Sexagesimal, Centesimal and Circular Systems

We know, Sexagesimal, Centesimal and Circular Systems are the three different systems of measuring angles. Sexagesimal system is also known as English system and centesimal system is known as French system.

To convert the one system to the other system its very necessary to know the relation among the Sexagesimal system, Centesimal system and Circular system.

The relation between Sexagesimal, Centesimal and Circular systems are discussed below:

Since 90° = 1 right angle, hence, 180° = 2 right angles.
Again, 100g = 1 right angle; hence, 200g = 2 right angles.
And, πc = 2 right angles.
Therefore, 180° = 200g = πc.

Let, D°, Gg and Rc be the sexagesimal, centesimal and circular measures respectively of a given angle.
Now, 90° = 1 right angle
Therefore, 1° = 1/90 right angle
Therefore, D° = D/90 right angle

Again, 100g = 1 right angle
Therefore, 1g = 1/100 right angle
Therefore, Gg = G/100 right angle.

And, 1c = 2/π right angle
Therefore, Rc = 2R/π right angle.

Therefore we have,
D/90 = G/100 = 2R/π
D/180 = G/200 = R/π

1. The circular measure of an angle is π/8; find its value in sexagesimal and centesimal systems.



= 180°/8, [Since, πc = 180°)

= 22°30'

Again, πc/8

= 200g/8 [Since, πc = 200g)

= 25g

Therefore, the sexagesimal and centesimal measures of the angle πc/8 are 22°30' and 25g respectively.

2. Find in sexagesimal, centesimal and circular units an internal angle of a regular Hexagon.


We know that the sum of the internal angles of a polygon of n sides = (2n - 4) rt. angles.

Therefore, the sum of the six internal angles of a regular pentagon = (2 ×  6 - 4) = 8 rt. angles.

Hence, each internal angle of the Hexagon = 8/6 rt. angles.  =  4/3 rt. angles.

Therefore, each internal angle of the regular Hexagon in sexagesimal system measures 4/3  ×   90°,  (Since, 1 rt. angle = 90°) = 120°;

In centesimal system measures

4/3 × 100g (Since, 1 rt. angle = 100g)

= (400/3)g

= 1331/3

and in circular system measures (4/3 × π/2)c, (Since, 1 rt. angle = πc/2)

= (2π/3)c.

3. The angles of a triangle are in A. P. If the greatest and the least are in the ratio 5 : 2, find the angles of the triangle in radian.


Let (a - d), a and (a + d) radians (which are in A. P.) be the angles of the triangle where a> 0 and d > 0.

Then, a - d + a + a + d = π, (Since, the sum of the three angles of a triangle = 180° = π radian)

or, 3a = π

or, a =  π/3.

By problem, we have,

(a + d)/(a – d) = 5/2

or, 5(a – d) = 2(a + d)

or, 5a - 5d = 2a + 2d.

or, 5a – 2a = 2d + 5d

or, 3a = 7d

or, 7d = 3a

or, d = (3/7)a

or, d = (3/7)  × ( π/3)

or, d = π/7

Therefore, the required angles of the triangle are (π/3- π/7), π/3 and (π/3 + π/7) radians

i.e.,  4π/21, π/3 and 10π/21 radians.

Measurement of Angles

11 and 12 Grade Math

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