Page 207 - Maths Class 06

B
Step 3. Sim i larly, tak ing E as the cen tre and with the same ra dius (as in step 2).
draw an arc in ter sect ing the pre vi ous arc at P. Join AP and pro duce it to Q
¾® D
get AQ. (See Fig. 14.42)
P
¾®
Thus, ray AQ is the required bisector of ÐCAB or ÐBAC.
A C
E
Fig. 14.42

Construction of Angles 60°, 90° and 120°
To construct an Angle of 60°.

¾® A B
Step 1. Draw any ray AB.
Fig. 14.43
P
¾®
Step 2. Tak ing A as the cen tre and with any suit able ra dius, draw an arc PQ
A B
that cuts AB at Q. Q
Fig. 14.44
P R

Step 3. Tak ing Q as the cen tre and ra dius equal to AQ, draw an arc cut ting the
pre vi ous arc PQ at R. A B
Q
C
Fig. 14.45

¾® R
Step 4. Join AR and pro duce it to get AC. P
Step 5. ÐBAC is the re quired an gle equal to 60°.
A 60° B
Q
Fig. 14.46
To Construct an Angle of 90°

¨ A B C
Step 1. Draw a line AC and mark a point B on it.
Fig. 14.47

Step 2. Tak ing B as the cen tre and with any suit able ra dius,
¨
draw an arc PQ cut ting AC at P and Q. A P B Q C
Fig. 14.48

D

Step 3. Tak ing P and Q as the cen tres and with any con ve nient
ra dius, draw arcs in ter sect ing each other at D.

A P B Q C
Fig. 14.49

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