Question

Construct the angles of the following measurements:

${30}^{\circ }$

${15}^{\circ }$

${22.5}^{\circ }$

Answer 1

(i) 30°

• Steps of Construction:
  

1.Taking O as centre and

some radius , draw an arc of a circle, which intersects OA, say at a point B.

2.Taking B as centre and

with the same radius as before, draw an arc intersecting the previously drawn

arc, say at a point C.

3.Draw the ray OE

passing through C. Then ∠EOA = 60°.

4. Taking B and C as

centres and with the radius more than 1/2 BC, draw arcs to intersect each other, at D.

5. Draw the ray OD. This

ray OD is the bisector of the ∠ EOA, i.e.,

∠EOD = ∠AOD =1/2 ∠EOA = 1/2(60°) = 30∘.

∠AOD =30∘.

(ii) 22½°

• Steps of Construction:
  

1. Taking O as centre and

some radius, draw an arc of a circle, which intersects OA, at a point B.

2.Taking B as centre and

with the same radius as before, draw an arc intersecting the previously drawn

arc,at a point C .

3.Taking C as centre and

with the same radius as before, draw an arc intersecting the arc drawn in step

1, at D.

4. Draw the ray OE

passing through C. Then ∠EOA = 60∘.

5. Draw the ray OF

passing through D. Then ∠ FOE = 60∘.

6. Next, taking C and D

as centres and with radius more than 1/2 CD, draw arcs to intersect each other, say at G.

7. Draw the ray OG. This

ray OG is the bisector of the ∠ FOE, i.e.,

∠FOG = ∠EOG = 1/2 ∠FOE = 1/2 (60∘) = 30∘.

Thus, ∠GOA = ∠GOE + ∠EOA = 30∘ + 60∘ = 90∘.

8. Now, taking O as

centre and any radius, draw an arc to intersect the rays OA and OG, say at H

and I.

9. Next, taking H and I

as centres and with the radius more than 1/2HI, draw arcs to intersect each other, at J.

10. Draw the ray OJ. This ray OJ is the bisector of the∠ GOA.

i.e.,

∠GOJ = ∠AOJ = 1/2 ∠GOA = 1/2( 90∘) = 45∘.

11. Now, taking O as centre and any radius, draw an arc to intersect

the rays OA and OJ, say at K and L.

12. Next, taking K and L as centres and with the radius more

than 1/2KL, draw arcs to

intersect each other, at M.

13. Draw the ray OM. This ray OM is the bisector of the angle AOJ,

i.e., ∠JOM = ∠AOM = 1/2 ∠AOJ = 1/2( 45∘ ) = 22½∘

∠AOM = 22½∘

(iii) 15∘

• Steps of construction:
  

1. Taking O as centre and

some radius, draw an arc of a circle, which intersects OA, at a point B.

2. Taking B as centre and

with the same radius as before, draw an arc intersecting the previously drawn

arc,at a point C.

3. Draw the ray OE

passing through C. Then ∠EOA =60∘.

4.Now, taking B and C as

centres and with the radius more than 1/2BC, draw arcs to intersect each other, at D.

5. Draw the ray OD

intersecting the arc drawn in step 1 at F. This ray OD is the bisector of the

angle EOA,

i.e., ∠EOD = ∠AOD = 1/2 ∠EOA = 1/2 (60∘) = 30∘.

6. Now, taking B and F as

centres and with the radius more than 1/2 BF, draw arcs to intersect each other, at G.

7. Draw the ray OG. This

ray OG is the bisector of the∠AOD, i.e.,

∠DOG = ∠AOG = 1/2 ∠AOD = 1/2 (30∘) = 15∘.

∠AOG = 15∘.