# Construction of Angles

#### Summary

**Construction of copy of an angle: **

An exact copy of a line segment can be constructed using a ruler and a compass.

- Construction of a copy of an angle YXZ.
- Draw a line AB.
- Mark any point O on AB.
- Place the compass pointer at vertex X of the given figure and draw an arc with a convenient radius, cutting rays XY and XZ at points E and F, respectively.
- Without changing the compass settings, draw an arc on line AB from point O. It cuts line AB at P.
- Set the compass to length EF.
- Without changing the compass settings, draw an arc from P cutting the previous arc at point Q.
- Join points O and Q.
- Hence, ∠POQ is the required copy of ∠YXZ.

**Construction of the bisector of an angle:**

- Construction of the bisector of an angle LMN.
- Place the compass pointer at vertex M of the given angle.
- Draw an arc cutting rays ML and MN at U and V respectively.
- Draw an arc with V as the centre and a radius more than half the length of UV in the interior of ∠LMN.
- Draw another arc with U as the centre and the same radius intersecting the previous arc.
- Name the point of intersection of the arcs as X.
- Join points M and X.
- Ray MX is the required bisector of ∠LMN.

**Construction of a 60° angle:**

- Draw a line.
- Mark a point P on the line.
- Draw an arc from point P with a convenient radius cutting the line at a point.
- Name the point of intersection of the arc and the line as Q.
- Draw another arc with Q as the centre and the same radius so that it passes through point P.
- Name the point of intersection of the two arcs as R.
- Join points P and R.
- Ray PR forms an angle with ray PQ at point P, which measures 60°.
- ∠QPR is the required angle measuring 60°.

**Construction of a 30° angle:**

- To obtain a 30° angle, we need to bisect a 60° angle.
- Draw an arc with Q as the centre and a radius more than half the length of QR.
- Draw another arc with R as the centre without changing the compass settings so that it intersects the previous arc.
- Name the point of intersection of the arcs as S.
- Join points P and S.
- ∠QPS is the required angle measuring 30°.

In a similar way, we can construct a 120° angle and 90° angle without using the protractor.

**Construction of a 120° angle**

- Draw line XY.
- Mark a point on the line and name it as P.
- Draw an arc with P as the centre and a convenient radius so that it cuts the line at Q.
- Draw another arc with Q as the centre without changing the compass settings so that it intersects the first arc at R.
- Draw another arc with R as the centre without changing the compass settings so that it intersects the first drawn arc at point S.
- Join points P and S.
- ∠SPQ is the required angle measuring 120°.

**Construction of a 90° angle**

- Draw line
*l*and mark point P on it. - Draw an arc with P as the centre and a convenient radius cutting line
*l*at Q. - Draw another arc with Q as the centre and the same radius cutting the first arc at R.
- Draw an arc with R as the centre and the same radius cutting the first arc at S.
- Join points P and R.
- Join points P and S.
- 90° lies to the centre of 60° and 120°.
- Draw an arc with R as the centre and a radius more than half the length of RS in the interior of ∠RPS.
- Draw another arc with S as the centre and the same radius so that it intersects the previous arc at T.
- Join points P and T.
- PT is the perpendicular line to PQ.
- ∠QPT is the required angle measuring 90°.

#### Videos
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#### Questions & Answers*arrow_upward*

### 1 . draw an angle of measure 153 degree

Answer.

- SME Approved

### 2 . how to draw an angle of 153 degree & 147 degree.

**To draw 147° using protractor.**

First draw a line segment or a ray OA

Makre O as centre and place t...

### 3 . what did you mean by bisector in maths

Bisector: If a line segment is divided into two equal parts by a line then it is called a bisector of the given line segment.

Angle...

### 4 . summative assesment 2

Soon we will be uploading these SA2 sample papers

### 5 . Prove that ∠ADC = ∠ABC.

Pl post each question individually.