Construct the angles of the following measurements:
(i) 30° (ii) \(22{\Large\frac{1}{2}}\)° (iii) 15°
Solution:
(i) We need to construct an angle of 60 degrees and then bisect it to get an angle measuring 30°.
Steps of Construction:
a) Draw a ray PQ.
b) To construct an angle of 60°, with P as a center and any radius, draw a wide arc to intersect PQ at R. With R as a center and same radius, draw an arc to intersect the initial arc at S. Then, ∠SPR = 60°
c)To bisect ∠SPR, with R and S as centers and radius greater than half of SR, draw two arcs to intersect at T. Join P and T
So, PT is the angle bisector. Hence, ∠TPR = 1/2 ∠SPR =30°
ii) We need to construct two adjacent angles of 60° and bisect the second one to get a 90° angle. This has to be bisected again to get a 45° angle. The 45° angle has to be further bisected to get \(22{\Large\frac{1}{2}}\)° angle.
\(22{\Large\frac{1}{2}}\)° = 45°/2
45° = 90°/2 = (30° + 60°)/2
Steps of Construction:
a) Draw a ray PQ.
b) To construct an angle of 60°, with P as a center and any radius, draw a wide arc to intersect PQ at R. With R as a center and same radius draw an arc to intersect the initial arc at S. Then, ∠SPR = 60°
c) To construct an adjacent angle of 60°, with S as the center and the same radius as before, draw an arc to intersect the initial arc at T. Then, ∠TPS = 60°
d) To bisect ∠TPS, with T and S as centers and radius greater than half of TS , draw arcs to intersect each other at Z. Join P and Z. Then, ∠ZPQ = 90°e) To bisect ∠ZPQ, with R and U as centers and radius more than half of RU, draw arcs to intersect each other at V. Join P and V. Then, ∠VPQ = 45°
e) To bisect ∠VPQ = 45°, with W and R as centers and radius greater than half of WR, draw arcs to intersect each other at X. Join P and X. PX bisects ∠VPQ.
Hence, ∠XPQ = 1/2 ∠WPQ
= 1/2 × 45°
= \(22{\Large\frac{1}{2}}\)°
(iii) We need to construct an angle of 60 degrees and then bisect it to get an angle measuring 30°. This has to be bisected again to get a 15° angle.
15° = 30°/2 = (60°/2)/2
Steps of Construction:
i) Draw a ray PQ.
ii) To construct an angle of 60°, with P as a center and any radius, draw a wide arc to intersect PQ at R. With R as a center and same radius draw an arc to intersect the initial arc at S. Then, ∠SPR = 60°
iv) Now, we will bisect ∠SPR. With R and S as centers and radius greater than half of RS, draw arcs to intersect each other at T. Join P and T. So, PT is the angle bisector of ∠SPR.
∠TPQ = 1/2 ∠SPR
= 1/2 × 60°
= 30°
iv) To bisect ∠TPQ, With R and W as centers and radius greater than half of arc RW, draw arcs to intersect each other at U. Join P and U. PU is the angle bisector of ∠TPQ .
∠UPQ = 1/2 ∠TPQ = 15°
☛ Check: NCERT Solutions for Class 9 Maths Chapter 11
Video Solution:
Construct the angles of the following measurements: (i) 30° (ii) \(22{\Large\frac{1}{2}}\)° (iii) 15°
Maths NCERT Solutions Class 9 Chapter 11 Exercise 11.1 Question 3
Summary:
It is given that we have to construct an angle of 30°, 22.5°, and 15° at the initial point of a given ray. We have drawn the angle using a compass and ruler and justified the construction.
☛ Related Questions:
- Construct an angle of 90° at the initial point of a given ray and justify the construction.
- Construct an angle of 45° at the initial point of a given ray and justify the construction.
- Construct the following angles and verify by measuring them by a protractor:(i) 75°(ii) 105°(iii) 135°
- Construct an equilateral triangle, given its side and justify the construction.