Sec 2 G2 Maths Trigonometric ratios worksheet with answers

A free Sec 2 G2 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 G2 Maths worksheet: Trigonometric ratios

Total: 30 marksSuggested time: 50 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

In the diagram,

\angle ABC = 90^\circ

. Find

x

, correct to 3 significant figures.

Answercm[2]

A ramp

4

m long rises to a platform

1

m above level ground. Find the angle the ramp makes with the ground,

x^\circ

, correct to 1 decimal place.

Answer°[2]

In the diagram,

\angle STU = 90^\circ

. Find

x

, correct to 1 decimal place.

Answer°[2]

The diagram shows the side view of a footbridge. The deck

AB

18

m long and

C

is the point on

AB

such that

AC : CB = 2 : 3

. Two steel struts,

AD

and

BD

, support the deck;

D

is vertically below

C

and

\angle BAD = 30^\circ

. Steel strut costs $85 per metre when the total length bought is 25 m or less, and $72 per metre for the whole order when more than 25 m is bought. Find the cost of the steel for the two struts, correct to the nearest $100.

Answer$[3]

In the diagram,

\angle PQR = 90^\circ

PR = 52

cm and

QR = 48

cm. Find the value of

\tan \angle RPQ

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

Answer[3]

Section B

(a)In the diagram,

\angle STU = 90^\circ

ST = 10

cm and

TU = 24

cm. Write down the value of

\tan \angle SUT

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

Answer (a)[2]

(b)In the diagram,

\angle XYZ = 90^\circ

. Find

x

, correct to 3 significant figures.

Answer (b)cm[2]

(a)In the diagram,

\angle KLM = 90^\circ

. Find

x

, correct to 3 significant figures.

Answer (a)m[2]

(b)A ramp

5

m long rises to a platform

1.2

m above level ground. Find the angle the ramp makes with the ground,

x^\circ

, correct to 1 decimal place.

Answer (b)°[2]

(c)In the diagram,

\angle EFG = 90^\circ

EG = 34

cm and

\sin \angle FEG = \dfrac{15}{17}

. The point

H

lies on

FG

such that

FH : HG = 3 : 1

. Find the length of

EH

, correct to 3 significant figures.

Answer (c)cm[3]

(a)In the diagram,

\angle LMN = 90^\circ

. Find

x

, correct to 3 significant figures.

Answer (a)cm[2]

(b)In the diagram,

\angle WXY = 90^\circ

XY = 12

cm and

\sin \angle XWY = \dfrac{4}{5}

. The point

Z

lies on

WY

such that

WZ : ZY = 1 : 2

. Find the length of

WZ

Answer (b)cm[2]

(c)From a point

P

on level ground in front of a block of flats, the angles of elevation of the top and the bottom of a window in the wall are

44^\circ

and

38^\circ

. The bottom of the window is

10

m above the ground. Find the height of the window,

h

m, correct to 3 significant figures.

Answer (c)m[3]

END OF PAPER

Answers and worked solutions

Question 1

Answer:

= 2.72

cm (3 s.f.)[2]

From the 63° angle: the side

x

cm is the adjacent and the side

6

cm is the hypotenuse.

\cos 63^\circ = \dfrac{x}{6}

x = 6 \cos 63^\circ

= 2.7239\ldots

= 2.72

cm (3 s.f.)

Question 2

Answer:

= 14.5^\circ

(1 d.p.) °[2]

From the angle at the bottom of the ramp: the height

1

m is the opposite and the ramp,

4

m, is the hypotenuse.

\sin x^\circ = \dfrac{1}{4}

x = \sin^{-1}\left(\dfrac{1}{4}\right)

= 14.4775\ldots

= 14.5^\circ

(1 d.p.)

Question 3

Answer:

= 28.1^\circ

(1 d.p.) °[2]

From the angle x°: the side

8

cm is the opposite and the side

15

cm is the adjacent.

\tan x^\circ = \dfrac{8}{15}

x = \tan^{-1}\left(\dfrac{8}{15}\right)

= 28.0724\ldots

= 28.1^\circ

(1 d.p.)

Question 4

Answer:

\phantom{\text{Cost}} = 1690.3291\ldots

, so about $1700 (to the nearest $100).[3]

AC = \dfrac{2}{5} \times 18

\phantom{AC} = 7.2

m, so

CB = 18 - 7.2

CB = 10.8

AD = \dfrac{7.2}{\cos 30^\circ}

\phantom{AD} = 8.3138\ldots

CD = 7.2 \tan 30^\circ

\phantom{CD} = 4.1569\ldots

BD = \sqrt{10.8^2 + (4.1569\ldots)^2}

\phantom{BD} = 11.5723\ldots

\text{Total} = 8.3138\ldots + 11.5723\ldots

\phantom{\text{Total}} = 19.8862\ldots

m, which is not more than

25

m, so the price is $85 per metre.

\text{Cost} = 19.8862\ldots \times 85

\phantom{\text{Cost}} = 1690.3291\ldots

, so about $1700 (to the nearest $100).

Question 5

Answer:

= \dfrac{12}{5}

[3]

PQ^2 = 52^2 - 48^2

= 400

PQ = \sqrt{400}

\phantom{PQ} = 20

cmFrom

\angle RPQ

QR

is the opposite and

PQ

is the adjacent.

\tan \angle RPQ = \dfrac{48}{20}

= \dfrac{12}{5}

Question 6

(a)Answer:

= \dfrac{5}{12}

[2]

From

\angle SUT

ST

is the opposite and

TU

is the adjacent.

\tan \angle SUT = \dfrac{10}{24}

= \dfrac{5}{12}

(b)Answer:

= 11.8

cm (3 s.f.)[2]

From the 46° angle: the side

x

cm is the hypotenuse and the side

8.5

cm is the opposite.

\sin 46^\circ = \dfrac{8.5}{x}

x = \dfrac{8.5}{\sin 46^\circ}

= 11.8163\ldots

= 11.8

cm (3 s.f.)

Question 7

(a)Answer:

= 16.4

m (3 s.f.)[2]

From the 35° angle: the side

x

m is the adjacent and the side

20

m is the hypotenuse.

\cos 35^\circ = \dfrac{x}{20}

x = 20 \cos 35^\circ

= 16.3830\ldots

= 16.4

m (3 s.f.)

(b)Answer:

= 13.9^\circ

(1 d.p.) °[2]

From the angle at the bottom of the ramp: the height

1.2

m is the opposite and the ramp,

5

m, is the hypotenuse.

\sin x^\circ = \dfrac{1.2}{5}

x = \sin^{-1}\left(\dfrac{1.2}{5}\right)

= 13.8865\ldots

= 13.9^\circ

(1 d.p.)

(c)Answer:

= 27.6

cm (3 s.f.)[3]

\dfrac{FG}{EG} = \dfrac{15}{17}

(from the sin ratio)

FG = \dfrac{15}{17} \times 34

= 30

FH = \dfrac{3}{4} \times 30

= 22.5

EH^2 = 16^2 + 22.5^2

\phantom{EH^2} = 762.25

= 27.6088\ldots

= 27.6

cm (3 s.f.)

Question 8

(a)Answer:

= 3.61

cm (3 s.f.)[2]

From the 67° angle: the side

x

cm is the adjacent and the side

8.5

cm is the opposite.

\tan 67^\circ = \dfrac{8.5}{x}

x = \dfrac{8.5}{\tan 67^\circ}

= 3.6080\ldots

= 3.61

cm (3 s.f.)

(b)Answer:

= 5

cm[2]

\dfrac{WY}{XY} = \dfrac{5}{4}

(from the sin ratio, turned upside down)

WY = \dfrac{5}{4} \times 12

= 15

WZ = \dfrac{1}{3} \times 15

= 5

(c)Answer:

= 2.36

m (3 s.f.)[3]

\tan 38^\circ = \dfrac{10}{d}

, so

d = \dfrac{10}{\tan 38^\circ}

d = 12.7994\ldots

mTop:

\tan 44^\circ = \dfrac{T}{12.7994\ldots}

, so

T = 12.7994\ldots \tan 44^\circ

T = 12.3602\ldots

h = 12.3602\ldots - 10

= 2.3602\ldots

= 2.36

m (3 s.f.)

Sec 2 G2 Maths Trigonometric ratios worksheet with answers

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

Or practise it online with step-by-step help when you get one wrong.

Section B

(a)In the diagram,

\angle STU = 90^\circ

ST = 10

cm and

TU = 24

cm. Write down the value of

\tan \angle SUT

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

Answer (a)[2]

(b)In the diagram,

\angle XYZ = 90^\circ

. Find

x

, correct to 3 significant figures.

Answer (b)cm[2]

Answers and worked solutions

Question 1

Answer:

= 2.72

cm (3 s.f.)[2]

From the 63° angle: the side

x

cm is the adjacent and the side

6

cm is the hypotenuse.

\cos 63^\circ = \dfrac{x}{6}

x = 6 \cos 63^\circ

= 2.7239\ldots

= 2.72

cm (3 s.f.)

Question 2

Answer:

= 14.5^\circ

(1 d.p.) °[2]

From the angle at the bottom of the ramp: the height

1

m is the opposite and the ramp,

4

m, is the hypotenuse.

\sin x^\circ = \dfrac{1}{4}

x = \sin^{-1}\left(\dfrac{1}{4}\right)

= 14.4775\ldots

= 14.5^\circ

(1 d.p.)

Question 3

Answer:

= 28.1^\circ

(1 d.p.) °[2]

From the angle x°: the side

8

cm is the opposite and the side

15

cm is the adjacent.

\tan x^\circ = \dfrac{8}{15}

x = \tan^{-1}\left(\dfrac{8}{15}\right)

= 28.0724\ldots

= 28.1^\circ

(1 d.p.)

Question 4

Answer:

\phantom{\text{Cost}} = 1690.3291\ldots

, so about $1700 (to the nearest $100).[3]

AC = \dfrac{2}{5} \times 18

\phantom{AC} = 7.2

m, so

CB = 18 - 7.2

CB = 10.8

AD = \dfrac{7.2}{\cos 30^\circ}

\phantom{AD} = 8.3138\ldots

CD = 7.2 \tan 30^\circ

\phantom{CD} = 4.1569\ldots

BD = \sqrt{10.8^2 + (4.1569\ldots)^2}

\phantom{BD} = 11.5723\ldots

\text{Total} = 8.3138\ldots + 11.5723\ldots

\phantom{\text{Total}} = 19.8862\ldots

m, which is not more than

25

m, so the price is $85 per metre.

\text{Cost} = 19.8862\ldots \times 85

\phantom{\text{Cost}} = 1690.3291\ldots

, so about $1700 (to the nearest $100).

Question 5

Answer:

= \dfrac{12}{5}

[3]

PQ^2 = 52^2 - 48^2

= 400

PQ = \sqrt{400}

\phantom{PQ} = 20

cmFrom

\angle RPQ

QR

is the opposite and

PQ

is the adjacent.

\tan \angle RPQ = \dfrac{48}{20}

= \dfrac{12}{5}

Question 6

(a)Answer:

= \dfrac{5}{12}

[2]

From

\angle SUT

ST

is the opposite and

TU

is the adjacent.

\tan \angle SUT = \dfrac{10}{24}

= \dfrac{5}{12}

(b)Answer:

= 11.8

cm (3 s.f.)[2]

From the 46° angle: the side

x

cm is the hypotenuse and the side

8.5

cm is the opposite.

\sin 46^\circ = \dfrac{8.5}{x}

x = \dfrac{8.5}{\sin 46^\circ}

= 11.8163\ldots

= 11.8

cm (3 s.f.)

Question 7

(a)Answer:

= 16.4

m (3 s.f.)[2]

From the 35° angle: the side

x

m is the adjacent and the side

20

m is the hypotenuse.

\cos 35^\circ = \dfrac{x}{20}

x = 20 \cos 35^\circ

= 16.3830\ldots

= 16.4

m (3 s.f.)

(b)Answer:

= 13.9^\circ

(1 d.p.) °[2]

From the angle at the bottom of the ramp: the height

1.2

m is the opposite and the ramp,

5

m, is the hypotenuse.

\sin x^\circ = \dfrac{1.2}{5}

x = \sin^{-1}\left(\dfrac{1.2}{5}\right)

= 13.8865\ldots

= 13.9^\circ

(1 d.p.)

(c)Answer:

= 27.6

cm (3 s.f.)[3]

\dfrac{FG}{EG} = \dfrac{15}{17}

(from the sin ratio)

FG = \dfrac{15}{17} \times 34

= 30

FH = \dfrac{3}{4} \times 30

= 22.5

EH^2 = 16^2 + 22.5^2

\phantom{EH^2} = 762.25

= 27.6088\ldots

= 27.6

cm (3 s.f.)

Question 8

(a)Answer:

= 3.61

cm (3 s.f.)[2]

From the 67° angle: the side

x

cm is the adjacent and the side

8.5

cm is the opposite.

\tan 67^\circ = \dfrac{8.5}{x}

x = \dfrac{8.5}{\tan 67^\circ}

= 3.6080\ldots

= 3.61

cm (3 s.f.)

(b)Answer:

= 5

cm[2]

\dfrac{WY}{XY} = \dfrac{5}{4}

(from the sin ratio, turned upside down)

WY = \dfrac{5}{4} \times 12

= 15

WZ = \dfrac{1}{3} \times 15

= 5

(c)Answer:

= 2.36

m (3 s.f.)[3]

\tan 38^\circ = \dfrac{10}{d}

, so

d = \dfrac{10}{\tan 38^\circ}

d = 12.7994\ldots

mTop:

\tan 44^\circ = \dfrac{T}{12.7994\ldots}

, so

T = 12.7994\ldots \tan 44^\circ

T = 12.3602\ldots

h = 12.3602\ldots - 10

= 2.3602\ldots

= 2.36

m (3 s.f.)