angle \mathrm{P}+\angle \mathrm{Q}+\angle \mathrm{R}+\angle \mathrm{S}+\angle \mathrm{T}=2\left(\angle 1+\angle 2+\angle 3+\angle 4+\angle 5\right)-900 \\ \mathrm{In} \mathrm{pentagon} \mathrm{ABCDE}, \\ \angle 1+\angle 2+\angle 3+\angle 4+\angle 5=540 \\ \Rightarrow \angle \mathrm{P}+\angle \mathrm{Q}+\angle \mathrm{R}+\angle \mathrm{S}+\angle \mathrm{T}=2\left(540\right)-900=1080-900 \\ \mathbf{\Rightarrow }\mathbf{\angle }\mathbf{P}\mathbf{+}\mathbf{\angle }\mathbf{Q}\mathbf{+}\mathbf{\angle }\mathbf{R}\mathbf{+}\mathbf{\angle }\mathbf{S}\mathbf{+}\mathbf{\angle }\mathbf{T}\mathbf{=}\mathbf{180}\mathbf{ }\left(\mathbf{Answer}\right) \end{gathered}$$

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