Construct an angle of 45° at the initial point of a given ray and justify the construction.

**Answer**

Steps of construction:

Step 1: A ray OY is drawn.

Step 2: With O as a centre and any radius, an arc ABC is drawn cutting OY at A.

Step 3: With A as a centre and the same radius, mark a point B on the arc ABC.

Step 4: With B as a centre and the same radius, mark a point C on the arc ABC.

Step 5: With A and B as centre, draw two arcs intersecting each other with the same radius at X.

Step 6: X and Y are joined and a ray making an angle 90° with YZ is formed.

Step 7: With A and E as centres, two arcs are marked intersecting each other at D and the bisector of ∠XOY is drawn.

Justification for construction:

By construction,

∠XOY = 90°

We constructed the bisector of ∠XOY as DOY.

Thus,

∠DOY = 1/2 ∠XOY

∠DOY = 1/2×90° = 45°

**Answer**

(i) 30°

Steps of constructions:

Step 1: A ray OY is drawn.

Step 2: With O as a centre and any radius, an arc AB is drawn cutting OY at A.

Step 3: With A and B as centres, two arcs are marked intersecting each other at X and the bisector of is drawn.

Thus, ∠XOY is the required angle making 30° with OY.

(ii) 22.5°

Steps of constructions:

Step 1: An angle ∠XOY = 90° is drawn.

Step 2: Bisector of ∠XOY is drawn such that ∠BOY = 45° is constructed.

Step 3: Again, ∠BOY is bisected such that ∠AOY is formed.

Thus, ∠AOY is the required angle making 22.5° with OY.

(iii) 15°

Steps of constructions:

Step 1: An angle ∠AOY = 60° is drawn.

Step 2: Bisector of ∠AOY is drawn such that ∠BOY = 30° is constructed.

Step 3: With C and D as centres, two arcs are marked intersecting each other at X and the bisector of ∠BOY is drawn.

Thus, ∠XOY is the required angle making 15° with OY.

Construct the following angles and verify by measuring them by a protractor:

(i) 75° (ii) 105° (iii) 135°

**Answer**

(i) 75°

Steps of constructions:

Step 1: A ray OY is drawn.

Step 2: An arc BAE is drawn with O as a centre.

Step 3: With E as a centre, two arcs are A and C are made on the arc BAE.

Step 4: With A and B as centres, arcs are made to intersect at X and ∠XOY = 90° is made.

Step 5: With A and C as centres, arcs are made to intersect at D

Step 6: OD is joined and and ∠DOY = 75° is constructed.

Thus, ∠DOY is the required angle making 75° with OY.

(ii) 105°

Steps of constructions:

Step 1: A ray OY is drawn.

Step 2: An arc ABC is drawn with O as a centre.

Step 3: With A as a centre, two arcs are B and C are made on the arc ABC.

Step 4: With B and C as centres, arcs are made to intersect at E and ∠EOY = 90° is made.

Step 5: With B and C as centres, arcs are made to intersect at X

Step 6: OX is joined and and ∠XOY = 105° is constructed.

Thus, ∠XOY is the required angle making 105° with OY.

(iii) 135°

Steps of constructions:Step 1: A ray DY is drawn.

Step 2: An arc ACD is drawn with O as a centre.

Step 3: With A as a centre, two arcs are B and C are made on the arc ACD.

Step 4: With B and C as centres, arcs are made to intersect at E and ∠EOY = 90° is made.

Step 5: With F and D as centres, arcs are made to intersect at X or bisector of ∠EOD is constructed.

Step 6: OX is joined and and ∠XOY = 135° is constructed.

Thus, ∠XOY is the required angle making 135° with DY.

**Answer**

Steps of constructions:

Step 1: A line segment AB=4 cm is drawn.

Step 2: With A and B as centres, two arcs are made.

Step 4: With D and E as centres, arcs are made to cut the previous arc respectively and forming angle of 60° each.

Step 5: Lines from A and B are extended to meet each other at C.

Thus, ABC is the required triangle formed.

Justification:

By construction,

AB = 4 cm, ∠A = 60° and ∠B = 60°

We know that,

∠A + ∠B + ∠C = 180° (Sum of the angles of a triangle)

⇒ 60° + 60° + ∠C = 180°

⇒ 120° + ∠C = 180°

⇒ ∠C = 60°

BC = CA = 4 cm (Sides opposite to equal angles are equal)

AB = BC = CA = 4 cm

∠A = ∠B = ∠C = 60°