The equilateral triangle ABP lies inside the square ABCD. Find angle  CPD also draw a digram .

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Please find below the solution to the asked query :


$$\begin{gathered} \mathrm{Here} \mathrm{PA}=\mathrm{PB}=\mathrm{AB}=\mathrm{BC}=\mathrm{AD}\hspace{0.17em} \\ \angle \mathrm{PAD}=90-60=30 = \angle \mathrm{PBC} \\ \mathrm{Now} \mathrm{BP}=\mathrm{BC} \mathrm{so} \\ \angle \mathrm{BCP}=\angle \mathrm{BPC}=\frac{180-30}{2}=75= \angle \mathrm{APD} \\ \mathrm{So} \mathrm{angle} \mathrm{around} \mathrm{point} \mathrm{P} \mathrm{is} \\ \angle \mathrm{CPD}+60+75+75=360 \\ \angle \mathrm{CPD}= 150 \mathrm{degree} \end{gathered}$$
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