Page 178 - ICSE Math 6

Step 3: Without changing the opening of the compass, draw an
arc with Q as centre to intersect ray QR at S. Q S R

P

Step 4: With S as centre, draw an arc equal to radius DE to Q R
intersect the previous arc at P. S

Step 5: Join Q and P. Thus, ∠PQR is the required angle as ∠PQR P
= ∠ABC. Q
S R

To Construct a Bisector of an Angle A
If ∠ABC is such that ∠ABD and ∠CBD are adjacent angles and D
∠ABD = ∠CBD, then ray BD is known as the bisector of ∠ABC.
Let PQR be the angle to be bisected. To draw the bisector of ∠PQR, B
follow the steps given below. P C

Step 1: Using a compass, draw an arc of any suitable radius with
centre at Q, such that it intersects ray QP at S and ray
QR at T. Q R P

Step 2: With centre at T, draw an arc with radius more than half S
of TS. Q
P T R

Step 3: With centre at S, draw an arc with the same radius to S
intersect the arc drawn in step 2 at O. Q T O R P

Step 4: Join Q and O. Thus, ray QO is the bisector of ∠PQR as
∠PQO = ∠RQO. S O
Q T R
To Construct Angle of Standard Magnitude

An angle of 60°

To construct an angle of 60°, follow the steps given below.
Step 1: Draw ray AB with A as its initial point.
A B
Step 2: Using a compass, draw an arc of any suitable radius with
centre at A such that it intersects ray AB at C. A
C B
D
Step 3: Without changing the opening of the compass, draw an
arc to intersect the arc drawn earlier at D. A C B

Step 4: Join A and D. Thus, ∠DAB is the required angle of D
magnitude 60°.
A
C B

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