Chapter 3.3 - Relation Between Degree and Radian Measure

In the previous section, we saw the details of radian measure. In this section, we will see the relation between degree and radian measure.

The relation can be explained in 5 steps:
1. Consider the rotation of a ray of length 1 meter. If the ray completes one revolution, it will reach back at it’s original position. We have seen this situation in previous sections. It is shown again in fig.3.10 below.

Fig.3.10

2. We can see that, the length of arc AB = 2π. So the amount of rotation (angle) in radian measure is 2π^c.
3. Also recall that, the amount of this rotation (angle) in degree measure is 360^o.
4. So we can write:
Amount of rotation = 2π^c = 360^o.
• From this we get: Eq.3.2: π^c = 180^o.
• Using this relation, we can convert any  given radian measure to degree measure and vice versa.
5. Let us see some examples:
Convert 30^o, 45^o, 60^o, 90^o, 180^o, 270^o and 360^o into radian measure
Solution:
• From Eq.3.2, we have: π^c = 180^o
$\Rightarrow 1^o ={\frac{\pi}{180}}^\rm{c}$
• Thus we get:
$\begin{eqnarray} 30^o &=& 30 \times \frac{\pi}{180}={\frac{\pi}{6}}^\rm{c} \nonumber \\ 45^o &=& 45 \times \frac{\pi}{180}={\frac{\pi}{4}}^\rm{c} \nonumber \\ 60^o &=& 60 \times \frac{\pi}{180}={\frac{\pi}{3}}^\rm{c} \nonumber \\ 90^o &=& 90 \times \frac{\pi}{180}={\frac{\pi}{2}}^\rm{c} \nonumber \\ 180^o &=& 180 \times \frac{\pi}{180}={\pi}^\rm{c} \nonumber \\ 270^o &=& 270 \times \frac{\pi}{180}={\frac{3\pi}{2}}^\rm{c} \nonumber \\ 360^o &=& 360 \times \frac{\pi}{180}={2\pi}^\rm{c} \nonumber \ \end{eqnarray}$

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Notation convention

This can be written in 3 steps:
1. In scientific and engineering activities, angles are measured either in degrees or in radians.
2. Since both degrees and radians are used, scientists have agreed upon a convention. It is shown in figs.3.10(b) and (c)above.
3. In fig.b, we see that, the angle is written as 𝜃. But in fig.c, it is β
    ♦ If we see 𝜃, that angle is measured in degrees.
    ♦ If we see β, that angle is measured in radians.

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Let us see some solved examples:

Solved example 3.11
Convert the following degree measures into radian measures:
(i) 40^o (ii) 40^o20'
Solution:
Part (i):
• We have: π^c = 180^o
$\rm{\Rightarrow 1^o =\frac{\pi}{180}}$ ^c
• Thus we get:
$\rm{40^o =40 \times \frac{\pi}{180}=\frac{40 \pi}{180}}$ ^c
Part (ii):
1. Angle = 40^o20' indicates that, the ray first rotated by 40^o. Then it rotated further by 20'. Since 20' is much smaller than 1^o, it is difficult to draw on paper. But for numerical calculations, minutes and seconds are crucial.
2. We have already converted 40^o in part (i). So we will convert 20'
• We have: 60' = 1^o
$\rm{\Rightarrow 1' ={\frac{1}{60}}^o}$
$\rm{\Rightarrow 20' =20 \times {\frac{1}{60}}^o = {\frac{1}{3}}^o}$
3. So our next task is to convert $\rm{{\frac{1}{3}}^o}$ into radians.
• We have: π^c = 180^o
$\rm{\Rightarrow 1^o =\frac{\pi}{180}}$ ^c
• Thus we get:
$\rm{{\frac{1}{3}}^o = {\frac{1}{3}} \times \frac{\pi}{180}={\frac{\pi}{3 \times 180}}^c}$
4. So total rotation in radians = $\rm{\frac{40 \pi}{180}+\frac{\pi}{3 \times 180}={\frac{121\pi}{540}}^c}$

Solved example 3.12
Convert 6 radians into degree measure
Solution:
1. We have: π^c = 180^o
$\rm{\Rightarrow 1^c ={\frac{180}{\pi}}^o}$
• Thus we get:
$\rm{6^c =6 \times \frac{180}{\pi}=\frac{6 \times 180}{22/7}={343 \frac{7}{11}}^o}$
2. We got $\rm{{343 \frac{7}{11}}^o}$
• That means:
First the ray rotated through 343^o. Then it further rotated through $\rm{{\frac{7}{11}}^o}$.
• $\rm{{\frac{7}{11}}^o}$ is less than 1^o. So it is better to express it as minutes.
3. We have: 60' = 1^o
$\rm{\Rightarrow {\frac{7}{11}}^o ={\frac{7}{11}} \times 60 = {\frac{420}{11}}={38 \frac{2}{11}}'}$
4. We got $\rm{{38 \frac{2}{11}}'}$
• That means:
First it rotated through 38'. Then it further rotated through $\rm{{\frac{2}{11}}'}$.
• $\rm{{\frac{2}{11}}'}$ is less than 1'. So it is better to express it as seconds.
5. We have: 1' = 60''.
$\rm{\Rightarrow {\frac{2}{11}}' ={\frac{2}{11}}' \times {60} = 10.9''}$
6. Now we can calculate the total:
    ♦ From (2), we have 343^o
    ♦ From (4), we have 38'
    ♦ From (5) we have 10.9''
• So total rotation = 6^c = 343^o 38' 10.9''

Solved example 3.13
Find the radius of the circle in which a central angle of 60^o intercepts an arc of length 37.4 cm (use $\mathbf\small{\rm{\pi=\frac{22}{7}}}$)
Solution:
• This problem can be easily solved using Eq.3.1.
• But first we will try to solve it using the basics. It can be written in 4 steps:
1. In fig.3.11(a) below, the given circle is shown in blue color. A central angle of 60^o is intercepting an arc of length l = 0.374 meter. We are asked to find the radius r of the blue circle.

Fig.3.11

2. To find arc lengths or radii, we can make use of a unit circle. It is shown in green color in fig.b.
• A unit circle will have a circumference of 6.28 meter. This is far greater than 0.374 meter. So the unit circle will be larger than the given blue circle.
3. For the blue circle, let PQ be a fraction k of the total circumference 2πr.
• Then for the green circle also, the same fraction k will be applicable. This is because, the angle 60^o is same for both the circles.
• So for the green circle, AB is the fraction k of the total circumference.
• Thus we get:
    ♦ PQ = l = 2πrk
    ♦ AB = l₁ = 2πk
Taking ratio, we get: $\mathbf\small{{\frac{l}{l_1}}=\frac{2\pi rk}{2\pi k}=r}$
4. But l₁ is the radian measure of 60^o, which is equal to π/3
So we get:
$\mathbf\small{{\frac{l}{l_1}}=\frac{0.374\;m}{\pi /3}=\frac{0.374\;m \times 3}{\pi}=\frac{0.374\;m \times 3}{22/7}=r}$
• Thus r = 0.357 meter

Alternate method using Eq.3.1:
• We have: Angle in radians = $\mathbf\small{{\frac{l}{r}}}$
    ♦ So from fig.a, we get: $\mathbf\small{{\frac{\pi}{3}=\frac{0.374 \;m}{r}}}$
    ♦ Thus $\mathbf\small{{r=\frac{0.374 \times 3 \;m}{22/7}}}$ = 0.357 meter.

Solved example 3.14
The minute hand of a watch is 1.5 cm long. How far does it's tip move in 40 minutes. (use π = 3.14)
Solution:
1. In fig.3.12(a) below, OP is the initial position of the minute hand. After 40 minutes, it reaches OQ.

Fig.3.12

2. In 60 minutes, the minute hand rotates through 2π^c.
• So in 40 minutes, the minute hand rotates through $\mathbf\small{\frac{40}{60}\times\,2\pi={\frac{4\pi}{3}}^{\rm{c}}}$
• So in the fig.a, we get: $\mathbf\small{\beta={\frac{4\pi}{3}}^{\rm{c}}}$
3. Using Eq.3.1, we can write: $\mathbf\small{\beta={\frac{l}{r}}^{\rm{c}}}$
$\mathbf\small{\Rightarrow \frac{4\pi}{3}=\frac{l}{0.015}}$
$\mathbf\small{\Rightarrow l}$ = 0.0628 meter = 6.28 cm.

Solved example 3.15
If the arcs of the same lengths in two circles subtend angles 65^o and 110^o at the center, find the ratio of their radii.
Solution:
1. In fig.3.12(b) above, we see two circles. Magenta and orange.
2. In the magenta circle, OP is the initial side and OQ is the terminal side.
• The angle of rotation is 65^o
    ♦ This is $\mathbf\small{65 \times \frac{\pi}{180}={\frac{13\pi}{36}}^{\rm{c}}}$
• Arc length is l_P meter.
3. In the orange circle, OM is the initial side and ON is the terminal side.
• The angle of rotation is 110^o
    ♦ This is $\mathbf\small{110 \times \frac{\pi}{180}={\frac{11\pi}{18}}^{\rm{c}}}$
• Arc length is l_M meter.
4. From Eq.3.1, we have: $\mathbf\small{\text{Angle (in radians)}=\frac{\text{Arc}}{\text{radius}}}$
    ♦ Applying this to the magenta circle, we get: $\mathbf\small{\frac{13\pi}{36}=\frac{l_P}{r_P}}$
    ♦ Applying this to the orange circle, we get: $\mathbf\small{\frac{11\pi}{18}=\frac{l_M}{r_M}}$
5. Taking ratios, we get: $\frac{\frac{13\pi}{36}}{\frac{11\pi}{18}}=\frac{\frac{l_P}{r_P}}{\frac{l_M}{r_M}}$
$\Rightarrow \frac{13}{22}=\frac{l_P \;r_M}{l_M \; r_P}$
6. But it is given that, l_P = l_M
• So we get: $\frac{13}{22}=\frac{r_M}{r_P}$
7. The circle with radius r_P is the first circle given to us. So the required ratio is:
$\frac{r_P}{r_M}=\frac{22}{13}$

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Link to a few more solved examples is given below:

Solved examples 3.16 to 3.22

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In the next section, we will see the general form of trigonometric ratios using the unit circle.

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