The construction of angles is an essential aspect of geometry and is one of the basic needs for learning geometric construction. The term construction in geometry means drawing the shapes, lines, or angles accurately with the help of some instruments. When you want to construct angles, you will require a pair of a compass or a protractor, a ruler (straightedge), and a pencil. In this article, let’s study the concept of Construction of Angles in detail.

Construction of Angles and Triangles

Acute Angle : An angle between \({{\rm{0}}^{\rm{o}}}\) to \({\rm{9}}{{\rm{0}}^{\rm{o}}}\) is known as an acute angle. The figure given below \(\left( {50^\circ } \right)\) is an acute angle.

If all the angles of a triangle are less than \({\rm{9}}{{\rm{0}}^{\rm{o}}},\) then such a triangle is known as an acute triangle. In equilateral triangles, the measurement of each angle is \({\rm{6}}{{\rm{0}}^{\rm{o}}},\) so equilateral triangles are acute triangles.

In the figure given below, the triangle with \({55^{\rm{o}}} – {55^{\rm{o}}} – {70^{\rm{o}}}\) is also an acute-angled triangle.

Obtuse Angle: An angle that is more than \({\rm{9}}{{\rm{0}}^{\rm{o}}}\) but less than \({\rm{18}}{{\rm{0}}^{\rm{o}}}\) is known as an obtuse angle. The figure shown below is \({\rm{12}}{{\rm{0}}^{\rm{o}}},\) which is an obtuse angle.

In an obtuse triangle, only one angle will be obtuse, and then the remaining two angles will be acute angles such that the sum of the angles in the triangle should be \({\rm{18}}{{\rm{0}}^{\rm{o}}}.\) Examples of two obtuse-angled triangles (\({130^{\rm{o}}} – {25^{\rm{o}}} – {25^{\rm{o}}}\) and \({110^{\rm{o}}} – {40^{\rm{o}}} – {30^{\rm{o}}}\)) are shown below.

Right Angle: An angle, whose measure is exactly \({\rm{9}}{{\rm{0}}^{\rm{o}}},\) is known as a right angle.

In a right triangle, one angle is \({\rm{9}}{{\rm{0}}^{\rm{o}}}.\) The longer side of the right triangle, which is opposite to the right angle, is known as hypotenuse. The right-angled triangles are shown below:

Learn How to Construct Triangles

Construction of Angles Using The Protractor

Protractor: It is an instrument for measuring or constructing an angle of a given measure. It is a semicircular piece of a geometrical instrument made of metal or plastic, which is marked in degrees from \({{\rm{0}}^{\rm{o}}}\) to \({\rm{18}}{{\rm{0}}^{\rm{o}}}\) from left to right as well as from right to left, as shown in the given diagram:

The centre \(0\) of this piece is also the midpoint of its baseline.

Example: Find the measure of each one of the angles given below by using a protractor.

Method: (i) To measure \(∠AOB,\) place the protractor so that its centre is precisely on the vertex \(O\) of the angle, and the baseline lies along the arm \(OA.\) Read off the mark through which the arm \(OB\) passes, starting from \({{\rm{0}}^{\rm{o}}}\) on the side of \(A,\) as shown in the given diagram:

Thus, we find that \(\angle AOB = {39^{\rm{o}}}\)

(ii) To measure \(∠PQR,\) place the protractor so that its centre is exactly on the vertex \(Q\) of the angle, and the baseline lies along the arm \(PQ.\) Read off the mark through which the arm \(QR\) passes, starting from \({{\rm{0}}^{\rm{o}}}\) on the side of \(P,\) as shown in the figure. Thus, we find that \(\angle PQR = {128^{\rm{o}}}.\)

Construction of Angles Using Compass and Ruler

The following steps must be followed for easy and error-construction of angles

Construction of \(90^\circ \) and \(45^\circ \) Using Compass and Ruler

For the construction of an angle using a compass and a ruler, you need to follow the given steps:

We will construct an angle of \({\rm{9}}{{\rm{0}}^{\rm{o}}}\) and then construct an angle bisector to bisect the \({\rm{9}}{{\rm{0}}^{\rm{o}}}\) such that we will get an angle of \({\rm{4}}{{\rm{5}}^{\rm{o}}}.\)

1. First, you have to draw a ray and name it \(AB.\)
2. Next, with \(A\) as the centre and any width as radius, you have to draw a semicircular arc that touches the line segment \(AB\) and mark it as \(Q.\)
3. Now, without changing the radius, you draw another arc with \(Q\) as the centre and label this point as \(D.\)
4. You have to repeat the same process with \(D\) as the centre and label the arc drawn as \(E.\)
5. Now, with the points \(D\) and \(E\) as the centre and taking the same radius, draw two arcs such that they intersect each other at the point \(F.\)
6. Now, join the point \(A\) and \(F,\) and this line is known as perpendicular line, and it makes \(90\) degrees with the line segment \(AB.\) So, \(\angle BAF = {90^{\rm{o}}}\)
7. You can label the point of intersection of the semicircular arc with the line \(AF\) as \(P.\)
8. Finally, using the same radius to draw an arc in step \(2,\) draw two arcs with \(P\) and \(Q\) as centres. Mark the point of intersection of these arcs as \(R.\) Now join the points \(A\) and \(R,\angle RAB\) and \(\angle FAR\) are equal to \({\rm{4}}{{\rm{5}}^{\rm{o}}}\) each.

Draw an Angle Equal to a Given Angle

Draw an angle equal to a given angle \(∠ABC.\)

Steps of construction:

1. Draw a ray named \(OX.\)
2. With \(B\) as a centre and any radius, draw an arc cutting \(BA\) and \(BC\) at \(P\) and \(Q,\) respectively.

3. With \(O\) as a centre and the same radius, draw an arc, cutting, \(OX\) at \(R.\)
4. With \(R\) as centre and radius as \(PQ,\) cut the arc through \(R\) at \(S.\)
5. Join \(OS\) and produce it to any point \(Y.\)
Then, \(∠XOY\) is the required angle equal to \(∠ABC.\)

Verification: Measure \(∠ABC\) and \(∠XOY.\)
You would find that \(∠XOY=∠ABC.\)

Bisect a Given Angle

An angle \(∠AOB\) is given. We have to bisect the angle. 

Steps of construction:

1. With \(O\) as a centre and any convenient radius, draw an arc, cutting \(OA\) and \(OB\) at \(P\) and \(Q.\)
2. With centre \(P\) and radius more than \(\frac{1}{2}\left( {PQ} \right),\) draw an arc.
3. With centre \(Q\) and the same radius as before, draw another arc, cutting the previously drawn arc at a point \(R.\)
4. Join \(OR\) and produce it to any point \(X.\) then, ray \(OX\) bisects \(∠AOB.\)

Verification: Measure \(∠AOX\) and \(∠BOX.\)

You would find that \(∠AOX=∠BOX.\)

Solved Examples on Construction of Angles

Q.1. Construct an angle of \({60^{\rm{o}}}\) using a compass.
Ans: Steps of construction
1. Draw a ray \(OA.\)
2. With \(O\) as a centre and any suitable radius, draw an arc cutting \(OA\) at a point \(B.\)
3. With \(B\) as a centre and the same radius as before, draw another arc to cut the previous arc at \(C.\)
4. Join \(OC\) and produce it to \(D.\)
Now, \(\angle AOD = {60^{\rm{o}}}\)

Q.2. Construct an angle of \({120^{\rm{o}}}\) using a pair of compasses.
Ans: Steps of construction
1. Draw a ray \(OA.\)
2. With \(O\) as a centre and any suitable radius, draw an arc cutting \(OA\) at \(B.\)
3. With \(B\) as a centre and the same radius, cut the arc at \(C.\) Again, with \(C\) as a centre and the same radius, cut the arc at \(D.\)
4. Join \(OD\) and produce it to \(E.\)
Then, \(\angle AOE = {120^{\rm{o}}}.\)

Q.3. Construct an angle of \({30^{\rm{o}}}\) using a pair of compasses.
Ans: Steps of construction
1. Construct an angle \(\angle AOD = {60^{\rm{o}}}\) as stated in \(Q.1\) above.
2. Draw the bisector \(OE\) of \(∠AOD.\)
Then, \(\angle AOE = {30^{\rm{o}}}.\)

Q.4. Construct an angle of \({90^{\rm{o}}}\) using a pair of compasses.
Ans: Steps of construction
1. Draw a ray \(OA.\)
2. With \(O\) as a centre and any suitable radius, draw an arc, cutting \(OA\) at \(B.\)
3. With \(B\) as a centre and the same radius, cut the previously drawn arc at \(C,\) with \(C\) as a centre and the same radius, cut the arc at \(D.\)
4. With \(C\) as a centre and radius more than half \(CD,\) draw an arc.
5. With \(D\) as a centre and the same radius, draw another arc to cut the previous arc at \(E.\)
6. Join \(OE.\)
Then, \(\angle AOE = {90^{\rm{o}}}.\)

Q.5. Construct an angle of \({45^{\rm{o}}}\) using a pair of compasses.
Ans: Steps of construction
Draw \(\angle AOE = {90^{\rm{o}}}\) as stated in \(Q.4\) above.
Draw the bisector \(OF\) of \(∠AOE.\)
Then, \(\angle AOF = {45^{\rm{o}}}.\)

Summary

In this article, we discussed about construction of angles with examples. Then, we have shown how to construct angles using a protractor. We also discussed the construction of angles using compass and ruler for few standard angles like \({30^{\rm{o}}},{45^{\rm{o}}},{60^{\rm{o}}},{90^{\rm{o}}},{120^{\rm{o}}},\) that are frequently used in geometry.

We also discussed how to construct an angle equal to a given angle and how to bisect an angle. Few solved examples are provided for a better understanding of the construction steps.

FAQs on Construction of Angles

Here we have provided the answers to the most frequently asked questions on construction of angles:

Q.1. How do you construct all angles? Ans: You can construct any angles by using the compass, ruler, or protractor known as mathematical instruments used to construct the angles.

Q.2. How do you construct an angle in geometry? Ans: In Euclidean geometry, an angle is a diagram that is formed by using two rays, known as the sides of the angle, which share a common endpoint, known as the vertex of the angle. Angles are formed by using two rays that lie in the plane that contains the rays. The intersection of two planes also forms angles.

Q.3. What are the steps to construct a 90-degree angle? Ans: Steps of construction 1. Draw a ray \(OA.\) 2. With \(O\) as a centre and any suitable radius, draw an arc, cutting \(OA\) at \(B.\) 3. With \(B\) as a centre and the same radius, cut the previously drawn arc at \(C,\) with \(C\) as a centre and the same radius, cut the arc at \(D.\) 4. With \(C\) as a centre and radius more than half \(CD,\) draw an arc. 5. With \(D\) as a centre and the same radius, draw another arc to cut the previous arc at \(E.\) 6. Join \(OE.\) Then, \(\angle AOE = {90^{\rm{o}}}.\)

Q.4. How do you construct a 30 degree angle? Ans: Steps of construction Construct an angle \(\angle AOD = {60^{\rm{o}}}.\) Draw the bisector \(OE\) of \(∠AOD.\) Then, \(\angle AOE = {30^{\rm{o}}}.\)

Q.5. Which type of angle is 60 degrees? Ans: A \(60\) degrees angle is known as an acute angle as it is less than \(90\) degrees. \({60^{\rm{o}}}\) in radians is \(\frac{\pi }{3}\) and the measure of each angle of an equilateral triangle that is \({60^{\rm{o}}}.\)