trigonometry

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, AB is a tangent to
the circle at B. If the circle radius is 40 mm and
AB = 150 mm, calculate the length AO.
14.3 Arc length and area of a sector
One radian is 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 length s,
θ radians = s/r or arc length, s = rθ (1)
where θ is in radians.
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,
where XY is a tangent to the circle at point X.
If the circle radius OX is 10 cm and length
OY is 40 cm, determine the length of the
connecting rod XY.
O
X
40 cm
Y
Figure 14.6
When s = whole circumference (= 2πr) then
θ = s/r = 2πr/r = 2π.
i.e. 2π rad = 360 ◦ or
π rad = 180 ◦
Thus 1 rad = 180 ◦ /π = 57.30 ◦ , correct to 2 decimal
places.
Since π rad = 180 ◦ , then π/2 = 90 ◦ , π/3 = 60 ◦ ,
π/4 = 45 ◦ , and so on.
Area of a sector =
θ
360 (πr2 )
when θ is in degrees
= θ
2π (πr2 ) = 1 2 r2 θ (2)
when θ is in radians
Problem 3.
(b) 69 ◦ 47 ′ .
Convert to radians: (a) 125 ◦
(a) Since 180 ◦ = π rad then 1 ◦ = π/180 rad,
therefore
( π
) c
125 ◦ = 125 = 2.182 rad
180
(Note that c means ‘circular measure’ and indicates
radian measure.)
(b) 69 ◦ 47 ′ = 69 47◦
60 = 69.783◦
( π
) c
69.783 ◦ = 69.783 = 1.218 rad
180
Figure 14.5
[38.73 cm]
Problem 4. Convert to degrees and minutes:
(a) 0.749 rad (b) 3π/4 rad.