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Trigonometry Practice 1 (Calculator) [123 marks]

A ship leaves port A on a bearing of 030∘ . It sails a distance of 25 km to point B. At B, the ship changes direction to a bearing of
100∘ . It sails a distance of 40 km to reach point C. This information is shown in the diagram below.

1a. Find the distance the second ship will travel.

[4 marks]

Markscheme
finding AB̂C = 110∘ (= 1.92 radians) (A1)
evidence of choosing cosine rule (M1)
e.g. AC2 = AB2 + BC2 − 2(AB)(BC) cos AB̂C
correct substitution A1
e.g. AC2 = 252 + 402 − 2(25)(40) cos 110∘
AC = 53.9 (km) A1

1b. Find the bearing of the course taken by the second ship. [3 marks]

Markscheme
METHOD 1
correct substitution into the sine rule A1
sin BÂC sin 110 ∘
e.g. 40
= 53.9
A1

BÂC = 44.2∘
bearing = 074∘ A1 N1
METHOD 2
correct substitution into the cosine rule A1
402 − 252 − 53.9 2
e.g. cos BÂC = −2(25)(53.9)
A1

BÂC = 44.3∘
bearing = 074∘ A1 N1
[3 marks]
The diagram below shows triangle PQR. The length of [PQ] is 7 cm , the length of [PR] is 10 cm , and PQ̂R is 75∘ .

2a. Find PR̂Q . [3 marks]

Markscheme
choosing sine rule (M1)
sin R sin 75∘
correct substitution 7
= 10
A1

sin R = 0.676148 …
PR̂Q = 42.5∘ A1 N2
[3 marks]

2b.
Find the area of triangle PQR. [3 marks]

Markscheme
P = 180 − 75 − R
P = 62.5 (A1)
substitution into any correct formula A1
1
e.g. area ΔPQR = × 7 × 10 × sin(their P)
2

= 31.0 (cm2) A1 N2
[3 marks]

The diagram below shows a circle centre O, with radius r. The length of arc ABC is 3π cm and AÔC = 9
.

3a. Find the value of r. [2 marks]

Markscheme
evidence of appropriate approach M1
e.g. 3π = r 2π9

r = 13.5 (cm) A1 N1
[2 marks]

3b. Find the perimeter of sector OABC. [2 marks]

Markscheme
perimeter = 27 + 3π (cm) (= 36.4) A1 N2
[2 marks]

3c.
Find the area of sector OABC. [2 marks]

Markscheme
evidence of appropriate approach M1
1 2 2π
e.g. 2
× 13.5 × 9

area = 20.25π (cm2 ) (= 63.6) A1 N1

[2 marks]

The following diagram shows a pole BT 1.6 m tall on the roof of a vertical building.
The angle of depression from T to a point A on the horizontal ground is 35∘ .
The angle of elevation of the top of the building from A is 30∘ .

4.
Find the height of the building. [7 marks]
Markscheme
METHOD 1
appropriate approach M1
e.g. completed diagram
attempt at set up A1
e.g. correct placement of one angle
tan 30 = h
x
, tan 35 = h+1.6
x
A1A1
attempt to set up equation M1
e.g. isolate x
correct equation A1
e.g. h
tan 30
= h+1.6
tan 35

h = 7.52 A1 N3
METHOD 2
sin 30 = h
l
A1

in triangle ATB, Â = 5∘ , Tˆ = 55∘ A1A1

choosing sine rule M1
correct substitution
h/ sin 30 1.6
e.g. sin 55
= sin 5
A1
1.6×sin 30×sin 55
h= sin 5
A1

h = 7.52 A1 N3
[7 marks]

A circle centre O and radius r is shown below. The chord [AB] divides the area of the circle into two parts. Angle AOB is θ .

5a.
Find an expression for the area of the shaded region. [3 marks]

Markscheme
substitution into formula for area of triangle A1
1
e.g. 2
r × r sin θ

evidence of subtraction M1
correct expression A1 N2
e.g. 12 r2 θ − 12 r2 sin θ , 12 r2 (θ − sin θ)

[3 marks]
5b.
The chord [AB] divides the area of the circle in the ratio 1:7. Find the value of θ . [5 marks]

Markscheme
1
evidence of recognizing that shaded area is 8
of area of circle M1
1
e.g. 8
seen anywhere

setting up correct equation A1

e.g. 12 r2 (θ − sin θ) = 18 πr2

eliminating 1 variable M1
e.g. 12 (θ − sin θ) = 18 π , θ − sin θ = π
4

attempt to solve M1
e.g. a sketch, writing sin x − x + π4 = 0

θ = 1.77 (do not accept degrees) A1 N1

[5 marks]

The following diagram shows a circle with centre O and radius 4 cm.

The points A, B and C lie on the circle. The point D is outside the circle, on (OC).

6a.

Markscheme
choosing sine rule (M1)
correct substitution A1
e.g. sin 0.8
= sin 0.3

AD = 9.71 (cm) A1 N2
[3 marks]

6b.
Find OD. [4 marks]
Markscheme
METHOD 1
finding angle OAD = π − 1.1 = (2.04) (seen anywhere) (A1)
choosing cosine rule (M1)
correct substitution A1
e.g. OD = 9.71 + 42 − 2 × 9.71 × 4 × cos(π − 1.1)
2 2

OD = 12.1 (cm) A1 N3
METHOD 2
finding angle OAD = π − 1.1 = (2.04) (seen anywhere) (A1)
choosing sine rule (M1)
correct substitution A1
OD 9.71 4
e.g. = sin 0.8
= sin 0.3
sin(π−1.1)

OD = 12.1 (cm) A1 N3
[4 marks]

6c.
Find the area of sector OABC. [2 marks]

Markscheme
correct substitution into area of a sector formula (A1)
e.g. area = 0.5 × 42 × 0.8
area = 6.4 (cm2 ) A1 N2
[2 marks]

6d.
Find the area of region ABCD. [4 marks]

Markscheme
substitution into area of triangle formula OAD (M1)
correct substitution A1
e.g. A= 12 × 4 × 12.1 × sin 0.8 , A= 12 × 4 × 9.71 × sin 2.04 , A= 12 × 12.1 × 9.71 × sin 0.3

subtracting area of sector OABC from area of triangle OAD (M1)

e.g. area ABCD = 17.3067 − 6.4
area ABCD = 10.9 (cm2 ) A1 N2
[4 marks]
The circle shown has centre O and radius 3.9 cm.

Points A and B lie on the circle and angle AOB is 1.8 radians.

7a.
Find AB. [3 marks]

Markscheme
METHOD 1
choosing cosine rule (M1)
substituting correctly A1
−−−−−−−−−−−−−−−−−−−−−−−−
e.g. AB = √3.92 + 3.92 − 2(3.9)(3.9) cos 1.8

AB = 6.11 (cm) A1 N2
METHOD 2
evidence of approach involving right-angled triangles (M1)
substituting correctly A1
e.g. sin 0.9 = x
3.9
, 12 AB = 3.9 sin 0.9

AB = 6.11 (cm) A1 N2
METHOD 3
choosing the sine rule (M1)
substituting correctly A1
sin 0.670… sin 1.8
e.g. 3.9
= AB

AB = 6.11 (cm) A1 N2
[3 marks]

7b.
Find the area of the shaded region. [4 marks]
Markscheme
METHOD 1

reflex AÔB = 2π − 1.8 (= 4.4832) (A2)

correct substitution A = 12 (3.9)2 (4.4832 …) A1

area =34.1 (cm2) A1 N2

METHOD 2
finding area of circle A = π(3.9)2 (= 47.78 …) (A1)
finding area of (minor) sector A = 12 (3.9)2 (1.8) (= 13.68 …) (A1)

subtracting M1
e.g. π(3.9)2 − 0.5(3.9)2 (1.8) , 47.8 − 13.7
area = 34.1 (cm2) A1 N2
METHOD 3

finding reflex AÔB = 2π − 1.8 (= 4.4832) (A2)

finding proportion of total area of circle A1
2π−1.8
e.g. 2π
× π(3.9)2 , θ

× πr2

area = 34.1 (cm2) A1 N2

[4 marks]

The diagram below shows a triangle ABD with AB =13 cm and AD = 6.5 cm.
Let C be a point on the line BD such that BC = AC = 7 cm.

8a.
Find the size of angle ACB. [3 marks]
Markscheme
METHOD 1
evidence of choosing the cosine formula (M1)
correct substitution A1
72 + 72 − 132
e.g. cos AĈB = 2×7×7

AĈB = 2.38 radians (= 136∘ ) A1 N2

METHOD 2
evidence of appropriate approach involving right-angled triangles (M1)
correct substitution A1

7

[3 marks]

Markscheme
METHOD 1

AĈD = π − 2.381 (180 − 136.4) (A1)

evidence of choosing the sine rule in triangle ACD (M1)
correct substitution A1
6.5 7
e.g. sin 0.760…
=

C ÂD = π − (0.760 … + 0.836 …) (180 − (43.5 … + 47.9 …))

= 1.54 (= 88.5∘ ) A1 N3
METHOD 2

evidence of choosing the sine rule in triangle ABD (M1)

correct substitution A1
6.5 13
e.g. sin 0.380…
=

C ÂD = π − 0.836 … − (π − 2.381 …) (= 180 − 47.9 … − (180 − 136.4))

= 1.54 (= 88.5∘ ) A1 N3

Note: Two triangles are possible with the given information. If candidate finds AD̂C = 2.31 (132∘ ) leading to C ÂD = 0.076
(4.35∘ ) , award marks as per markscheme.
[5 marks]
The following diagram shows the triangle ABC.

9a. Find AĈB .

[4 marks]

Markscheme
correct substitution into the formula for the area of a triangle A1
1 1
e.g. 2
× 5 × 13.6 × sin C = 20 , 2
× 5 × h = 20

attempt to solve (M1)

8
e.g. sin C = 0.5882 … , sin C = 13.6

AĈB = 144∘ (2.51 radians) A1 N3

[4 marks]

9b.
Find AB. [3 marks]

Markscheme
evidence of choosing the cosine rule (M1)
correct substitution A1
e.g. (AB)2 = 52 + 13.62 − 2(5)(13.6) cos 143.968 …
AB = 17.9 A1 N2
[3 marks]

The diagram below shows a quadrilateral ABCD with obtuse angles AB̂C and AD̂C .

AB = 5 cm, BC = 4 cm, CD = 4 cm, AD = 4 cm , BÂC = 30∘ , AB̂C = x∘ , AD̂C = y∘ .

−−−−−−−−−−
10a.
Use the cosine rule to show that AC = √41 − 40 cos x . [1 mark]
Markscheme
correct substitution A1
e.g. 25 + 16 − 40 cos x , 52 + 42 − 2 × 4 × 5 cos x
−−−−−−−−−−
AC = √41 − 40 cos x AG
[1 mark]

10b.
Use the sine rule in triangle ABC to find another expression for AC. [2 marks]

Markscheme
correct substitution A1
AC 4 1
e.g. sin x
= sin 30
, 2
AC = 4 sin x
4 sin x
AC = 8 sin x (accept sin 30
) A1 N1

[2 marks]

10c.
(i) Hence, find x, giving your answer to two decimal places. [6 marks]

(ii) Find AC .

Markscheme
(i) evidence of appropriate approach using AC M1
−−−−−−−−−−
e.g. 8 sin x = √41 − 40 cos x , sketch showing intersection
correct solution 8.682 …, 111.317 … (A1)
obtuse value 111.317 … (A1)
x = 111.32 to 2 dp (do not accept the radian answer 1.94 ) A1 N2
(ii) substituting value of x into either expression for AC (M1)
e.g. AC = 8 sin 111.32
AC = 7.45 A1 N2
[6 marks]

(ii) Hence, or otherwise, find the area of triangle ACD.

Markscheme
(i) evidence of choosing cosine rule (M1)
a2+ c2 − b2
e.g. cos B = 2ac

correct substitution A1
42 + 42 − 7.45 2
e.g. 2×4×4
, 7.452 = 32 − 32 cos y , cos y = −0.734 …

y = 137 A1 N2
(ii) correct substitution into area formula (A1)
1
e.g. 2
× 4 × 4 × sin 137 , 8 sin 137

area = 5.42 A1 N2
[5 marks]

The diagram below shows a circle with centre O and radius 8 cm.

The points A, B, C, D, E and F are on the circle, and [AF] is a diameter. The length of arc ABC is 6 cm.

11a.
Find the size of angle AOC . [2 marks]

Markscheme
appropriate approach (M1)
e.g. 6 = 8θ

AÔC = 0.75 A1 N2
[2 marks]

11b.
Hence find the area of the shaded region. [6 marks]
Markscheme
evidence of substitution into formula for area of triangle (M1)
1
e.g. area = × 8 × 8 × sin(0.75)
2

area = 21.8 … (A1)

evidence of substitution into formula for area of sector (M1)
1
e.g. area = × 64 × 0.75
2

area of sector = 24 (A1)

evidence of substituting areas (M1)
e.g. 12 r2 θ − 12 ab sin C , area of sector − area of triangle

area of shaded region = 2.19 cm2 A1 N4

[6 marks]

11c.
The area of sector OCDE is 45 cm2 . [2 marks]

Find the size of angle COE .

Markscheme
attempt to set up an equation for area of sector (M1)
e.g. 45 = 12 × 82 × θ

C ÔE = 1.40625 (1.41 to 3 sf) A1 N2

[2 marks]

11d.
Find EF . [5 marks]
Markscheme
METHOD 1
attempting to find angle EOF (M1)
e.g. π − 0.75 − 1.41

EÔF = 0.985 (seen anywhere) A1

evidence of choosing cosine rule (M1)
correct substitution A1
−−−−−−−−−−−−−−−−−−−−−−−−
e.g. EF = √82 + 82 − 2 × 8 × 8 × cos 0.985
EF = 7.57 cm A1 N3
METHOD 2
attempting to find angles that are needed (M1)
e.g. angle EOF and angle OEF

EÔF = 0.9853 … and OÊF (or OF̂E) = 1.078 … A1

evidence of choosing sine rule (M1)
correct substitution (A1)
EF 8
e.g. sin 0.985
= sin 1.08

EF = 7.57 cm A1 N3
METHOD 3
attempting to find angle EOF (M1)
e.g. π − 0.75 − 1.41

EÔF = 0.985 (seen anywhere) A1

evidence of using half of triangle EOF (M1)
e.g. x = 8 sin 0.985
2

correct calculation A1
e.g. x = 3.78
EF = 7.57 cm A1 N3
[5 marks]

Consider the following circle with centre O and radius 6.8 cm.

The length of the arc PQR is 8.5 cm.

12a.
Find the value of θ . [2 marks]
Markscheme
correct substitution (A1)
8.5
e.g. 8.5 = θ(6.8) , θ = 6.8

θ = 1.25 (accept 71.6∘ ) A1 N2

[2 marks]

12b.
Find the area of the shaded region. [4 marks]

Markscheme
METHOD 1
correct substitution into area formula (seen anywhere) (A1)
e.g. A = π(6.8)2 , 145.267 …
correct substitution into area formula (seen anywhere) (A1)
e.g. A = 12 (1.25)(6.82 ) , 28.9

valid approach M1
e.g. π(6.8)2 − 12 (1.25)(6.82 ) ; 145.267 … − 28.9 ; πr2 − 12 r2 sin θ

A = 116 (cm2 ) A1 N2
METHOD 2
attempt to find reflex angle (M1)
e.g. 2π − θ , 360 − 1.25
correct reflex angle (A1)

AÔB = 2π − 1.25 (= 5.03318 …)

correct substitution into area formula A1
1
e.g. A = 2
(5.03318 …)(6.82 )

A = 116 (cm2 ) A1 N2
[4 marks]

13a. Find the two possible values of AĈB .

[4 marks]

Markscheme
Note: accept answers given in degrees, and minutes.
evidence of choosing sine rule (M1)
sin A sin B
e.g. a
= b

correct substitution A1
sin θ sin 30∘ 5
e.g. 10
= 7
, sin θ = 7

AĈB = 45.6∘ ​ , AĈB = 134∘ ​ A1A1 N1N1

Note: If candidates only find the acute angle in part (a), award no marks for (b).
[4 marks]

AˆC
13b. Hence, find AB̂C , given that it is acute.
[2 marks]

Markscheme
attempt to substitute their larger value into angle sum of triangle (M1)
e.g. 180∘ − (134.415…∘ + 30∘ )

AB̂C = 15.6∘ ​ A1 N2
[2 marks]

AB = 7 cm, BC = 9 cm and AB̂C = 120∘ .

14a.
Find AC . [3 marks]

Markscheme
evidence of choosing cosine rule (M1)
e.g. a2 + b2 − 2ab cos C
correct substitution A1
e.g. 72 + 92 − 2(7)(9) cos 120∘
−−−
AC = 13.9 (= √193 ) A1 N2

[3 marks]

14b. Find BÂC .

[3 marks]
Markscheme
METHOD 1
evidence of choosing sine rule (M1)
sin Â sin B̂
e.g. BC
= AC

correct substitution A1
sin Â sin 120
e.g. 9
= 13.9

Â = 34.1∘ A1 N2
METHOD 2
evidence of choosing cosine rule (M1)
2 2 2
AB +AC −BC
e.g. cos Â = 2(AB)(AC)

correct substitution A1
7 + 13.9 − 92
2 2
e.g. cos Â = 2(7)(13.9)

Â = 34.1∘ A1 N2
[3 marks]

There is a vertical tower TA of height 36 m at the base A of a hill. A straight path goes up the hill from A to a point U. This
information is represented by the following diagram.

The path makes a 4∘ angle with the horizontal.

The point U on the path is 25 m away from the base of the tower.
The top of the tower is fixed to U by a wire of length x m.

15a.
Complete the diagram, showing clearly all the information above. [3 marks]
Markscheme

A1A1A1 N3

Note: Award A1 for labelling 4∘ with horizontal, A1 for labelling [AU] 25 metres, A1 for drawing [TU].
[3 marks]

15b.
Find x . [4 marks]

Markscheme
T ÂU = 86∘ (A1)
evidence of choosing cosine rule (M1)
correct substitution A1
e.g. x2 = 25 + 36 − 2(25)(36) cos 86∘
2 2

x = 42.4 A1 N3
[4 marks]
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