Conversion from
Sexagesimal to Circular System

Worked-out problems on the conversion from sexagesimal to circular system:

1. Express 40° 16’ 24” is radian.

Solution:

40° 16’ 24”

= 40° + 16’ + 24”

We know 1° = 60”

= 40° + 16’ + (24/60)’

= 40° + (16 + 2/5)’

= 40° + (82/5)’

We know 1° = 60’

= 40° + (82/5 × 60)°

= (40 + 41/150)°

= (6041/150)°

We know 180° = πc

Therefore, 6041°/150 = (πc/180) × (6041/150) = 6041/27000 πc

Therefore, 40° 16’ 24” = 6041/27000 πc


2. Show that 1° < 1c

Solution:

We know 180° = πc

or, 1° = (π/180)c

or, 1° = (22/7 × 180) c < 1c

Therefore, 1° < 1c

3. Two angles of a triangle are 75° and 45°. Find the value of the third angle in circular measure.

In ∆ABC, ∠ABC = 75° and ∠ACB = 45°; ∠BAC = ?

You know that the sum of the three angles of a triangle is 180°

Therefore, ∠BAC = 180° - (75° + 45°)

= 180° - 120°

= 60°

Again, we know: 180° = π

Therefore, 60° = 60 π/180 = π/3

In ΔABC, ∠BAC = π/3


4. A rotating ray revolves in the anticlockwise direction and makes two complete revolutions from its initial position and moves further to trace an angle of 30°. What are the sexagesimal and circular measures of the angle with reference to trigonometrical measure?

As the rotating ray does in the anti-clockwise direction, the angle formed is positive. We know, in one complete revolution the rotating ray traces an angle of 360°. So in two complete revolutions it makes an angle of 360° × 2 i.e. 720°. It has moved further to trace an angle of 30°. So the magnitude of the angle formed is (720° + 30°) i.e. 750°

Now, 180° = π

Therefore, 750° = 750 π/180 = 25 π/6


5. The ratio of the angles subtended at the centre by two unequal arcs of a circle is 5 : 3. If the magnitude of the second angle is 45°, find the sexagesimal and circular measures of the first angle.

Let the measure of the first angle be θ°

Then, according to the given condition, θ°/45° = 5/3

Therefore, θ° = 5/3 × 45° = 75°

Again we know, 180° = π

Therefore, 75° = 75 π/180 = 5 π/12

Therefore, the sexagesimal measure of the first angle is 75° and circular measure is 5 π/12.


6. ABC is an equilateral triangle in which AD is the line segment that joins the vertex A to the mid point of the side BC. What is the circular measure of ∠BAD?

Solution:

As ∆ABC is equilateral

Therefore, ∠BAC = 60°

We also know that the median of an equilateral triangle bisects the corresponding vertiealange. Therefore, ∠BAD = 30°

Therefore, the circular measure of ∠BAD = 30 π/180 = π/6

The above solved problems help us to learn in trigonometry, about the conversion from sexagesimal to circular system.

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