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Peper Referenc:e(s)

1380/4H Examiner's use only

Edexcel GCSE Mathematics (Linear)- 1380 Paper 4 (Calculator)

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Trigonometry Past Paper Questions

Arranged by Topic Model Answers

Materials required for examination Ruler graduated in centimetres and millimetres, protractor, compasses, pen, HB pencil, eraser, calculator. Tracing paper may be used.

Instructions to Candidates

Items included with question papers Nil

In the boxes above, write your centre number, candidate number, your surname, initials and signature. Check that you have the correct question paper. Answer ALL the questions. Write your answers in the spaces provided in this question paper. You must NOT write on the formulae page. Anything you write on the formulae page will gain NO credit. If you need more space to complete your answer to any question, use additional answer sheets.

Information for Candidates The marks for individual questions and the parts of questions are shown in round brackets: e.g. (2). There are 26 questiorn in this question paper. The total mark for this paper is 100. There are 24 pages in this question paper. Any blank pages are indicated. Calculators may be used. If your calculator does not have an button, take the value ofn to be 3.142 unless the question instructs otherwise.

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GCSE Mathematics (Linear) 1380

Fonnul ae: Higher Tier

You must not write on this for·mulae page. Anything you write on this formulae page will gain NO credit.

Volume of a prism = area of cross section x length

Volume of sphere = 4nr3

.)

Surface area of sphere = 4nr2

In any triangJe ABC

S. R I a b c me u e --= --= --sin A sin B sin C

Area of triangle =!ab sin C

Volume of cone =+nr2h .)

Curved sur·face area of cone = nrl

T he Quadratic Equation

The solutions of ax2 + bx + c = 0

where a:~- 0, are given by

- b ±~(b2 - 4ac) X = -----''-'----'-

2a

_ I

AdjacentcosHypotenuse

8.2cos10.6

39.323

x

x

=

=

=

39.3

51.3

AdjacentcosHypotenuse

5cos8

51.3178

x

x

=

=

=

OppositetanAdjacent

=

tan 4012.5

12.5 tan4010.488

y

yy

=

==

o

o

10.5

OppositetanAdjacent

=

tan4012.5

4tan10

21.8014

y

RPQ

RPQ

=

=

∠ =

o

21.8

121 8 7228

A bh

A

A

=

= × ×

=

2 2 2

2 2 2

2

2

7 849 64113

11310.6301

AC AB BCACACAC

ACAC

= += += +=

==

10.63

28

OppositetanAdjacent

=

32tan46

34.8244

y

y

=

=

34.8

121 14 6242

A bh

A

A

=

= × ×

=

42

2 2 2

2

2

14 6196 36232

23215.2315

ABABAB

ABAB

= += +=

==

15.23

AdjacentcosHypotenuse

=

cos5816

16cos588.4787

AB

ABAB

=

==

o

o

8.48

1 sin21 9 15sin110264.4292

A ac B

A

A

=

= × ×

=

o

64.4

OppositetanAdjacent

=

2.4tan4.5

28.072

x

x

=

=

28.1

180 90 28.07261.928

AA

∠ = − −∠ =

Bearing of C from A180 61.928

118.072−

= 118

2 2 2

2 2 2

cos2

10 12 15cos2 10 12

19cos240

85.459

b c aAbc

A

A

A

+ −=

+ −=× ×

=

= (use or INV cos key)1cos−

85.5

2 2 2

2 2 2

2

2

8 364 955

7.42

CB CA ABCBCBCBCB

= −= −= −=

=Oppositesin

Hypotenuse=

(Pythagoras)

7.42sin50

sin50 7.427.42 9.686

sin50

CDCD

CD

=

=

= =

o

o

o

9.69

( )( )

2 2 2

2 2 2

2

2 cos

9.69 19 2 9.69 19cos 40

172.823

172.82313.1462

d e c ec D

d

d

dd

= + −

= + − × ×

=

==

o

13.2

- 1

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