Sec 3 G2 Maths Applications of trigonometry worksheet with answers

A free Sec 3 G2 Maths worksheet on applications of trigonometry: 7 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 3 G2 Maths worksheet: Applications of trigonometry

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

B

9

km east and

10

km north of

A

. Find the bearing of

B

from

A

, correct to 1 decimal place.

Answer°[2]

A lamp post

15

m tall stands on level ground. Find the angle of elevation of its top from a point

12

m from its foot, correct to 1 decimal place.

Answer°[2]

ABCD

is a rectangular field on level ground, with

AB = 40

m and

BC = 18

m. A vertical pole

CT

of height

6

m stands at

C

. Find the angle of elevation of

T

from

A

, correct to 1 decimal place.

Answer°[3]

The diagram shows a cuboid with

AB = 10

cm,

BC = 5

cm and

CG = 3

cm. Find the angle between

AG

and the base

ABCD

, correct to 1 decimal place.

Answer°[3]

Section B

Two ships leave port

P

. Ship

A

sails

15

km on a bearing of

028^\circ

. Ship

B

sails

9

km on a bearing of

145^\circ

(a)Find the bearing of

P

from

A

Answer (a)°[1]

(b)Find the distance

AB

, correct to 3 significant figures.

Answer (b)km[3]

(c)Find how far north of

P

ship

A

is, correct to 3 significant figures.

Answer (c)km[2]

Two ships leave port

P

. Ship

A

sails

10

km on a bearing of

064^\circ

. Ship

B

sails

11

km on a bearing of

152^\circ

(a)Find the bearing of

P

from

A

Answer (a)°[1]

(b)Find the distance

AB

, correct to 3 significant figures.

Answer (b)km[3]

(c)Find how far north of

P

ship

A

is, correct to 3 significant figures.

Answer (c)km[2]

FT

is a vertical flagpole standing on level ground.

A

B

F

and

C

lie in a straight line on the ground, with

AB = 30

m. The angles of elevation of

T

from

A

and from

B

are

28^\circ

and

67^\circ

. The angle of depression of

C

from

T

40^\circ

(a)Find angle

ATB

Answer (a)°[1]

(b)Find the length of

BT

, correct to 3 significant figures.

Answer (b)m[2]

(c)Find the height of the flagpole,

FT

, correct to 3 significant figures.

Answer (c)m[2]

(d)Find the distance

AC

, correct to 3 significant figures.

Answer (d)m[3]

END OF PAPER

Answers and worked solutions

Question 1

Answer:

= 042.0^\circ

(1 d.p.) °[2]

\tan \alpha = \dfrac{9}{10}

\alpha = \tan^{-1}\left(\dfrac{9}{10}\right)

\alpha = 41.9872\ldots^\circ

Bearing

= 41.9872\ldots^\circ

= 042.0^\circ

(1 d.p.)

Question 2

Answer:

\phantom{\theta} = 51.3^\circ

(1 d.p.) °[2]

\tan \theta = \dfrac{15}{12}

\theta = \tan^{-1}\left(\dfrac{15}{12}\right)

\theta = 51.3401\ldots^\circ

\phantom{\theta} = 51.3^\circ

(1 d.p.)

Question 3

Answer:

\phantom{\theta} = 7.8^\circ

(1 d.p.) °[3]

AC^2 = 40^2 + 18^2

AC^2 = 1924

AC = \sqrt{1924}

\tan \theta = \dfrac{CT}{AC}

\theta = \tan^{-1}\left(\dfrac{6}{\sqrt{1924}}\right)

\theta = 7.7890\ldots^\circ

\phantom{\theta} = 7.8^\circ

(1 d.p.)

Question 4

Answer:

\phantom{\theta} = 15.0^\circ

(1 d.p.) °[3]

AC^2 = 10^2 + 5^2

AC^2 = 125

AC = \sqrt{125}

The angle between

AG

and the base is

\angle GAC

(

G

is directly above

C

\tan \theta = \dfrac{CG}{AC}

\tan \theta = \dfrac{3}{\sqrt{125}}

\theta = \tan^{-1}\left(\dfrac{3}{\sqrt{125}}\right)

\theta = 15.0202\ldots^\circ

\phantom{\theta} = 15.0^\circ

(1 d.p.)

Question 5

(a)Answer:

208^\circ

°[1]

The north lines at

P

and

A

are parallel, so the bearing back is

180^\circ

more.Bearing of

P

from

A

= 28^\circ + 180^\circ

= 208^\circ

(b)Answer:

20.7

km[3]

\angle APB = 145^\circ - 28^\circ

\phantom{\angle APB} = 117^\circ

AB^2 = 15^2 + 9^2 - 2(15)(9) \cos 117^\circ

AB^2 = 428.5774\ldots

AB = 20.7021\ldots

\phantom{AB} = 20.7

(c)Answer:

13.2

km[2]

\cos 28^\circ = \dfrac{y}{15}

y = 15 \cos 28^\circ

\phantom{y} = 13.2442\ldots

\phantom{y} = 13.2

Question 6

(a)Answer:

244^\circ

°[1]

The north lines at

P

and

A

are parallel, so the bearing back is

180^\circ

more.Bearing of

P

from

A

= 64^\circ + 180^\circ

= 244^\circ

(b)Answer:

14.6

km[3]

\angle APB = 152^\circ - 64^\circ

\phantom{\angle APB} = 88^\circ

AB^2 = 10^2 + 11^2 - 2(10)(11) \cos 88^\circ

AB^2 = 213.3221\ldots

AB = 14.6055\ldots

\phantom{AB} = 14.6

(c)Answer:

4.38

km[2]

\cos 64^\circ = \dfrac{y}{10}

y = 10 \cos 64^\circ

\phantom{y} = 4.3837\ldots

\phantom{y} = 4.38

Question 7

(a)Answer:

39^\circ

°[1]

Angle

ABT = 180^\circ - 67^\circ

ABT = 113^\circ

(angles on a straight line)Angle

ATB = 180^\circ - 28^\circ - 113^\circ

ATB = 39^\circ

(b)Answer:

22.4

m[2]

\dfrac{BT}{\sin 28^\circ} = \dfrac{30}{\sin 39^\circ}

BT = \dfrac{30 \sin 28^\circ}{\sin 39^\circ}

= 22.3799\ldots

\phantom{} = 22.4

(c)Answer:

20.6

m[2]

\sin 67^\circ = \dfrac{FT}{BT}

FT = 22.3799\ldots \times \sin 67^\circ

= 20.6008\ldots

\phantom{} = 20.6

(d)Answer:

63.3

m[3]

BF = 22.3799\ldots \times \cos 67^\circ

\phantom{BF} = 8.7445\ldots

Angle

TCF = 40^\circ

(alternate angles).

FC = \dfrac{20.6008\ldots}{\tan 40^\circ}

\phantom{FC} = 24.5511\ldots

AC = 30 + 8.7445\ldots + 24.5511\ldots

= 63.2956\ldots

\phantom{} = 63.3

Sec 3 G2 Maths Applications of trigonometry worksheet with answers

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

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

Section B

Two ships leave port

P

. Ship

A

sails

15

km on a bearing of

028^\circ

. Ship

B

sails

9

km on a bearing of

145^\circ

(a)Find the bearing of

P

from

A

Answer (a)°[1]

(b)Find the distance

AB

, correct to 3 significant figures.

Answer (b)km[3]

(c)Find how far north of

P

ship

A

is, correct to 3 significant figures.

Answer (c)km[2]

Answers and worked solutions

Question 1

Answer:

= 042.0^\circ

(1 d.p.) °[2]

\tan \alpha = \dfrac{9}{10}

\alpha = \tan^{-1}\left(\dfrac{9}{10}\right)

\alpha = 41.9872\ldots^\circ

Bearing

= 41.9872\ldots^\circ

= 042.0^\circ

(1 d.p.)

Question 2

Answer:

\phantom{\theta} = 51.3^\circ

(1 d.p.) °[2]

\tan \theta = \dfrac{15}{12}

\theta = \tan^{-1}\left(\dfrac{15}{12}\right)

\theta = 51.3401\ldots^\circ

\phantom{\theta} = 51.3^\circ

(1 d.p.)

Question 3

Answer:

\phantom{\theta} = 7.8^\circ

(1 d.p.) °[3]

AC^2 = 40^2 + 18^2

AC^2 = 1924

AC = \sqrt{1924}

\tan \theta = \dfrac{CT}{AC}

\theta = \tan^{-1}\left(\dfrac{6}{\sqrt{1924}}\right)

\theta = 7.7890\ldots^\circ

\phantom{\theta} = 7.8^\circ

(1 d.p.)

Question 4

Answer:

\phantom{\theta} = 15.0^\circ

(1 d.p.) °[3]

AC^2 = 10^2 + 5^2

AC^2 = 125

AC = \sqrt{125}

The angle between

AG

and the base is

\angle GAC

(

G

is directly above

C

\tan \theta = \dfrac{CG}{AC}

\tan \theta = \dfrac{3}{\sqrt{125}}

\theta = \tan^{-1}\left(\dfrac{3}{\sqrt{125}}\right)

\theta = 15.0202\ldots^\circ

\phantom{\theta} = 15.0^\circ

(1 d.p.)

Question 5

(a)Answer:

208^\circ

°[1]

The north lines at

P

and

A

are parallel, so the bearing back is

180^\circ

more.Bearing of

P

from

A

= 28^\circ + 180^\circ

= 208^\circ

(b)Answer:

20.7

km[3]

\angle APB = 145^\circ - 28^\circ

\phantom{\angle APB} = 117^\circ

AB^2 = 15^2 + 9^2 - 2(15)(9) \cos 117^\circ

AB^2 = 428.5774\ldots

AB = 20.7021\ldots

\phantom{AB} = 20.7

(c)Answer:

13.2

km[2]

\cos 28^\circ = \dfrac{y}{15}

y = 15 \cos 28^\circ

\phantom{y} = 13.2442\ldots

\phantom{y} = 13.2

Question 6

(a)Answer:

244^\circ

°[1]

The north lines at

P

and

A

are parallel, so the bearing back is

180^\circ

more.Bearing of

P

from

A

= 64^\circ + 180^\circ

= 244^\circ

(b)Answer:

14.6

km[3]

\angle APB = 152^\circ - 64^\circ

\phantom{\angle APB} = 88^\circ

AB^2 = 10^2 + 11^2 - 2(10)(11) \cos 88^\circ

AB^2 = 213.3221\ldots

AB = 14.6055\ldots

\phantom{AB} = 14.6

(c)Answer:

4.38

km[2]

\cos 64^\circ = \dfrac{y}{10}

y = 10 \cos 64^\circ

\phantom{y} = 4.3837\ldots

\phantom{y} = 4.38

Question 7

(a)Answer:

39^\circ

°[1]

Angle

ABT = 180^\circ - 67^\circ

ABT = 113^\circ

(angles on a straight line)Angle

ATB = 180^\circ - 28^\circ - 113^\circ

ATB = 39^\circ

(b)Answer:

22.4

m[2]

\dfrac{BT}{\sin 28^\circ} = \dfrac{30}{\sin 39^\circ}

BT = \dfrac{30 \sin 28^\circ}{\sin 39^\circ}

= 22.3799\ldots

\phantom{} = 22.4

(c)Answer:

20.6

m[2]

\sin 67^\circ = \dfrac{FT}{BT}

FT = 22.3799\ldots \times \sin 67^\circ

= 20.6008\ldots

\phantom{} = 20.6

(d)Answer:

63.3

m[3]

BF = 22.3799\ldots \times \cos 67^\circ

\phantom{BF} = 8.7445\ldots

Angle

TCF = 40^\circ

(alternate angles).

FC = \dfrac{20.6008\ldots}{\tan 40^\circ}

\phantom{FC} = 24.5511\ldots

AC = 30 + 8.7445\ldots + 24.5511\ldots

= 63.2956\ldots

\phantom{} = 63.3