Sec 4 G1 Maths Trigonometry worksheet with answers

A free Sec 4 G1 Maths worksheet on trigonometry: 11 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 4 G1 Maths worksheet: 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

In the diagram,

\angle ABC = 90^\circ

. Find

x

, correct to 3 significant figures.

Answerm[2]

In the diagram,

\angle XYZ = 90^\circ

. Find

x

, correct to 1 decimal place.

Answer°[2]

A car is

25

m from the foot of a lighthouse

15

m high. Find the angle of depression of the car from the top of the lighthouse, correct to 1 decimal place.

Answer°[2]

A building

60

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

30

m from its foot, correct to 1 decimal place.

Answer°[2]

In the diagram,

\angle DEF = 90^\circ

. Find

x

, correct to 3 significant figures.

Answerm[2]

In the diagram,

\angle ABC = 90^\circ

. Find

x

, correct to 1 decimal place.

Answer°[2]

In the diagram,

\angle DEF = 90^\circ

. Find

x

, correct to 3 significant figures.

Answercm[2]

From a point

25

m from the foot of a building on level ground, the angle of elevation of the top is

41^\circ

. Find the height of the building, correct to 3 significant figures.

Answerm[2]

In the diagram,

\angle LMN = 90^\circ

. Find

x

, correct to 3 significant figures.

Answerm[2]

Section B

(a)In the diagram,

\angle ABC = 90^\circ

. Find

x

, correct to 1 decimal place.

Answer (a)°[2]

(b)From the top of a tower

12

m high, the angle of depression of a buoy is

32^\circ

. Find the distance of the buoy from the foot of the tower, correct to 3 significant figures.

Answer (b)m[2]

(c)In the diagram,

\angle DEF = 90^\circ

. Find

x

, correct to 3 significant figures.

Answer (c)m[2]

(a)In the diagram,

\angle STU = 90^\circ

. Find

x

, correct to 3 significant figures.

Answer (a)m[2]

(b)In the diagram,

\angle DEF = 90^\circ

. Find

x

, correct to 1 decimal place.

Answer (b)°[2]

(c)A tree

40

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

60

m from its foot, correct to 1 decimal place.

Answer (c)°[2]

END OF PAPER

Answers and worked solutions

Question 1

Answer:

= 33.7

m (3 s.f.)[2]

From the 24° angle: the side

x

m is the adjacent and the side

15

m is the opposite.

\tan 24^\circ = \dfrac{15}{x}

x = \dfrac{15}{\tan 24^\circ}

= 33.6905\ldots

= 33.7

m (3 s.f.)

Question 2

Answer:

= 66.4^\circ

(1 d.p.) °[2]

From the angle x°: the side

15

m is the hypotenuse and the side

6

m is the adjacent.

\cos x^\circ = \dfrac{6}{15}

x = \cos^{-1}\left(\dfrac{6}{15}\right)

= 66.4218\ldots

= 66.4^\circ

(1 d.p.)

Question 3

Answer:

\phantom{\theta} = 31.0^\circ

(1 d.p.) °[2]

Angle

TBF = \theta

(alt. ∠s)

\tan \theta = \dfrac{15}{25}

\theta = \tan^{-1}\left(\dfrac{15}{25}\right)

\theta = 30.9637\ldots^\circ

\phantom{\theta} = 31.0^\circ

(1 d.p.)

Question 4

Answer:

\phantom{\theta} = 63.4^\circ

(1 d.p.) °[2]

\tan \theta = \dfrac{60}{30}

\theta = \tan^{-1}\left(\dfrac{60}{30}\right)

\theta = 63.4349\ldots^\circ

\phantom{\theta} = 63.4^\circ

(1 d.p.)

Question 5

Answer:

= 28.6

m (3 s.f.)[2]

From the 35° angle: the side

x

m is the adjacent and the side

20

m is the opposite.

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

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

= 28.5629\ldots

= 28.6

m (3 s.f.)

Question 6

Answer:

= 49.7^\circ

(1 d.p.) °[2]

From the angle x°: the side

17

m is the hypotenuse and the side

11

m is the adjacent.

\cos x^\circ = \dfrac{11}{17}

x = \cos^{-1}\left(\dfrac{11}{17}\right)

= 49.6797\ldots

= 49.7^\circ

(1 d.p.)

Question 7

Answer:

= 5.03

cm (3 s.f.)[2]

From the 57° angle: the side

x

cm is the opposite and the side

6

cm is the hypotenuse.

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

x = 6 \sin 57^\circ

= 5.0320\ldots

= 5.03

cm (3 s.f.)

Question 8

Answer:

\phantom{h} = 21.7

m (3 s.f.)[2]

\tan 41^\circ = \dfrac{h}{25}

h = 25 \tan 41^\circ

h = 21.7321\ldots

\phantom{h} = 21.7

m (3 s.f.)

Question 9

Answer:

= 16.8

m (3 s.f.)[2]

From the 63° angle: the side

x

m is the hypotenuse and the side

15

m is the opposite.

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

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

= 16.8348\ldots

= 16.8

m (3 s.f.)

Question 10

(a)Answer:

= 59.7^\circ

(1 d.p.) °[2]

From the angle x°: the side

12

cm is the opposite and the side

7

cm is the adjacent.

\tan x^\circ = \dfrac{12}{7}

x = \tan^{-1}\left(\dfrac{12}{7}\right)

= 59.7435\ldots

= 59.7^\circ

(1 d.p.)

(b)Answer:

\phantom{x} = 19.2

m (3 s.f.)[2]

Angle

TBF = 32^\circ

(alt. ∠s)

\tan 32^\circ = \dfrac{12}{x}

x = \dfrac{12}{\tan 32^\circ}

x = 19.2040\ldots

\phantom{x} = 19.2

m (3 s.f.)

(c)Answer:

= 14.0

m (3 s.f.)[2]

From the 35° angle: the side

x

m is the opposite and the side

20

m is the adjacent.

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

x = 20 \tan 35^\circ

= 14.0041\ldots

= 14.0

m (3 s.f.)

Question 11

(a)Answer:

= 25.4

m (3 s.f.)[2]

From the 38° angle: the side

x

m is the hypotenuse and the side

20

m is the adjacent.

\cos 38^\circ = \dfrac{20}{x}

x = \dfrac{20}{\cos 38^\circ}

= 25.3803\ldots

= 25.4

m (3 s.f.)

(b)Answer:

= 44.4^\circ

(1 d.p.) °[2]

From the angle x°: the side

20

m is the hypotenuse and the side

14

m is the opposite.

\sin x^\circ = \dfrac{14}{20}

x = \sin^{-1}\left(\dfrac{14}{20}\right)

= 44.4270\ldots

= 44.4^\circ

(1 d.p.)

(c)Answer:

\phantom{\theta} = 33.7^\circ

(1 d.p.) °[2]

\tan \theta = \dfrac{40}{60}

\theta = \tan^{-1}\left(\dfrac{40}{60}\right)

\theta = 33.6900\ldots^\circ

\phantom{\theta} = 33.7^\circ

(1 d.p.)

Sec 4 G1 Maths Trigonometry 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 ABC = 90^\circ

. Find

x

, correct to 1 decimal place.

Answer (a)°[2]

(b)From the top of a tower

12

m high, the angle of depression of a buoy is

32^\circ

. Find the distance of the buoy from the foot of the tower, correct to 3 significant figures.

Answer (b)m[2]

(c)In the diagram,

\angle DEF = 90^\circ

. Find

x

, correct to 3 significant figures.

Answer (c)m[2]

Answers and worked solutions

Question 1

Answer:

= 33.7

m (3 s.f.)[2]

From the 24° angle: the side

x

m is the adjacent and the side

15

m is the opposite.

\tan 24^\circ = \dfrac{15}{x}

x = \dfrac{15}{\tan 24^\circ}

= 33.6905\ldots

= 33.7

m (3 s.f.)

Question 2

Answer:

= 66.4^\circ

(1 d.p.) °[2]

From the angle x°: the side

15

m is the hypotenuse and the side

6

m is the adjacent.

\cos x^\circ = \dfrac{6}{15}

x = \cos^{-1}\left(\dfrac{6}{15}\right)

= 66.4218\ldots

= 66.4^\circ

(1 d.p.)

Question 3

Answer:

\phantom{\theta} = 31.0^\circ

(1 d.p.) °[2]

Angle

TBF = \theta

(alt. ∠s)

\tan \theta = \dfrac{15}{25}

\theta = \tan^{-1}\left(\dfrac{15}{25}\right)

\theta = 30.9637\ldots^\circ

\phantom{\theta} = 31.0^\circ

(1 d.p.)

Question 4

Answer:

\phantom{\theta} = 63.4^\circ

(1 d.p.) °[2]

\tan \theta = \dfrac{60}{30}

\theta = \tan^{-1}\left(\dfrac{60}{30}\right)

\theta = 63.4349\ldots^\circ

\phantom{\theta} = 63.4^\circ

(1 d.p.)

Question 5

Answer:

= 28.6

m (3 s.f.)[2]

From the 35° angle: the side

x

m is the adjacent and the side

20

m is the opposite.

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

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

= 28.5629\ldots

= 28.6

m (3 s.f.)

Question 6

Answer:

= 49.7^\circ

(1 d.p.) °[2]

From the angle x°: the side

17

m is the hypotenuse and the side

11

m is the adjacent.

\cos x^\circ = \dfrac{11}{17}

x = \cos^{-1}\left(\dfrac{11}{17}\right)

= 49.6797\ldots

= 49.7^\circ

(1 d.p.)

Question 7

Answer:

= 5.03

cm (3 s.f.)[2]

From the 57° angle: the side

x

cm is the opposite and the side

6

cm is the hypotenuse.

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

x = 6 \sin 57^\circ

= 5.0320\ldots

= 5.03

cm (3 s.f.)

Question 8

Answer:

\phantom{h} = 21.7

m (3 s.f.)[2]

\tan 41^\circ = \dfrac{h}{25}

h = 25 \tan 41^\circ

h = 21.7321\ldots

\phantom{h} = 21.7

m (3 s.f.)

Question 9

Answer:

= 16.8

m (3 s.f.)[2]

From the 63° angle: the side

x

m is the hypotenuse and the side

15

m is the opposite.

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

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

= 16.8348\ldots

= 16.8

m (3 s.f.)

Question 10

(a)Answer:

= 59.7^\circ

(1 d.p.) °[2]

From the angle x°: the side

12

cm is the opposite and the side

7

cm is the adjacent.

\tan x^\circ = \dfrac{12}{7}

x = \tan^{-1}\left(\dfrac{12}{7}\right)

= 59.7435\ldots

= 59.7^\circ

(1 d.p.)

(b)Answer:

\phantom{x} = 19.2

m (3 s.f.)[2]

Angle

TBF = 32^\circ

(alt. ∠s)

\tan 32^\circ = \dfrac{12}{x}

x = \dfrac{12}{\tan 32^\circ}

x = 19.2040\ldots

\phantom{x} = 19.2

m (3 s.f.)

(c)Answer:

= 14.0

m (3 s.f.)[2]

From the 35° angle: the side

x

m is the opposite and the side

20

m is the adjacent.

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

x = 20 \tan 35^\circ

= 14.0041\ldots

= 14.0

m (3 s.f.)

Question 11

(a)Answer:

= 25.4

m (3 s.f.)[2]

From the 38° angle: the side

x

m is the hypotenuse and the side

20

m is the adjacent.

\cos 38^\circ = \dfrac{20}{x}

x = \dfrac{20}{\cos 38^\circ}

= 25.3803\ldots

= 25.4

m (3 s.f.)

(b)Answer:

= 44.4^\circ

(1 d.p.) °[2]

From the angle x°: the side

20

m is the hypotenuse and the side

14

m is the opposite.

\sin x^\circ = \dfrac{14}{20}

x = \sin^{-1}\left(\dfrac{14}{20}\right)

= 44.4270\ldots

= 44.4^\circ

(1 d.p.)

(c)Answer:

\phantom{\theta} = 33.7^\circ

(1 d.p.) °[2]

\tan \theta = \dfrac{40}{60}

\theta = \tan^{-1}\left(\dfrac{40}{60}\right)

\theta = 33.6900\ldots^\circ

\phantom{\theta} = 33.7^\circ

(1 d.p.)