Sec 2 Maths Trigonometric ratios worksheet with answers

A free Sec 2 Maths worksheet on trigonometric ratios: 8 questions, 30 marks, short questions first and then structured ones, in the school exam style. The answers and full worked solutions are on the last page.

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Sec 2 Maths worksheet: Trigonometric ratios

Total: 30 marksSuggested time: 45 minutes

Name: ______________________________Class: ____________ Date: ____________

Answer all questions. Show all your working. Give non-exact answers correct to 3 significant figures, or 1 decimal place for angles in degrees, unless the question says otherwise.

Section A

A lamp post

32

m tall stands on level ground. A point

A

on the ground is

50

m from its foot. Find the angle of elevation,

x^\circ

, of its top

T

from

A

, correct to 1 decimal place.

Answer°[2]

In the diagram,

\angle KLM = 90^\circ

. Find

x

, correct to 3 significant figures.

Answerm[2]

In the diagram,

\angle LMN = 90^\circ

. Find

x

, correct to 1 decimal place.

Answer°[2]

In the diagram,

\angle PQR = 90^\circ

PQ = 12

cm and

\cos \angle QPR = \dfrac{3}{5}

. The point

S

lies on

PR

such that

PS : SR = 1 : 2

. Find the length of

PS

Answercm[3]

Section B

(a)In the diagram,

\angle ABC = 90^\circ

BC = 20

cm and

\sin \angle BAC = \dfrac{5}{13}

. The point

D

lies on

AC

such that

AD : DC = 1 : 3

. Find the length of

AD

Answer (a)cm[2]

(b)In the diagram,

\angle KLM = 90^\circ

. Find

x

, correct to 3 significant figures.

Answer (b)cm[2]

(a)In the diagram,

\angle XYZ = 90^\circ

. Find

x

, correct to 3 significant figures.

Answer (a)m[2]

(b)An aircraft takes off and climbs in a straight line at

8^\circ

to the horizontal, at a constant speed of

288

km/h. Find the height it gains in

40

seconds,

h

m, correct to 3 significant figures.

Answer (b)m[3]

(a)In the diagram,

\angle ABC = 90^\circ

AB = 10

cm and

AC = 26

cm. Write down the value of

\cos \angle CAB

, giving your answer as a fraction in its simplest form.

Answer (a)[2]

(b)From a point

P

on level ground

15

m from the foot of an office building, the angles of elevation of the top and the bottom of a window in the wall are

48^\circ

and

37^\circ

. Find the height of the window,

h

m, correct to 3 significant figures.

Answer (b)m[3]

(a)In the diagram,

\angle LMN = 90^\circ

. Find

x

, correct to 3 significant figures.

Answer (a)cm[2]

(b)A ladder

6

m long leans against an upright wall, making an angle of

78^\circ

with level ground. Find the distance of its foot from the wall,

d

m, correct to 3 significant figures.

Answer (b)m[2]

(c)In the diagram,

\angle ABC = 90^\circ

AC = 34

cm and

\sin \angle BAC = \dfrac{8}{17}

. The point

D

lies on

BC

such that

BD : DC = 1 : 2

. Find the length of

BD

Answer (c)cm[3]

END OF PAPER

Answers and worked solutions

Question 1

Answer:

= 32.6^\circ

(1 d.p.) °[2]

From the angle of elevation: the height

32

m is the opposite and the distance

50

m along the ground is the adjacent.

\tan x^\circ = \dfrac{32}{50}

x = \tan^{-1}\left(\dfrac{32}{50}\right)

= 32.6192\ldots

= 32.6^\circ

(1 d.p.)

Question 2

Answer:

= 27.5

m (3 s.f.)[2]

From the 57° angle: the side

x

m is the hypotenuse and the side

15

m is the adjacent.

\cos 57^\circ = \dfrac{15}{x}

x = \dfrac{15}{\cos 57^\circ}

= 27.5411\ldots

= 27.5

m (3 s.f.)

Question 3

Answer:

= 53.1^\circ

(1 d.p.) °[2]

From the angle x°: the side

12

m is the adjacent and the side

20

m is the hypotenuse.

\cos x^\circ = \dfrac{12}{20}

x = \cos^{-1}\left(\dfrac{12}{20}\right)

= 53.1301\ldots

= 53.1^\circ

(1 d.p.)

Question 4

Answer:

= 6.67

cm (3 s.f.)[3]

\dfrac{PR}{PQ} = \dfrac{5}{3}

(from the cos ratio, turned upside down)

PR = \dfrac{5}{3} \times 12

= 20

PS = \dfrac{1}{3} \times 20

= 6.6666\ldots

= 6.67

cm (3 s.f.)

Question 5

(a)Answer:

= 13

cm[2]

\dfrac{AC}{BC} = \dfrac{13}{5}

(from the sin ratio, turned upside down)

AC = \dfrac{13}{5} \times 20

= 52

AD = \dfrac{1}{4} \times 52

= 13

(b)Answer:

= 13.9

cm (3 s.f.)[2]

From the 57° angle: the side

x

cm is the opposite and the side

9

cm is the adjacent.

\tan 57^\circ = \dfrac{x}{9}

x = 9 \tan 57^\circ

= 13.8587\ldots

= 13.9

cm (3 s.f.)

Question 6

(a)Answer:

= 33.0

m (3 s.f.)[2]

From the 63° angle: the side

x

m is the hypotenuse and the side

15

m is the adjacent.

\cos 63^\circ = \dfrac{15}{x}

x = \dfrac{15}{\cos 63^\circ}

= 33.0403\ldots

= 33.0

m (3 s.f.)

(b)Answer:

= 445

m (3 s.f.)[3]

Speed

= \dfrac{288}{3.6}

= 80

m/sDistance

= 80 \times 40

= 3200

\sin 8^\circ = \dfrac{h}{3200}

h = 3200 \sin 8^\circ

= 445.3539\ldots

= 445

m (3 s.f.)

Question 7

(a)Answer:

= \dfrac{5}{13}

[2]

From

\angle CAB

AB

is the adjacent and

AC

is the hypotenuse.

\cos \angle CAB = \dfrac{10}{26}

= \dfrac{5}{13}

(b)Answer:

= 5.36

m (3 s.f.)[3]

Bottom:

\tan 37^\circ = \dfrac{B}{15}

, so

B = 15 \tan 37^\circ

B = 11.3033\ldots

mTop:

\tan 48^\circ = \dfrac{T}{15}

, so

T = 15 \tan 48^\circ

T = 16.6591\ldots

h = 16.6591\ldots - 11.3033\ldots

= 5.3558\ldots

= 5.36

m (3 s.f.)

Question 8

(a)Answer:

= 11.0

cm (3 s.f.)[2]

From the 57° angle: the side

x

cm is the hypotenuse and the side

6

cm is the adjacent.

\cos 57^\circ = \dfrac{6}{x}

x = \dfrac{6}{\cos 57^\circ}

= 11.0164\ldots

= 11.0

cm (3 s.f.)

(b)Answer:

= 1.25

m (3 s.f.)[2]

From the angle at the foot of the ladder: the distance from the wall,

d

m, is the adjacent and the ladder,

6

m, is the hypotenuse.

\cos 78^\circ = \dfrac{d}{6}

d = 6 \cos 78^\circ

= 1.2474\ldots

= 1.25

m (3 s.f.)

(c)Answer:

= 5.33

cm (3 s.f.)[3]

\dfrac{BC}{AC} = \dfrac{8}{17}

(from the sin ratio)

BC = \dfrac{8}{17} \times 34

= 16

BD = \dfrac{1}{3} \times 16

= 5.3333\ldots

= 5.33

cm (3 s.f.)

Practice questions by skill

Sec 2 Maths Trigonometric ratios worksheet with answers

In the print window, choose Save as PDF to download it.

Need the formulas first? Read the Trigonometric ratios notes. Or practise it online with step-by-step help when you get one wrong.

Section B

(a)In the diagram,

\angle ABC = 90^\circ

BC = 20

cm and

\sin \angle BAC = \dfrac{5}{13}

. The point

D

lies on

AC

such that

AD : DC = 1 : 3

. Find the length of

AD

Answer (a)cm[2]

(b)In the diagram,

\angle KLM = 90^\circ

. Find

x

, correct to 3 significant figures.

Answer (b)cm[2]

Answers and worked solutions

Question 1

Answer:

= 32.6^\circ

(1 d.p.) °[2]

From the angle of elevation: the height

32

m is the opposite and the distance

50

m along the ground is the adjacent.

\tan x^\circ = \dfrac{32}{50}

x = \tan^{-1}\left(\dfrac{32}{50}\right)

= 32.6192\ldots

= 32.6^\circ

(1 d.p.)

Question 2

Answer:

= 27.5

m (3 s.f.)[2]

From the 57° angle: the side

x

m is the hypotenuse and the side

15

m is the adjacent.

\cos 57^\circ = \dfrac{15}{x}

x = \dfrac{15}{\cos 57^\circ}

= 27.5411\ldots

= 27.5

m (3 s.f.)

Question 3

Answer:

= 53.1^\circ

(1 d.p.) °[2]

From the angle x°: the side

12

m is the adjacent and the side

20

m is the hypotenuse.

\cos x^\circ = \dfrac{12}{20}

x = \cos^{-1}\left(\dfrac{12}{20}\right)

= 53.1301\ldots

= 53.1^\circ

(1 d.p.)

Question 4

Answer:

= 6.67

cm (3 s.f.)[3]

\dfrac{PR}{PQ} = \dfrac{5}{3}

(from the cos ratio, turned upside down)

PR = \dfrac{5}{3} \times 12

= 20

PS = \dfrac{1}{3} \times 20

= 6.6666\ldots

= 6.67

cm (3 s.f.)

Question 5

(a)Answer:

= 13

cm[2]

\dfrac{AC}{BC} = \dfrac{13}{5}

(from the sin ratio, turned upside down)

AC = \dfrac{13}{5} \times 20

= 52

AD = \dfrac{1}{4} \times 52

= 13

(b)Answer:

= 13.9

cm (3 s.f.)[2]

From the 57° angle: the side

x

cm is the opposite and the side

9

cm is the adjacent.

\tan 57^\circ = \dfrac{x}{9}

x = 9 \tan 57^\circ

= 13.8587\ldots

= 13.9

cm (3 s.f.)

Question 6

(a)Answer:

= 33.0

m (3 s.f.)[2]

From the 63° angle: the side

x

m is the hypotenuse and the side

15

m is the adjacent.

\cos 63^\circ = \dfrac{15}{x}

x = \dfrac{15}{\cos 63^\circ}

= 33.0403\ldots

= 33.0

m (3 s.f.)

(b)Answer:

= 445

m (3 s.f.)[3]

Speed

= \dfrac{288}{3.6}

= 80

m/sDistance

= 80 \times 40

= 3200

\sin 8^\circ = \dfrac{h}{3200}

h = 3200 \sin 8^\circ

= 445.3539\ldots

= 445

m (3 s.f.)

Question 7

(a)Answer:

= \dfrac{5}{13}

[2]

From

\angle CAB

AB

is the adjacent and

AC

is the hypotenuse.

\cos \angle CAB = \dfrac{10}{26}

= \dfrac{5}{13}

(b)Answer:

= 5.36

m (3 s.f.)[3]

Bottom:

\tan 37^\circ = \dfrac{B}{15}

, so

B = 15 \tan 37^\circ

B = 11.3033\ldots

mTop:

\tan 48^\circ = \dfrac{T}{15}

, so

T = 15 \tan 48^\circ

T = 16.6591\ldots

h = 16.6591\ldots - 11.3033\ldots

= 5.3558\ldots

= 5.36

m (3 s.f.)

Question 8

(a)Answer:

= 11.0

cm (3 s.f.)[2]

From the 57° angle: the side

x

cm is the hypotenuse and the side

6

cm is the adjacent.

\cos 57^\circ = \dfrac{6}{x}

x = \dfrac{6}{\cos 57^\circ}

= 11.0164\ldots

= 11.0

cm (3 s.f.)

(b)Answer:

= 1.25

m (3 s.f.)[2]

From the angle at the foot of the ladder: the distance from the wall,

d

m, is the adjacent and the ladder,

6

m, is the hypotenuse.

\cos 78^\circ = \dfrac{d}{6}

d = 6 \cos 78^\circ

= 1.2474\ldots

= 1.25

m (3 s.f.)

(c)Answer:

= 5.33

cm (3 s.f.)[3]

\dfrac{BC}{AC} = \dfrac{8}{17}

(from the sin ratio)

BC = \dfrac{8}{17} \times 34

= 16

BD = \dfrac{1}{3} \times 16

= 5.3333\ldots

= 5.33

cm (3 s.f.)