Question

Draw an angle of ${135^\circ }$ and bisect it.

Answer 1

Hint: First draw a line AB and taking O as center and a radius (whatever is convenient), mark an arc. Now draw two arcs that are intersecting each other above point O and join the point to O. Then we will draw the bisector of $\angle ROA$ named OD which will make $\angle DOB = {135^ \circ }$ . Then draw the bisector of this newly drawn angle.

Complete step-by-step answer:
We have to draw a given angle and bisect it.
So the steps of construction of angle of ${135^\circ }$ are given as-
1.First we will draw a line AB and take its middle point as O. The line is given as-

2.Now taking O as center and a radius (whatever is convenient, suppose $4$ cm) , mark an arc which will intersect the line AB at P and Q. The diagram is given as-

3.Now taking P and Q as centers and radius more than half the length of PQ, draw two arcs that are intersecting each other above point O at R.

4.Then join the points O and R. This makes \[\angle ROB = \angle ROA = {90^ \circ }\]

5.Now we will draw the bisector of $\angle ROA$ named OD. This bisector will divide the angle into two equal parts so each angle will become ${45^ \circ }$ and then we can get, $\angle DOB = \angle DOR + \angle ROB = {90^ \circ } + {45^ \circ } = {135^ \circ }$.
To draw the bisector first mark points P and T. Then by taking P and T as centers and radius more than half the length of PQ, draw an arc. Name the point of their intersection D and join OD.

6.Now we also have to bisect this angle so we will draw the bisector OE which bisects the angle ${135^ \circ }$ into two equal parts using the process in step five.

This is the required diagram.

Note: We can also draw the angle of ${135^ \circ }$ by using a protractor. Follow these steps-
1.First follow step one given in the question then place the protractor on point O so that the line OB and protector line coincide.
2.Mark point D on the angle ${135^ \circ }$ and join OD.
3.Then draw the bisector of the angle using the process of step five given in the question.