Higher Engineering Mathematics

138 GEOMETRY AND TRIGONOMETRY

Problem 1. If the diameter of a circle is 75 mm,

find its circumference.

Circumference,c=π×diameter=πd
=π(75)=235.6 mm.

Problem 2. In Fig. 14.4,ABis a tangent to

the circle atB. If the circle radius is 40 mm and

AB=150 mm, calculate the lengthAO.

Figure 14.4

A tangent to a circle is at right angles to a radius
drawn from the point of contact, i.e.ABO= 90 ◦.
Hence, using Pythagoras’ theorem:

AO^2 =AB^2 +OB^2

AO=

√

(AB^2 +OB^2 )=

√

[(150)^2 +(40)^2 ]

= 155 .2mm

Now try the following exercise.

Exercise 63 Further problems on properties

of circles

1. If the radius of a circle is 41.3 mm, calculate
   the circumference of the circle.
   [259.5 mm]


2. Find the diameter of a circle whose perimeter
   is 149.8 cm. [47.68 cm]


3. A crank mechanism is shown in Fig. 14.5,
   whereXYis a tangent to the circle at pointX.
   If the circle radiusOXis 10 cm and length
   OY is 40 cm, determine the length of the
   connecting rodXY.

O 40 cm Y

X

Figure 14.5 [38.73 cm]

14.3 Arc length and area of a sector

Oneradianis defined as the angle subtended at the

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

radius. With reference to Fig. 14.6, for arc lengths,

θradians=s/rorarc length, s=rθ (1)

whereθis in radians.

Figure 14.6

Whens=whole circumference (= 2 πr) then

θ=s/r= 2 πr/r= 2 π.

i.e. 2πrad= 360 ◦or πrad= 180 ◦

Thus 1 rad= 180 ◦/π= 57. 30 ◦, correct to 2 decimal

places.

Sinceπrad= 180 ◦, thenπ/ 2 = 90 ◦,π/ 3 = 60 ◦,

π/ 4 = 45 ◦, and so on.

Area of a sector=

θ

360

(πr^2 )

whenθis in degrees

=

θ

2 π

(πr^2 )=

1

2

r^2 θ (2)

whenθis in radians

Problem 3. Convert to radians: (a) 125◦

(b) 69◦ 47 ′.

(a) Since 180◦=π rad then 1◦=π/180 rad,

therefore

125 ◦= 125

( π

180

)c

=2.182 rad

(Note thatcmeans ‘circular measure’ and indi-

cates radian measure.)

(b) 69◦ 47 ′= 69

47 ◦

60

= 69. 783 ◦

69. 783 ◦= 69. 783

( π

180

)c

=1.218 rad

Problem 4. Convert to degrees and minutes:

(a) 0.749 rad (b) 3π/4 rad.