Sec 3 G2 Maths Further trigonometry worksheet with answers

A free Sec 3 G2 Maths worksheet on further trigonometry: 6 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: Further 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 triangle

ABC

BC = 10

cm,

CA = 12

cm and angle

ACB = 28^\circ

. Find the area of the triangle, correct to 3 significant figures.

Answercm²[2]

In triangle

ABC

, angle

BAC = 57^\circ

, angle

ABC = 53^\circ

and

BC = 6

cm. Find the length of

AC

, correct to 3 significant figures.

Answercm[3]

(a)Given that

\cos 71^\circ = 0.326

, write down the value of

\cos 109^\circ

Answer (a)[2]

(b)Given that

\sin x = 0.65

and

x

is obtuse, find

x

, correct to 1 decimal place.

Answer (b)°[3]

Section B

In triangle

ABC

AB = 10

cm,

AC = 12

cm and angle

BAC = 75^\circ

(a)Find the length of

BC

, correct to 3 significant figures.

Answer (a)cm[2]

(b)Find angle

ACB

, correct to 1 decimal place.

Answer (b)°[2]

(c)Find the area of triangle

ABC

, correct to 3 significant figures.

Answer (c)cm²[2]

In triangle

ABC

AB = 4

cm,

AC = 7

cm and angle

BAC = 106^\circ

(a)Find the length of

BC

, correct to 3 significant figures.

Answer (a)cm[2]

(b)Find angle

ACB

, correct to 1 decimal place.

Answer (b)°[2]

(c)Find the area of triangle

ABC

, correct to 3 significant figures.

Answer (c)cm²[2]

In triangle

PQR

PQ = 12

cm,

QR = 9

cm and angle

QPR = 32^\circ

. Angle

PRQ

is obtuse.

(a)Find angle

PRQ

, correct to 1 decimal place.

Answer (a)°[3]

(b)Find angle

PQR

, correct to 1 decimal place.

Answer (b)°[1]

(c)Find the length of

PR

, correct to 3 significant figures.

Answer (c)cm[2]

(d)Find the area of triangle

PQR

, correct to 3 significant figures.

Answer (d)cm²[2]

END OF PAPER

Answers and worked solutions

Question 1

Answer:

\phantom{\text{Area}} = 28.2

cm² (3 s.f.)[2]

\text{Area} = \dfrac{1}{2}ab \sin C

a = 10

b = 12

C = 28^\circ

\text{Area} = \dfrac{1}{2} \times 10 \times 12 \times \sin 28^\circ

\text{Area} = 28.1682\ldots

\phantom{\text{Area}} = 28.2

cm² (3 s.f.)

Question 2

Answer:

\phantom{x} = 5.71

cm (3 s.f.)[3]

\text{Sine rule}

BC = 6

is opposite angle

A = 57^\circ

AC = x

is opposite angle

B = 53^\circ

\dfrac{a}{\sin A} = \dfrac{b}{\sin B}

\dfrac{6}{\sin 57^\circ} = \dfrac{x}{\sin 53^\circ}

x = \dfrac{6 \sin 53^\circ}{\sin 57^\circ}

x = 5.7135\ldots

\phantom{x} = 5.71

cm (3 s.f.)

Question 3

(a)Answer:

\phantom{\cos 109^\circ} = -0.326

[2]

\cos 109^\circ = -\cos(180^\circ - 109^\circ)

\phantom{\cos 109^\circ} = -\cos 71^\circ

\phantom{\cos 109^\circ} = -0.326

(b)Answer:

\phantom{x} = 139.5^\circ

(1 d.p.) °[3]

\sin x = 0.65

Acute angle

= \sin^{-1}(0.65)

= 40.5416\ldots^\circ

\sin x = \sin(180^\circ - 40.5416\ldots^\circ)

x = 180^\circ - 40.5416\ldots^\circ

\phantom{x} = 139.4583\ldots^\circ

\phantom{x} = 139.5^\circ

(1 d.p.)

Question 4

(a)Answer:

13.5

cm[2]

BC^2 = 12^2 + 10^2 - 2(12)(10) \cos 75^\circ

BC^2 = 181.8834\ldots

BC = 13.4864\ldots

\phantom{BC} = 13.5

(b)Answer:

45.7^\circ

°[2]

\dfrac{\sin C}{10} = \dfrac{\sin 75^\circ}{13.4864\ldots}

(use the unrounded

BC

)

\sin C = 0.7162\ldots

C = \sin^{-1}(0.7162\ldots)

C = 45.7433\ldots

\phantom{C} = 45.7^\circ

(acute: it is opposite the shorter side)

(c)Answer:

58.0

cm²[2]

\text{Area} = \dfrac{1}{2} \times 12 \times 10 \times \sin 75^\circ

= 57.9555\ldots

\phantom{} = 58.0

cm²

Question 5

(a)Answer:

8.97

cm[2]

BC^2 = 7^2 + 4^2 - 2(7)(4) \cos 106^\circ

BC^2 = 80.4356\ldots

BC = 8.9685\ldots

\phantom{BC} = 8.97

(b)Answer:

25.4^\circ

°[2]

\dfrac{\sin C}{4} = \dfrac{\sin 106^\circ}{8.9685\ldots}

(use the unrounded

BC

)

\sin C = 0.4287\ldots

C = \sin^{-1}(0.4287\ldots)

C = 25.3865\ldots

\phantom{C} = 25.4^\circ

(acute: it is opposite the shorter side)

(c)Answer:

13.5

cm²[2]

\text{Area} = \dfrac{1}{2} \times 7 \times 4 \times \sin 106^\circ

= 13.4576\ldots

\phantom{} = 13.5

cm²

Question 6

(a)Answer:

135.0^\circ

°[3]

\dfrac{\sin PRQ}{12} = \dfrac{\sin 32^\circ}{9}

\sin PRQ = 0.7065\ldots

The calculator gives

\sin^{-1}(0.7065\ldots) = 44.9556\ldots^\circ

, an acute angle.Angle

PRQ

is obtuse, so it is

180^\circ

minus this.

PRQ = 180^\circ - 44.9556\ldots^\circ

\phantom{PRQ} = 135.0443\ldots^\circ

\phantom{PRQ} = 135.0^\circ

(b)Answer:

13.0^\circ

°[1]

Angle

PQR = 180^\circ - 32^\circ - 135.0443\ldots^\circ

PQR = 12.9556\ldots^\circ

PQR = 13.0^\circ

(c)Answer:

3.81

cm[2]

\dfrac{PR}{\sin 12.9556\ldots^\circ} = \dfrac{9}{\sin 32^\circ}

PR = \dfrac{9 \sin 12.9556\ldots^\circ}{\sin 32^\circ}

= 3.8076\ldots

\phantom{} = 3.81

(d)Answer:

12.1

cm²[2]

\text{Area} = \dfrac{1}{2} \times 12 \times 3.8076\ldots \times \sin 32^\circ

= 12.1066\ldots

\phantom{} = 12.1

cm²

Sec 3 G2 Maths Further trigonometry worksheet with answers

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

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

Section B

In triangle

ABC

AB = 10

cm,

AC = 12

cm and angle

BAC = 75^\circ

(a)Find the length of

BC

, correct to 3 significant figures.

Answer (a)cm[2]

(b)Find angle

ACB

, correct to 1 decimal place.

Answer (b)°[2]

(c)Find the area of triangle

ABC

, correct to 3 significant figures.

Answer (c)cm²[2]

Answers and worked solutions

Question 1

Answer:

\phantom{\text{Area}} = 28.2

cm² (3 s.f.)[2]

\text{Area} = \dfrac{1}{2}ab \sin C

a = 10

b = 12

C = 28^\circ

\text{Area} = \dfrac{1}{2} \times 10 \times 12 \times \sin 28^\circ

\text{Area} = 28.1682\ldots

\phantom{\text{Area}} = 28.2

cm² (3 s.f.)

Question 2

Answer:

\phantom{x} = 5.71

cm (3 s.f.)[3]

\text{Sine rule}

BC = 6

is opposite angle

A = 57^\circ

AC = x

is opposite angle

B = 53^\circ

\dfrac{a}{\sin A} = \dfrac{b}{\sin B}

\dfrac{6}{\sin 57^\circ} = \dfrac{x}{\sin 53^\circ}

x = \dfrac{6 \sin 53^\circ}{\sin 57^\circ}

x = 5.7135\ldots

\phantom{x} = 5.71

cm (3 s.f.)

Question 3

(a)Answer:

\phantom{\cos 109^\circ} = -0.326

[2]

\cos 109^\circ = -\cos(180^\circ - 109^\circ)

\phantom{\cos 109^\circ} = -\cos 71^\circ

\phantom{\cos 109^\circ} = -0.326

(b)Answer:

\phantom{x} = 139.5^\circ

(1 d.p.) °[3]

\sin x = 0.65

Acute angle

= \sin^{-1}(0.65)

= 40.5416\ldots^\circ

\sin x = \sin(180^\circ - 40.5416\ldots^\circ)

x = 180^\circ - 40.5416\ldots^\circ

\phantom{x} = 139.4583\ldots^\circ

\phantom{x} = 139.5^\circ

(1 d.p.)

Question 4

(a)Answer:

13.5

cm[2]

BC^2 = 12^2 + 10^2 - 2(12)(10) \cos 75^\circ

BC^2 = 181.8834\ldots

BC = 13.4864\ldots

\phantom{BC} = 13.5

(b)Answer:

45.7^\circ

°[2]

\dfrac{\sin C}{10} = \dfrac{\sin 75^\circ}{13.4864\ldots}

(use the unrounded

BC

)

\sin C = 0.7162\ldots

C = \sin^{-1}(0.7162\ldots)

C = 45.7433\ldots

\phantom{C} = 45.7^\circ

(acute: it is opposite the shorter side)

(c)Answer:

58.0

cm²[2]

\text{Area} = \dfrac{1}{2} \times 12 \times 10 \times \sin 75^\circ

= 57.9555\ldots

\phantom{} = 58.0

cm²

Question 5

(a)Answer:

8.97

cm[2]

BC^2 = 7^2 + 4^2 - 2(7)(4) \cos 106^\circ

BC^2 = 80.4356\ldots

BC = 8.9685\ldots

\phantom{BC} = 8.97

(b)Answer:

25.4^\circ

°[2]

\dfrac{\sin C}{4} = \dfrac{\sin 106^\circ}{8.9685\ldots}

(use the unrounded

BC

)

\sin C = 0.4287\ldots

C = \sin^{-1}(0.4287\ldots)

C = 25.3865\ldots

\phantom{C} = 25.4^\circ

(acute: it is opposite the shorter side)

(c)Answer:

13.5

cm²[2]

\text{Area} = \dfrac{1}{2} \times 7 \times 4 \times \sin 106^\circ

= 13.4576\ldots

\phantom{} = 13.5

cm²

Question 6

(a)Answer:

135.0^\circ

°[3]

\dfrac{\sin PRQ}{12} = \dfrac{\sin 32^\circ}{9}

\sin PRQ = 0.7065\ldots

The calculator gives

\sin^{-1}(0.7065\ldots) = 44.9556\ldots^\circ

, an acute angle.Angle

PRQ

is obtuse, so it is

180^\circ

minus this.

PRQ = 180^\circ - 44.9556\ldots^\circ

\phantom{PRQ} = 135.0443\ldots^\circ

\phantom{PRQ} = 135.0^\circ

(b)Answer:

13.0^\circ

°[1]

Angle

PQR = 180^\circ - 32^\circ - 135.0443\ldots^\circ

PQR = 12.9556\ldots^\circ

PQR = 13.0^\circ

(c)Answer:

3.81

cm[2]

\dfrac{PR}{\sin 12.9556\ldots^\circ} = \dfrac{9}{\sin 32^\circ}

PR = \dfrac{9 \sin 12.9556\ldots^\circ}{\sin 32^\circ}

= 3.8076\ldots

\phantom{} = 3.81

(d)Answer:

12.1

cm²[2]

\text{Area} = \dfrac{1}{2} \times 12 \times 3.8076\ldots \times \sin 32^\circ

= 12.1066\ldots

\phantom{} = 12.1

cm²