I.G.C.S.E. Trigonometry Index: Please click on the question number you want

Question 1 Question 2 Question 3 Question 4 Question 5 Question 6 You can access the solutions from the end of each question

Question 1 1. For the triangles below calculate the missing lengths.

a. b.

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x 3 cm

5 cm

y

9 m

12 m

Solution to question 1 a.

b. Click here to read the question again Click here to return to the index

x 3 cm

5 cm

y

9 m

12 m

Using Pythagoras’ theorem we have

2 2 2

2 2

3 5

3 5

5.83cm

9 25

34

x

x

= +

= +

= +

==

Using Pythagoras’ theorem, remembering that 12 m is the hypotenuse.

2 2 2

2 2 2

12 912 9

144 81

7.94m63

yy

y

= += −

= −

==

Question 2 For the triangles below find the missing letters. a. b. c. d. Click here to re Click here to re

34! 1.2 cm

a 22! b

7.5 cm

7.3 cm

4.6 cm

17.5!

3.45 cm

d

c

ad the solution to this question

turn to the index

Solution to question 2 First label the sides of each triangle, adjacent, opposite and hypotenuse. a. b. c. d. Click here Click here

34

7.5 c

1

3.45

j

7.3 c

hyp

to

to

! 1.2 cm

a

j

p

m

7.

cm

m

tan341.2

1

opptana.2 tan 0.809cm34

dja

a

θ = ⇒ =

⇒ = × =

!

!

ad

read the ques

return to the i

22! b

5!

d

4.6 cm

j

p

op

The missing side is the adjacent side and we have the hypotenuse and angle hence

cos227.5

adjcoshyp

bθ ⇒ == !

!

hyp

67 .95c.5 os22 mcb⇒ = =×

opp

adj

The angle is and we know the opposite side and hypotenuse hence

1 3

4.

9.1

6sin7.3

4.6sin7.3

oppsinhyp

c

c

θ

−

⇒ =

⇒ = =

=

!

The missing side is the hypotenuse side and we have the adjacent side and angle hence

hyp

p

3.45adjco

3

cos17.5

3.45

shyp

2c.6 m

d

d

θ ⇒ =

⇒ =

=

=

!

!

op

cos17.5 ad

c

hyp

ad

op

tion aga

ndex

The missing side is the opposite side and we have the adjacent and angle hence

in

Question 3 In the diagram below find the following lengths a. AC b. AD c. CD. Click here to read the s Click here to return to t

B

A

5.6 cm

!

olution to this q

he index

D

3227!

C

uestion

Solution to question 3 a. First draw out triangle ABC separately

5.6

6.29

adjcos coh

s2

cm

7

5.6cos27

yp AC

AC

θ

=

= ⇒ =

=

!

!

b. First draw out triangle ACD separately

cos32 5.6cos27

5.6

adjcoshy

cos32 5.33cm

p

cos27

AD

AD

θ ⇒ =

×

=

⇒ = =

!

!

!

!

c. Considering triangle ACD.

sin32 5.6cos27

5.6

oppsinhy

sin32 3.33cm

p

cos27

CD

AD

θ ⇒ =

×

=

⇒ = =

!

!

!

!

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p

32!

C

A

5.6cos27!

D

B

A

5.6 cm

32!27!

opp

27!

hyp

op

D

adj

hyp

C

C

B

5.6 cm

A

adj

Question 4 A ship sails from Porthampton on a bearing of 120! towards a buoy marker for 60 km and then changes course to a bearing of 150! for another 80 km until it reaches Littlehampton. a. Draw a scale diagram using a scale of 1 cm to 10 km and find the

distance and bearing of Littlehampton from Porthampton. b. Using trigonometry, calculate how far east and how far south the ship

has travelled. c. Calculate the distance and bearing of Littlehampton from Porthampton. Click here to read the solution to this question Click here to return to the index

Solution to question 4 a. Scale drawing scale 1 cm to 10 km. b.

Click

Bearing to be measured

Distance to be measured

m

D

C 120!

150!

A

B

Distance 135= km, measured with a ruler. Bearing is137! measured with a protractor.

Drawing out triangles PAB and BDL separa

adjcos cos30hyp 60

oppsin sin30hyp

6

60 sin30 (So

0 cos30

60

(Ea

u h)

)

t

st

PA

PAAB

AB

θ

θ

= ⇒ =

⇒ =

= ⇒

⇒

×

×

=

= !

!

!

!

cos

sin

PA

AB

θ

θ

⇒

⇒

here to continue with solution or go to

30!

m

P

p

B

80 k

60 km

N P

N

tely we have L

60!D

hyp opp

adj

60 k

A

B

hyp

op

adj

=

n

adj cos60hyp 80

opp sin60hyp

8

80 sin60 (So

0 cos60

80

(Ea

u h)

)

t

st

PA

AB

= ⇒ =

=

⇒

×

×

=

= !

!

!

!

ext page

80 km

L

The total distance east is 60cos30 80cos60 91.96 92.0k1 m= = =+! ! … The total distance south is 60sin30 80sin60 99.28 99.3k2 m= = =+! ! … c. The distance PL is found by Pythagoras’ theorem

( ) ( )2 260sin30 80s60cos30 80

135.328

in60

135

s6

km

co 0PL + += +

==

! !! !

…

We need to find angle ˆCPL so we use triangle CPL

opptanadj

θ = ⇒ ˆ tan

47.2

CPL −=

= But we need to find the beawhich is given by

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60cos30P

hyp

j

ad N

1

60co60sin30 80sin6

s30

0 0c s8 o 60 +

+! !

! !

!

ring of Littlehampton from Portshampton,

90 47.2 137.2+ =! ! !

again

x

60sin30 80sin60+! !

80sin60+! !

C

L

opp

Question 5 The height of an eye of a man is 175 cm. He is standing 20 m from a building, which has a flagpole on it. He looks up at an angle of elevation 23! and sees the top of the building. He then looks up at the top of the flagpole, which has an angle of elevation of 28! . a. Calculate the height of the building. b. Calculate the height of the flagpole. Click here to read the solution to this question Click here to return to the index

Solution to question 5 First draw a diagram a. Consider triangle ABC b. Consider triangle ABD Click here to read the question again Click here to return to the index

28!

23!m

A

p

23!

20 m

opptanadj

20 tan2BC

θ = ⇒

= × The height of th

1.75 8.49= + =

A B

D

opp hyp

28!

20 m

opptanadj

20 tan2BC

θ = ⇒

= × The height of thground is

1.75 10.63= + Height of flagpo

12.38 10.24= −

D

C

B

A

20 m

1.75

tan2320

3 8.49m

BC=

=

!

!

e wall is 10.239 10.2m=…

B

C

op

hyp

adj

tan2820

8 10.63m

BC=

=

!

!

e flagpole from the

12.384m=

le is CD 2 4.1 m=

adj

Question 6 A rectangular box ABCDEFGH has 5cmAB = , 6cmBC = and 3cmAE = Calculate a. AC, b. AG, c. The angle ˆCAG . Click here to read the solution to this questio Click here to return to the index

3 cm

G

E

A

H

F

D

B

5 cm

n

C

6 cm

Solution to question 6 a. Consider triangle ABC b. Consider triangle ACG.

c. Consider triangle ACG. t

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3 cm

G

E

A

A B

C

6 c

5 cm

(2 Note 661 3

8.37cm

70

AG = +

==

H

F

D

B

5 cm

opp ˆan tanadj

CAGθ = ⇒ =

on again

ex

A

m

By Pythagoras’

2 2

61

5

7

6

.81cm

AC = +

==

)21 61=

C

6 cm

1 361

21.0− =

!

G

p hyp

61 m

j

theorem

3 cm

op

C

ad

By Pythagoras’ theorem