Sec 3 E-Math Applications of trigonometry worksheet with answers

A free Sec 3 E-Math 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 E-Math worksheet: Applications of trigonometry

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

The bearing of

B

from

A

038^\circ

. Find the bearing of

A

from

B

Answer°[2]

From the top of a lighthouse

20

m high, the angle of depression of a car is

53^\circ

. Find the distance of the car from the foot of the lighthouse, correct to 3 significant figures.

Answerm[2]

The diagram shows a cuboid with

AB = 9

cm,

BC = 8

cm and

CG = 8

cm. Find the length of

AG

, correct to 3 significant figures.

Answercm[3]

A

and

B

are points on level ground in a straight line with the foot

F

of a tree, with

AB = 15

m. The angles of elevation of the top

T

from

A

and

B

are

35^\circ

and

45^\circ

respectively. Find the height of the tree, correct to 3 significant figures.

Answerm[3]

Section B

Two ships leave port

P

. Ship

A

sails

15

km on a bearing of

042^\circ

. Ship

B

sails

6

km on a bearing of

090^\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

15

km on a bearing of

057^\circ

. Ship

B

sails

6

km on a bearing of

174^\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]

A ship sails

7

km from port

A

B

on a bearing of

107^\circ

, then

12

km from

B

C

on a bearing of

137^\circ

(a)Find angle

ABC

Answer (a)°[2]

(b)Find the distance

AC

, correct to 3 significant figures.

Answer (b)km[2]

(c)Find angle

BAC

, correct to 1 decimal place.

Answer (c)°[2]

(d)Find the bearing of

C

from

A

Answer (d)°[1]

(e)Find the bearing of

A

from

C

Answer (e)°[1]

END OF PAPER

Answers and worked solutions

Question 1

Answer:

= 218^\circ

°[2]

Angle from south at

B

BA

= 38^\circ

(alt. ∠s)Bearing of

A

from

B

= 180^\circ + 38^\circ

= 218^\circ

Question 2

Answer:

\phantom{x} = 15.1

m (3 s.f.)[2]

Angle

TBF = 53^\circ

(alt. ∠s)

\tan 53^\circ = \dfrac{20}{x}

x = \dfrac{20}{\tan 53^\circ}

x = 15.0710\ldots

\phantom{x} = 15.1

m (3 s.f.)

Question 3

Answer:

\phantom{AG} = 14.5

cm (3 s.f.)[3]

AC^2 = 9^2 + 8^2

AC^2 = 145

AC = \sqrt{145}

AG^2 = AC^2 + CG^2

AG^2 = 145 + 8^2

AG^2 = 209

AG = \sqrt{209}

AG = 14.4568\ldots

\phantom{AG} = 14.5

cm (3 s.f.)

Question 4

Answer:

\phantom{h} = 35.0

m (3 s.f.)[3]

In triangle

BFT

h = x \tan 45^\circ

In triangle

AFT

h = (x + 15) \tan 35^\circ

x \tan 45^\circ = (x + 15) \tan 35^\circ

x(\tan 45^\circ - \tan 35^\circ) = 15 \tan 35^\circ

x = \dfrac{15 \tan 35^\circ}{\tan 45^\circ - \tan 35^\circ}

x = 35.0346\ldots

h = 35.0346\ldots \times \tan 45^\circ

h = 35.0346\ldots

\phantom{h} = 35.0

m (3 s.f.)

Question 5

(a)Answer:

222^\circ

°[1]

The north lines at

P

and

A

are parallel, so the bearing back is

180^\circ

more.Bearing of

P

from

A

= 42^\circ + 180^\circ

= 222^\circ

(b)Answer:

11.9

km[3]

\angle APB = 90^\circ - 42^\circ

\phantom{\angle APB} = 48^\circ

AB^2 = 15^2 + 6^2 - 2(15)(6) \cos 48^\circ

AB^2 = 140.5564\ldots

AB = 11.8556\ldots

\phantom{AB} = 11.9

(c)Answer:

11.1

km[2]

\cos 42^\circ = \dfrac{y}{15}

y = 15 \cos 42^\circ

\phantom{y} = 11.1471\ldots

\phantom{y} = 11.1

Question 6

(a)Answer:

237^\circ

°[1]

The north lines at

P

and

A

are parallel, so the bearing back is

180^\circ

more.Bearing of

P

from

A

= 57^\circ + 180^\circ

= 237^\circ

(b)Answer:

18.5

km[3]

\angle APB = 174^\circ - 57^\circ

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

AB^2 = 15^2 + 6^2 - 2(15)(6) \cos 117^\circ

AB^2 = 342.7182\ldots

AB = 18.5126\ldots

\phantom{AB} = 18.5

(c)Answer:

8.17

km[2]

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

y = 15 \cos 57^\circ

\phantom{y} = 8.1695\ldots

\phantom{y} = 8.17

Question 7

(a)Answer:

150^\circ

°[2]

Bearing of

A

from

B

= 107^\circ + 180^\circ

= 287^\circ

Angle

ABC = 287^\circ - 137^\circ

ABC = 150^\circ

(b)Answer:

18.4

km[2]

AC^2 = 7^2 + 12^2 - 2(7)(12) \cos 150^\circ

AC^2 = 338.4922\ldots

AC = 18.3981\ldots

\phantom{AC} = 18.4

(c)Answer:

19.0^\circ

°[2]

\dfrac{\sin BAC}{12} = \dfrac{\sin 150^\circ}{18.3981\ldots}

\sin BAC = 0.3261\ldots

BAC = 19.0334\ldots^\circ

\phantom{BAC} = 19.0^\circ

(d)Answer:

126.0^\circ

°[1]

Bearing of

C

from

A

= 107^\circ + 19.0334\ldots^\circ

= 126.0^\circ

(e)Answer:

306.0^\circ

°[1]

Bearing of

A

from

C

= 126.0^\circ + 180^\circ

= 306.0^\circ

Practice questions by skill

Sec 3 E-Math 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

042^\circ

. Ship

B

sails

6

km on a bearing of

090^\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:

= 218^\circ

°[2]

Angle from south at

B

BA

= 38^\circ

(alt. ∠s)Bearing of

A

from

B

= 180^\circ + 38^\circ

= 218^\circ

Question 2

Answer:

\phantom{x} = 15.1

m (3 s.f.)[2]

Angle

TBF = 53^\circ

(alt. ∠s)

\tan 53^\circ = \dfrac{20}{x}

x = \dfrac{20}{\tan 53^\circ}

x = 15.0710\ldots

\phantom{x} = 15.1

m (3 s.f.)

Question 3

Answer:

\phantom{AG} = 14.5

cm (3 s.f.)[3]

AC^2 = 9^2 + 8^2

AC^2 = 145

AC = \sqrt{145}

AG^2 = AC^2 + CG^2

AG^2 = 145 + 8^2

AG^2 = 209

AG = \sqrt{209}

AG = 14.4568\ldots

\phantom{AG} = 14.5

cm (3 s.f.)

Question 4

Answer:

\phantom{h} = 35.0

m (3 s.f.)[3]

In triangle

BFT

h = x \tan 45^\circ

In triangle

AFT

h = (x + 15) \tan 35^\circ

x \tan 45^\circ = (x + 15) \tan 35^\circ

x(\tan 45^\circ - \tan 35^\circ) = 15 \tan 35^\circ

x = \dfrac{15 \tan 35^\circ}{\tan 45^\circ - \tan 35^\circ}

x = 35.0346\ldots

h = 35.0346\ldots \times \tan 45^\circ

h = 35.0346\ldots

\phantom{h} = 35.0

m (3 s.f.)

Question 5

(a)Answer:

222^\circ

°[1]

The north lines at

P

and

A

are parallel, so the bearing back is

180^\circ

more.Bearing of

P

from

A

= 42^\circ + 180^\circ

= 222^\circ

(b)Answer:

11.9

km[3]

\angle APB = 90^\circ - 42^\circ

\phantom{\angle APB} = 48^\circ

AB^2 = 15^2 + 6^2 - 2(15)(6) \cos 48^\circ

AB^2 = 140.5564\ldots

AB = 11.8556\ldots

\phantom{AB} = 11.9

(c)Answer:

11.1

km[2]

\cos 42^\circ = \dfrac{y}{15}

y = 15 \cos 42^\circ

\phantom{y} = 11.1471\ldots

\phantom{y} = 11.1

Question 6

(a)Answer:

237^\circ

°[1]

The north lines at

P

and

A

are parallel, so the bearing back is

180^\circ

more.Bearing of

P

from

A

= 57^\circ + 180^\circ

= 237^\circ

(b)Answer:

18.5

km[3]

\angle APB = 174^\circ - 57^\circ

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

AB^2 = 15^2 + 6^2 - 2(15)(6) \cos 117^\circ

AB^2 = 342.7182\ldots

AB = 18.5126\ldots

\phantom{AB} = 18.5

(c)Answer:

8.17

km[2]

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

y = 15 \cos 57^\circ

\phantom{y} = 8.1695\ldots

\phantom{y} = 8.17

Question 7

(a)Answer:

150^\circ

°[2]

Bearing of

A

from

B

= 107^\circ + 180^\circ

= 287^\circ

Angle

ABC = 287^\circ - 137^\circ

ABC = 150^\circ

(b)Answer:

18.4

km[2]

AC^2 = 7^2 + 12^2 - 2(7)(12) \cos 150^\circ

AC^2 = 338.4922\ldots

AC = 18.3981\ldots

\phantom{AC} = 18.4

(c)Answer:

19.0^\circ

°[2]

\dfrac{\sin BAC}{12} = \dfrac{\sin 150^\circ}{18.3981\ldots}

\sin BAC = 0.3261\ldots

BAC = 19.0334\ldots^\circ

\phantom{BAC} = 19.0^\circ

(d)Answer:

126.0^\circ

°[1]

Bearing of

C

from

A

= 107^\circ + 19.0334\ldots^\circ

= 126.0^\circ

(e)Answer:

306.0^\circ

°[1]

Bearing of

A

from

C

= 126.0^\circ + 180^\circ

= 306.0^\circ