Ex 11.1, 4
Construct the following angles and verify by measuring them by a Protractor :
(i) 75°
75° = 60° + 15°
75° = 60° + (30°)/2
So, to we make 75° , we make 60° and then bisector of 30°
Steps of construction
 Draw a ray OA.
 Taking O as centre and any radius, draw an arc cutting OA at B.
 Now, taking B as centre and with the same radius as before, draw an arc intersecting the previously drawn arc at point C.

Now, with C as center, and same radius, draw another arc intersecting the previously drawn arc at point D.
Thus, ∠ AOE = 60°
and ∠ EOF = 60°  Taking C and D as Centre , with radius more than 1/2CD, draw arcs intersecting at P.

Join OP
Thus, ∠ EOP = 30°  Taking Q and C as Centre , with radius more than 1/2QC, draw arcs intersecting at R.
 Join OR
Thus, ∠ AOR = 75°
On measuring the ∠ AOR by protractor, we find that ∠ AOR = 75°
Thus, the construction is verified
(ii) 105°
105° = 90° + 15°
105° = 90° + (30°)/2
So, to make 105° , we make 90° and then bisector of 30°
Steps of construction
 Draw a ray OA.
 Taking O as centre and any radius, draw an arc cutting OA at B.
 Now, taking B as centre and with the same radius as before, draw an arc intersecting the previously drawn arc at point C.
 With C as centre and the same radius, draw an arc cutting the arc at D.
 With C and D as centres and radius more than 1/2 CD draw two arcs intersecting at P.
 Join OP. Thus, ∠ AOP = 90°
 Join OD ∠ POD = 30° as OP is bisector of ∠ COD.
 With Q and D as centres and radius more than 1/2 DQ draw two arcs intersecting at R.
Join OR
Thus, ∠ POR = 15°
∴ ∠ AOR = ∠ AOP + ∠ POR = 90° + 15°
∴ ∠ AOR = 105°
(iii) 135°
135° = 90° + 45°
So, to make 135° , we make 90° and then 45°
Steps of construction
 Draw a line OAA’ .
 Taking O as centre and any radius, draw an arc cutting OA at B.
 Now, taking B as centre and with the same radius as before, draw an arc intersecting the previously drawn arc at point C.
 With C as centre and the same radius, draw an arc cutting the arc at D.
 With C and D as centres and radius more than 1/2 CD draw two arcs intersecting at P.
 Join OP.
Thus, ∠ AOP = 90°
Also, ∠ A’OP = 90°
So, to we bisect ∠ A’OP
7.With B’ and Q as centres and radius more than 1/2 B’Q draw two arcs intersecting at R.
8.Join OR.
∴ ∠ A’OR = 45°
⇒ ∠ POR = 45°
Thus, ∠ AOR = ∠ AOP + ∠ POR = 90° + 35°
∴ ∠ AOR = 135°