TRIGONOMETRIC QUESTIONS FOR PRACTICE BY ARIHANT. VERY HELPFUL IN COMPETITIVE EXAMS.
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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.
3 cm
y
x
5 cm
12 m 9m
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Solution to question 1 Using Pythagoras’ theorem we have
a.
x 2
x
3 cm
=
32
+
52
x = 32 + 52 =
9 + 25
=
34
5 cm =
b.
5.83cm
Using Pythagoras’ theorem, remembering that 12 m is the hypotenuse. 122 = y 2 + 92
y 12 m 9m
y 2
=
=
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−
92
y = 144 − 81 =
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122
63 7.94m
Question 2 For the triangles below find the missing letters.
a.
b. 22
a 34
!
7.5 cm
!
1.2 cm
c.
d. 4.6 cm
d 17.5
7.3 cm
c
3.45 cm
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!
b
Solution to question 2 First label the sides of each triangle, adjacent, opposite and hypotenuse.
a. hyp
a opp
The missing side is the opposite side and we have the adjacent and angle hence a opp ⇒ tan34 = tanθ = adj 1.2 !
34
!
1.2 cm
⇒ a = 1.2 × tan 34
adj
b. 22
b
!
!
=
0.809cm
The missing side is the adjacent side and we have the hypotenuse and angle hence b adj ⇒ cos22 = cosθ = hyp 7.5
adj
!
7.5 cm hyp
⇒ b = 7.5 × cos 22
!
=
6.95cm
opp
c.
opp 4.6 cm hyp 7.3 cm
The angle is and we know the opposite side and hypotenuse hence opp 4.6 ⇒ sin c = sinθ = hyp 7.3
4.6 ⇒ c = sin 1 = 39.1 7.3 −
adj
c
d.
d 17.5
The missing side is the hypotenuse side and we have the adjacent side and angle hence adj 3.45 ⇒ cos17.5 = cosθ = d hyp
hyp
!
!
3.45 cm adj
!
opp
⇒ d =
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3.45 cos17.5
!
=
3.62 cm
Question 3 In the diagram below find the following lengths a. AC b. AD c. CD . A
32
!
27
!
5.6 cm
D
B
C
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Solution to question 3 A
27
32
!
!
5.6 cm
D
B
a.
A
First draw out triangle ABC separately cosθ =
AC =
b.
C
adj hyp
⇒ cos 27
5.6 cos27
!
=
!
=
27
5.6 AC
hyp
adj 5.6 cm
6.29 cm
First draw out triangle ACD separately cosθ =
adj ⇒ cos32 hyp
!
=
!
AD 5.6 cos27
B
C
opp
A
!
5.6 × cos32
32
!
adj
!
⇒ AD = c.
cos27
!
=
5.33cm
opp ⇒ sin32 hyp
!
=
cos27
CD 5.6 cos27
!
5.6 × sin32
!
⇒ AD =
cos27
!
=
3.33cm
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D
5.6
Considering triangle ACD . sinθ =
hyp
!
opp
C
Question 4 A ship sails from Porthampton on a bearing of 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 Litt lehampton 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.
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Solution to question 4 a.
Scale drawing scale 1 cm to 10 km. N
120
!
A
P N
60 km
Bearing to be measured
C
150
!
D
B
Distance to be measured
80 km
Distance = 135 km km,, measured with a ruler. Bearing is 137 measured with a protractor. !
L
b.
Drawing out triangles PAB and BDL separately we have adj
P 30
A
adj
B 60
!
hyp
D
!
opp 60 km
hyp
opp
B 80 km
L
cosθ =
adj ⇒ cos 30 hyp
!
=
PA 60
cosθ =
adj ⇒ cos 60 hyp
!
=
PA 80
⇒ PA = 60 × cos 30 (East) AB opp ⇒ sin 30 = sinθ =
⇒ PA = 80 × cos 60 (East) AB opp ⇒ sin 60 = sinθ =
⇒ AB = 60 × sin 30 (South)
⇒ AB = 80 × sin 60 60 (South)
!
!
hyp
60
!
!
!
hyp
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80
!
The total distance east is
The total distance south is
c.
60 co cos 30 30
=
!
60 si sin 30 30
=
+ !
80 co cos 60 60
!
80 si sin 60 60
!
+
=
91.961… = 92.0km
=
99.282… = 99.3km
The distance PL is found by Pythagoras’ theorem
( 60 cos 30
PL = =
135.328 …
=
135 km
!
+
80 cos 60
!
2
) ( 60 sin 30 +
!
+
80 sin60
!
)
2
ˆ so we use triangle CPL We need to find angle CPL N
P
adj 60 co cos 30 30
!
+
80 si sin 60 60
!
C
opp 60 si sin 30 30
hyp
!
+
80 si sin 60 60
!
L
tanθ =
opp sin 30 ˆ = tan 1 60 si ⇒ CPL adj 60cos 30
!
−
=
!
80 si sin 60 60 + 80 cos 60 !
+
!
47.2
!
But we need to find the bearing of Littlehampton from Portshampton, which is given by 90
!
+
47.2
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!
=
137.2
!
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 t he top of the flagpole, which has an angle of elevation of of 28 . !
!
a.
Calculate the height of the building.
b.
Calculate the height of the flagpole.
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Solution to question 5 First draw a diagram
D
C
28
!
A 23
B
!
1.75 m 20 m
a.
Consider triangle ABC C hyp
A
23
opp ⇒ tan 23 adj
BC = 20 × tan 23
!
=
!
=
BC 20
8.49 m
opp
!
20 m
tanθ =
B
The height of the wall is = 1.75 + 8.49 = 10.239 … = 10.2m
adj
b.
Consider triangle ABD D
opp
28
!
20 m
opp ⇒ tan 28 adj
BC = 20 × tan 28
hyp
A
tanθ =
B
adj
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!
=
!
=
BC 20
10.63 m
The height of the flagpole from the ground is = 1.75 + 10.63 = 12.384 m Height of flagpole is CD = 12. 38 − 10.24 = 2.1 4 m
Question 6 A rectangular box ABCDEFGH has AB = 5 cm , BC = 6 cm and AE = 3 cm Calculate
a.
AC ,
b.
AG ,
c.
ˆ . The angle CAG
H E
3 cm
A
F
D
5 cm
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G
C
B
6 cm
Solution to question 6 H
G
E
F
3 cm
D
A
a.
C 6 cm
B
5 cm
Consider triangle ABC C
By Pythagoras’ theorem
AC = 52 6 cm = =
A
b.
5 cm
+
62
61 7.81cm
B
Consider triangle ACG .
G
By Pythagoras’ theorem hyp
AG = 61 + 32 Note = =
(
)
61
2 =
opp
61
3 cm
70 8.37cm
A
C
61 m adj
c.
Consider triangle ACG . tanθ =
opp ˆ = tan 1 3 = 21.0 ⇒ CAG adj 61
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