Construct an angle of 270 degree & justify the construction. Is it same as 90 degrees or not? Justify your answer.

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$$\begin{gathered} \mathrm{Draw} \mathrm{a} \mathrm{line} \mathrm{segment} \mathrm{AB}. \\ \mathrm{Mark} \mathrm{the} \mathrm{centre} \mathrm{of} \mathrm{line} \mathrm{segment} \mathrm{AB} \mathrm{as} \mathrm{O}. \\ \mathrm{Using} \mathrm{compass}, \mathrm{measure} \mathrm{a} \mathrm{length} \mathrm{less} \mathrm{than} \mathrm{AO}. \\ \mathrm{Place} \mathrm{the} \mathrm{compass} \mathrm{at} \mathrm{A} \mathrm{and} \mathrm{cut} \mathrm{an} \mathrm{arc} \mathrm{downwards} \\ \mathrm{Place} \mathrm{the} \mathrm{compass} \mathrm{at} \mathrm{B} \mathrm{and} \mathrm{cut} \mathrm{an} \mathrm{arc} \mathrm{downwards} \mathrm{intersecting} \mathrm{previous} \mathrm{arc}. \\ \mathrm{The} \mathrm{perpendicular} \mathrm{bisector} \mathrm{of} \mathrm{the} \mathrm{line} \mathrm{segment} \mathrm{AB} \mathrm{is} \mathrm{constructed}. \\ \mathrm{Keeping} \mathrm{AB} \mathrm{as} \mathrm{a} \mathrm{diameter}, \mathrm{and} \mathrm{O} \mathrm{as} \mathrm{centre} , \mathrm{draw} \mathrm{a} 3 \mathrm{quarter} \mathrm{circle} \mathrm{as} \mathrm{shown}. \\ \mathrm{Mark} \mathrm{the} \mathrm{point} \mathrm{where} \mathrm{circle} \mathrm{meets} \mathrm{the} \mathrm{bisector} \mathrm{as} \mathrm{P}. \\ \mathrm{The} \mathrm{required} \mathrm{angle} \mathrm{measured} \mathrm{in} \mathrm{anticlockwise} \mathrm{direction} \mathrm{is} 270°. \\ \mathrm{It} \mathrm{is} \mathrm{the} \mathrm{same} \mathrm{as} 90°, \mathrm{since} \mathrm{angle} \mathrm{measured} \mathrm{in} \mathrm{clockwise} \mathrm{directio} \mathrm{is} 90° \mathrm{as} \mathrm{you} \mathrm{can} \mathrm{see} \mathrm{from} \mathrm{figure}. \end{gathered}$$

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