question_answer 1)

Draw an angle of measure \[{{147}^{o}}\] and construct its bisector.

Answer:

                Here, we firstly draw an angle of \[{{147}^{o}}\] and then construct its bisector. So, we use the following steps: Step I Draw \[{{l}_{1}},{{l}_{2}},{{l}_{3}},{{l}_{4}},{{l}_{5}}\] of any length place the centre of the protractor at A and the zero edge along AB. Step II Start with zero near B. Mark a point C at \[{{147}^{o}}\]. Step III Join AC. Then, \[{{l}_{6}}\]is an angle of measure \[{{147}^{o}}\]. Step IV Now, with A as centre and using compasses, draw an arc that cuts both rays of \[{{l}_{1}},{{l}_{2}},{{l}_{3}},{{l}_{4}}\] at P and Q. Step V   With P as centre, draw (in the interior of \[{{l}_{5}}\]) an arc whose radius is more than half of the length of PQ. Step VI With the same radius and with Q as centre, draw another arc in the interior of \[{{l}_{1}},{{l}_{2}}\] Let the two arcs intersect at D. Join \[{{l}_{3}}\] Then, \[{{l}_{1}},{{l}_{2}},{{l}_{3}}\] is the required bisector of \[{{l}_{4}}\].