Trigonometry
Worked examples
(AQA C2 Jan 2009) The diagram shows a sector OAB of a circle with
centre O and radius 10 cm.
The angle AOB is 0.8 radians.
(a) Find the area of the sector. (2 marks)
(b) (i) Find the perimeter of the sector OAB. (3 marks)
(ii) The perimeter of the sector OAB is equal to the perimeter of a square. Find the area of the square.
(2 marks)
Answer: (watch for units – the diagram has lengths in cms.)
a) sector area = 1/2 r2θ = ½ 100 0.8 = 40 sq. cms.
b) i) The perimeter is the arc length, plus 2 radii.
So perimeter = r θ + 2r = 8+20 = 28 cms.
ii) The square has a perimeter of 28 cms, so the length of the side of the square is 7 cms, and the
area is 49 sq. cms.
(AQA C2 Jan 2008
Answer: a) Area of the rectangle = 18 sq cm., so the area of the sector =9sq. cm.
Area of a sector is ½r2θ, so ½ 36θ=9 θ=9/18 = 0.5 radians
b) Length of the arc = rθ = 3 cm. So perimeter=6+6+3 = 15 cm.
(AQA C2 Jan 2009)
Answer:
a) Area = ½ X 7.4 X 5.26 X sin63° = 6.04 sq cms to 3 sig figs
b) (Cosine rule) a2 = 7.42 + 5.262-2 X 7.4 X 5.26 Xcos63°
=54.76 + 27.6676-35.342252
=47.085348
so a = 6.86 cm to 3 sig figs.
c) (Sine Rule) 5.26/sin B = 6.86/sin63
so sin B = 5.26 sin 63 /6.86 = 0.68 to 2 sig figs.
Answer: a) (Sine rule) a/sin72=18.7/sin50
so a = 23.2 cm
b) Area = ½ b c sinA = 1/2 18.7 23.2 sin 72 = 206 sq cm
AQA C2 Jan 2010
Answer: a)i) Area = ½ r2 θ = 0.5 225 1.2 = 135 sq cm.
ii) Arc length = rθ = 15 1.2 = 18 cm.
b) Use cosine rule to find AP=x
x2 = 152+102 – 2 15 10 cos 1.2 = 325 – 300 0.3624
=216.29
x=14.71 cm
PB=5cm
so perimeter = AP + PB + arc length = 14.71 + 5 + 18 = 37.7 cms to 3 sig figs
Edexcel C2 January 2005
4.
(a)
Show that the equation
5 cos2 x = 3(1 + sin x)
can be written as
5 sin2 x + 3 sin x – 2 = 0.
(b)
Hence solve, for 0  x < 360, the equation
5 cos2 x = 3(1 + sin x),
giving your answers to 1 decimal place where appropriate.
Answer:
a) cos2x+sin2x=1
so cos2x =1- sin2x
Substitute this in given equation:
5(1-sin2x)=3(1+sinx)
5-5 sin2x = 3 + 3sinx
5 sin2x + 3sinx - 2=0
b) (5sinx -2)(sinx +1)=0
so sinx=2/5 or sinx=-1
if sinx=-1, x=270°
if sinx=2/5, x=26.2°
or 180-26.2 = 153.8°
7.
Figure 1
B
8 cm
R
C
D
0.7 rad
11 cm
A
Figure 1 shows the triangle ABC, with AB = 8 cm, AC = 11 cm and  BAC = 0.7 radians. The arc BD,
where D lies on AC, is an arc of a circle with centre A and radius 8 cm. The region R, shown shaded in
Figure 1, is bounded by the straight lines BC and CD and the arc BD.
Find (a) the length of the arc BD,
(b)
the perimeter of R, giving your answer to 3 significant figures,
(c)
the area of R, giving your answer to 3 significant figures.
Answer: a) Arc length = rθ = 8 x 0.7 = 5.6 cms
b) CD = 11 – 8 = 3 cms
Find BC = x by Cosine Rule
x2 = 64 + 121 -2.88.cos0.7
=185-134.6 =50.39
x= 7.098
so perimeter = 5.6 + 3 +7.098 = 15.7 cms to 3 sig figs
c) Area of sector = ½ r θ2 = 0.5 64 0.49 = 15.68
Area of triangle = ½ b c sin A = 0.5 88 sin 0.7 = 28.35
So area of R = 28.35-15.68 = 12.7 sq cms
Edexcel C2 Jan 2006
5.
Figure 2
B
A
6m
5m
5m
O
In Figure 2 OAB is a sector of a circle, radius 5 m. The chord AB is 6 m long.
(a)
7
ˆ
Show that cos A O B = 25 .
(b)
ˆ
Hence find the angle A O B in radians, giving your answer to 3 decimal places.
(c)
Calculate the area of the sector OAB.
(d)
Hence calculate the shaded area
Answer: a) Use Cosine Rule
36 = 25+25-50 cos AOB
50 cos AOB = 50-36 = 14
cos AOB=14/50=7/25
b) AOB = cos-17/25 = 1.29 radians
c) Area of sector = ½ r2θ = 16.125 sq cms
d) Area of triangle = ½ base x height = ½ 6 4 = 12 sq. cms.
So area of shaded region = 16.125-12 = 4.125 sq cms
8.
(a)
Find all the values of  , to 1 decimal place, in the interval 0   < 360 for which
5 sin ( + 30) = 3.
(b)
Find all the values of  , to 1 decimal place, in the interval 0   < 360 for which
tan2  = 4.
Answer:
a) sin(θ+30)=0.6
so θ+30 = 36.87 or 180-36.87 =143.1
so θ = 6.9° or 113.1°
b) tan θ = ±2
θ = 63.4° or 270+63.4= 333.4°
OCR C2 Jan 2010
Answer:a) sin2x+cos2x=1
so sin2x=1-cos2x
Substitute in the given equation
2(1-cos2 x)=5cos x-1
2cos2x+5cos x-3=0
b) let cos x=c
(2c-1)(c+3)=0
so cos x = ½ or cos x = -3 (no real roots for this)
so x= π/4 or 7π/4 radians
Answer: i) Use Cosine Rule
132=102+142-2.10.14.cosA
280.cosA = 296-169 = 127
cos A=127/280 = 1.10 radians
ii) Arc length = rθ = 4 x 1.10 =4.40 cms.
So perimeter =4.40+8=12.4 cms
iii) Area = ½ r2 θ = ½ 16 X 1.10 = 8.80 sq cms
OCR C2 June 2009
Answer: i) Use Cosine Rule – angle will be opposite longest side
a2=b2+c2-2bccosA
2 x 6.4 x 7.0 cos A = 6.42+7.02 -11.32
89.6 cos A = 40.96 +49 – 127.69 = -37.73
cos A = -0.4211
A = 114.9°
ii) Area = ½ bc sinA = 1/2 x 6.4 x 7.0 x sin 114.9 = 20.32 sq cms.
Answer: i) 2x=45° or 180-45 = 135°
so x=22.5° or 67.5°
ii) sin2x=1-cos2x
so 2(1-cos2x)=2-√3 cos x
2cos2x-√3 cos x = 0
cos x(2cos x -√3)=0
cos x=0 or cos x = √3/2
x = π/2 or π/6 radians
OCR C2 Jan 2008
Answer: i) (-90,-2), (90,2)
ii)a) 180-α
b) –α
iii) 2 sin x = 2-3cos2x
2sin x =2-3(1-sin2x)
3sin2x-2sinx-1=0
(3sinx +1)(sinx-1)=0
sinx=-1/3 or sinx=1
x = -19.47° or -160.53° or 90°