Sec 3 E-Math Further trigonometry worksheet with answers

A free Sec 3 E-Math 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 E-Math worksheet: Further 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

In triangle

ABC

BC = 11

cm,

CA = 13

cm and the area is

25

cm². Given that angle

ACB

is acute, find angle

ACB

, correct to 1 decimal place.

Answer°[3]

In triangle

ABC

AB = 13

cm,

AC = 11

cm and angle

BAC = 48^\circ

. Find the length of

BC

, correct to 3 significant figures.

Answercm[3]

In triangle

ABC

BC = 8

cm,

CA = 8

cm and the area is

24

cm². Given that angle

ACB

is acute, find angle

ACB

, correct to 1 decimal place.

Answer°[3]

(a)Given that

\cos 44^\circ = 0.719

, write down the value of

\cos 136^\circ

Answer (a)[2]

(b)Given that

\sin \theta = \dfrac{8}{17}

and that

\theta

is obtuse, find, without using a calculator, the value of

\sin \theta + \cos \theta

Answer (b)[3]

Section B

PQR

is a triangular car park on level ground, with

PQ = 38

QR = 30

m and

PR = 55

(a)Find angle

PQR

, correct to 1 decimal place.

Answer (a)°[2]

(b)Find the area of the car park, correct to 3 significant figures.

Answer (b)m²[2]

(c)Find the shortest distance from

Q

PR

, correct to 3 significant figures.

Answer (c)m[2]

(d)Find angle

QPR

, correct to 1 decimal place.

Answer (d)°[2]

In the quadrilateral

ABCD

AB = 8

cm,

BC = 13

cm, angle

ABC = 56^\circ

and angle

CAD = 44^\circ

(a)Find the length of

AC

, correct to 3 significant figures.

Answer (a)cm[2]

(b)Find the area of triangle

ABC

, correct to 3 significant figures.

Answer (b)cm²[2]

(c)The area of triangle

ACD

26

cm². Find the length of

AD

, correct to 3 significant figures.

Answer (c)cm[2]

(d)Find the length of

CD

, correct to 3 significant figures.

Answer (d)cm[2]

END OF PAPER

Answers and worked solutions

Question 1

Answer:

\phantom{\theta} = 20.5^\circ

(1 d.p.) °[3]

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

25 = \dfrac{1}{2} \times 11 \times 13 \times \sin \theta

\sin \theta = \dfrac{2 \times 25}{11 \times 13}

\phantom{\sin \theta} = 0.3496\ldots

\theta = \sin^{-1}(0.3496\ldots)

\theta = 20.4659\ldots^\circ

\phantom{\theta} = 20.5^\circ

(1 d.p.)

Question 2

Answer:

\phantom{x} = 9.93

cm (3 s.f.)[3]

\text{Cosine rule}

a^2 = b^2 + c^2 - 2bc \cos A

b = 11

c = 13

A = 48^\circ

x^2 = 11^2 + 13^2 - 2(11)(13) \cos 48^\circ

x^2 = 98.6286\ldots

x = 9.9311\ldots

\phantom{x} = 9.93

cm (3 s.f.)

Question 3

Answer:

\phantom{\theta} = 48.6^\circ

(1 d.p.) °[3]

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

24 = \dfrac{1}{2} \times 8 \times 8 \times \sin \theta

\sin \theta = \dfrac{2 \times 24}{8 \times 8}

\phantom{\sin \theta} = 0.75

\theta = \sin^{-1}(0.75)

\theta = 48.5903\ldots^\circ

\phantom{\theta} = 48.6^\circ

(1 d.p.)

Question 4

(a)Answer:

\phantom{\cos 136^\circ} = -0.719

[2]

\cos 136^\circ = -\cos(180^\circ - 136^\circ)

\phantom{\cos 136^\circ} = -\cos 44^\circ

\phantom{\cos 136^\circ} = -0.719

(b)Answer:

\sin \theta + \cos \theta = -\dfrac{7}{17}

[3]

Opposite

= 8

, hypotenuse

= 17

: adjacent

= \sqrt{17^2 - 8^2}

Adjacent

= \sqrt{225}

= 15

(Pythagoras' theorem)

\cos \theta = -\cos \alpha

(

\theta

obtuse)

\cos \theta

= -\dfrac{15}{17}

\sin \theta + \cos \theta = \dfrac{8}{17} - \dfrac{15}{17}

\phantom{\sin \theta + \cos \theta} = \dfrac{8 - 15}{17}

\sin \theta + \cos \theta = -\dfrac{7}{17}

Question 5

(a)Answer:

107.4^\circ

°[2]

\cos PQR = \dfrac{30^2 + 38^2 - 55^2}{2(30)(38)}

\cos PQR = -0.2986\ldots

(negative, so the angle is obtuse)

PQR = 107.3785\ldots^\circ

\phantom{PQR} = 107.4^\circ

(b)Answer:

544

m²[2]

\text{Area} = \dfrac{1}{2} \times 30 \times 38 \times \sin 107.3785\ldots^\circ

= 543.9806\ldots

\phantom{} = 544

m²

(c)Answer:

19.8

m[2]

The shortest distance is the perpendicular height

h

from

Q

PR

\dfrac{1}{2} \times 55 \times h = 543.9806\ldots

h = \dfrac{2 \times 543.9806\ldots}{55}

= 19.7811\ldots

\phantom{} = 19.8

(d)Answer:

31.4^\circ

°[2]

\dfrac{\sin QPR}{30} = \dfrac{\sin 107.3785\ldots^\circ}{55}

\sin QPR = 0.5205\ldots

QPR = 31.3695\ldots^\circ

\phantom{QPR} = 31.4^\circ

(acute:

Q

is the obtuse angle)

Question 6

(a)Answer:

10.8

cm[2]

AC^2 = 8^2 + 13^2 - 2(8)(13) \cos 56^\circ

AC^2 = 116.6878\ldots

AC = 10.8022\ldots

\phantom{AC} = 10.8

(b)Answer:

43.1

cm²[2]

\text{Area} = \dfrac{1}{2} \times 8 \times 13 \times \sin 56^\circ

= 43.1099\ldots

\phantom{} = 43.1

cm²

(c)Answer:

6.93

cm[2]

\dfrac{1}{2} \times AC \times AD \times \sin 44^\circ = 26

(use the unrounded

AC

)

AD = \dfrac{2 \times 26}{10.8022\ldots \times \sin 44^\circ}

= 6.9297\ldots

\phantom{} = 6.93

(d)Answer:

7.55

cm[2]

CD^2 = (10.8022\ldots)^2 + (6.9297\ldots)^2 - 2(10.8022\ldots)(6.9297\ldots) \cos 44^\circ

CD^2 = 57.0145\ldots

CD = 7.5507\ldots

\phantom{CD} = 7.55

Practice questions by skill

Sec 3 E-Math 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

PQR

is a triangular car park on level ground, with

PQ = 38

QR = 30

m and

PR = 55

(a)Find angle

PQR

, correct to 1 decimal place.

Answer (a)°[2]

(b)Find the area of the car park, correct to 3 significant figures.

Answer (b)m²[2]

(c)Find the shortest distance from

Q

PR

, correct to 3 significant figures.

Answer (c)m[2]

(d)Find angle

QPR

, correct to 1 decimal place.

Answer (d)°[2]

Answers and worked solutions

Question 1

Answer:

\phantom{\theta} = 20.5^\circ

(1 d.p.) °[3]

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

25 = \dfrac{1}{2} \times 11 \times 13 \times \sin \theta

\sin \theta = \dfrac{2 \times 25}{11 \times 13}

\phantom{\sin \theta} = 0.3496\ldots

\theta = \sin^{-1}(0.3496\ldots)

\theta = 20.4659\ldots^\circ

\phantom{\theta} = 20.5^\circ

(1 d.p.)

Question 2

Answer:

\phantom{x} = 9.93

cm (3 s.f.)[3]

\text{Cosine rule}

a^2 = b^2 + c^2 - 2bc \cos A

b = 11

c = 13

A = 48^\circ

x^2 = 11^2 + 13^2 - 2(11)(13) \cos 48^\circ

x^2 = 98.6286\ldots

x = 9.9311\ldots

\phantom{x} = 9.93

cm (3 s.f.)

Question 3

Answer:

\phantom{\theta} = 48.6^\circ

(1 d.p.) °[3]

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

24 = \dfrac{1}{2} \times 8 \times 8 \times \sin \theta

\sin \theta = \dfrac{2 \times 24}{8 \times 8}

\phantom{\sin \theta} = 0.75

\theta = \sin^{-1}(0.75)

\theta = 48.5903\ldots^\circ

\phantom{\theta} = 48.6^\circ

(1 d.p.)

Question 4

(a)Answer:

\phantom{\cos 136^\circ} = -0.719

[2]

\cos 136^\circ = -\cos(180^\circ - 136^\circ)

\phantom{\cos 136^\circ} = -\cos 44^\circ

\phantom{\cos 136^\circ} = -0.719

(b)Answer:

\sin \theta + \cos \theta = -\dfrac{7}{17}

[3]

Opposite

= 8

, hypotenuse

= 17

: adjacent

= \sqrt{17^2 - 8^2}

Adjacent

= \sqrt{225}

= 15

(Pythagoras' theorem)

\cos \theta = -\cos \alpha

(

\theta

obtuse)

\cos \theta

= -\dfrac{15}{17}

\sin \theta + \cos \theta = \dfrac{8}{17} - \dfrac{15}{17}

\phantom{\sin \theta + \cos \theta} = \dfrac{8 - 15}{17}

\sin \theta + \cos \theta = -\dfrac{7}{17}

Question 5

(a)Answer:

107.4^\circ

°[2]

\cos PQR = \dfrac{30^2 + 38^2 - 55^2}{2(30)(38)}

\cos PQR = -0.2986\ldots

(negative, so the angle is obtuse)

PQR = 107.3785\ldots^\circ

\phantom{PQR} = 107.4^\circ

(b)Answer:

544

m²[2]

\text{Area} = \dfrac{1}{2} \times 30 \times 38 \times \sin 107.3785\ldots^\circ

= 543.9806\ldots

\phantom{} = 544

m²

(c)Answer:

19.8

m[2]

The shortest distance is the perpendicular height

h

from

Q

PR

\dfrac{1}{2} \times 55 \times h = 543.9806\ldots

h = \dfrac{2 \times 543.9806\ldots}{55}

= 19.7811\ldots

\phantom{} = 19.8

(d)Answer:

31.4^\circ

°[2]

\dfrac{\sin QPR}{30} = \dfrac{\sin 107.3785\ldots^\circ}{55}

\sin QPR = 0.5205\ldots

QPR = 31.3695\ldots^\circ

\phantom{QPR} = 31.4^\circ

(acute:

Q

is the obtuse angle)

Question 6

(a)Answer:

10.8

cm[2]

AC^2 = 8^2 + 13^2 - 2(8)(13) \cos 56^\circ

AC^2 = 116.6878\ldots

AC = 10.8022\ldots

\phantom{AC} = 10.8

(b)Answer:

43.1

cm²[2]

\text{Area} = \dfrac{1}{2} \times 8 \times 13 \times \sin 56^\circ

= 43.1099\ldots

\phantom{} = 43.1

cm²

(c)Answer:

6.93

cm[2]

\dfrac{1}{2} \times AC \times AD \times \sin 44^\circ = 26

(use the unrounded

AC

)

AD = \dfrac{2 \times 26}{10.8022\ldots \times \sin 44^\circ}

= 6.9297\ldots

\phantom{} = 6.93

(d)Answer:

7.55

cm[2]

CD^2 = (10.8022\ldots)^2 + (6.9297\ldots)^2 - 2(10.8022\ldots)(6.9297\ldots) \cos 44^\circ

CD^2 = 57.0145\ldots

CD = 7.5507\ldots

\phantom{CD} = 7.55