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Construction of Some Special Angles (75°, 135° and 150°)

Construction of an angle of 75°:
Steps of construction:
1. Draw a ray OA.

2. By taking centre at O and with the radius more than half of ray OA, draw an arc cutting OA
at D.

3. Taking D as the centre, draw another arc of the same radius as before cutting the previous
arc at P.
4. Join OP and extend it to E. Then, ∠AOE is the angle of 60°.
B C
5. Taking P as the centre draw another arc of the same radius as E
before to cut the first arc obtained in step 2 at E.
6. Draw two arcs of the same radius but greater than arc DE taking
respectively P and E as centres to intersect at B.

7. Join OB. Then, ∠AOB is the angle of 90°. E P
8. Now, bisect the ∠BOE. Let OC is the bisector.
Thus, ∠AOC is 60° + 15° = 75°. O D A

∠AOB = 90°, ∠AOP = 60°, \ ∠POB = 90° – 60° = 30°.

1 1
\ ∠POC = ∠POB = × 30° = 15°, \ ∠AOC = ∠AOP + ∠POC = 60° + 15° = 75°.
2 2
Construction of an angle of 135°:

Steps of construction: C
1. Draw a line segment AB and take a point O on it.
2. Draw ∠BOC = 90° by previous method. D

Thus, ∠AOC is also 90°.
3. Bisect ∠AOC. Let OD be the bisector of ∠AOC.
Therefore, ∠COD = 45°. A O B
Thus, ∠BOD is the required angle of 135°.

∠BOD = ∠BOC + ∠COD = 90° + 45° = 135°.

Construction of an angle of 150°:
Steps of construction:
1. Draw a line segment AB and take a point O on it. C

2. Draw ∠BOC = 120°
D
Thus, ∠AOC = 60° (180° – 120° = 60°) Q
3. Bisect ∠AOC. Let OD is the bisector of ∠AOC. 30° 120°

Therefore, ∠COD = 30°.
A O B
Thus, ∠BOD is the required angle of 150°.

∠BOD = ∠BOC + ∠COD = 120° + 30° = 150°.

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