Basic Engineering Mathematics, Fifth Edition

232 Basic Engineering Mathematics

7. Calculate the length of wire in the paper clip
   shown in Figure 26.6. The dimensions are in
   millimetres.

2.5rad

2.5rad

3rad

12

6

32

Figure 26.6

26.3 Radians and degrees

Oneradianis defined as the angle subtended at the

centre of a circle by an arc equal in length to the radius.

With reference to Figure 26.7, for arc lengths,

θradians=

s

r

s

r

O r



Figure 26.7

Whens=whole circumference(= 2 πr)then

θ=

s

r

=

2 πr

r

= 2 π

i.e. 2 πradians= 360 ◦orπradians= 180 ◦

Thus,1rad=

180 ◦

π

= 57. 30 ◦,correct to 2 decimal

places.

Sinceπrad= 180 ◦,then

π

2

= 90 ◦,

π

3

= 60 ◦,

π

4

= 45 ◦,

and so on.

Problem 5. Convert to radians: (a) 125◦

(b) 69◦ 47 ′

(a) Since 180◦=πrad, 1◦=

180

π

rad, therefore

125 ◦= 125

(π

180

)

rad=2.182 radians.

(b) 69◦ 47 ′= 69

47 ◦

60

= 69. 783 ◦(or, with your calcu-

lator, enter 69◦ 47 ′using◦’ ’ ’ function, press=

and press◦’ ’ ’ again).

and 69. 783 ◦= 69. 783

(π

180

)

rad

=1.218 radians.

Problem 6. Convert to degrees and minutes:

(a) 0.749 radians (b) 3π/4radians

(a) Sinceπrad= 180 ◦,1rad=

180 ◦

π

therefore 0.749rad= 0. 749

(

180

π

)◦

= 42. 915 ◦

0. 915 ◦=( 0. 915 × 60 )′= 55 ′, correct to the nea-

rest minute,

Hence, 0 .749 radians= 42 ◦ 55 ′

(b) Since 1rad=

(

180

π

)o

then

3 π

4

rad=

3 π

4

(

180

π

)o

=

3

4

( 180 )◦= 135 ◦

Problem 7. Express in radians, in terms ofπ,

(a) 150◦(b) 270◦(c) 37. 5 ◦

Since 180◦=πrad, 1 ◦=

π

180

rad

(a) 150◦= 150

(π

180

)

rad=

5 π

6

rad

(b) 270◦= 270

(π

180

)

rad=

3 π

2

rad

(c) 37. 5 ◦= 37. 5

(π

180

)

rad=

75 π

360

rad=

5 π

24

rad

Now try the following Practice Exercise

PracticeExercise 102 Radians and degrees

(answers on page 351)

1. Convert to radians in terms ofπ:
   (a) 30◦ (b) 75◦ (c) 225◦


2. Convert to radians, correct to 3 decimal
   places:
   (a) 48◦ (b) 84◦ 51 ′ (c) 232◦ 15 ′