Igcse Trigonometry 705b14
This document was ed by and they confirmed that they have the permission to share it. If you are author or own the copyright of this book, please report to us by using this report form. Report 2z6p3t
Overview 5o1f4z
& View Igcse Trigonometry as PDF for free.
More details 6z3438
- Words: 1,386
- Pages: 14
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
Click here to read the solution to this question Click here to return to the index
12 m 9m
Solution to question 1 Using Pythagoras’ theorem we have
a. x
3 cm
x 2 = 3 2 + 52 x = 32 + 52
5 cm
= 9 + 25 = 34 = 5.83cm
b.
Using Pythagoras’ theorem, ing that 12 m is the hypotenuse. 122 = y 2 + 92
y 12 m 9m
y 2 = 122 − 92 y = 144 − 81 = 63 = 7.94m
Click here to read the question again Click here to return to the index
Question 2 For the triangles below find the missing letters. a.
b. 22!
a
7.5 cm
34!
1.2 cm
c.
d. 4.6 cm d 17.5
7.3 cm c
3.45 cm
Click here to read the solution to this question Click here to return to the index
!
b
Solution to question 2 First label the sides of each triangle, adjacent, opposite and hypotenuse. a. hyp
a
opp
34!
1.2 cm adj
⇒ a = 1.2 × tan 34! = 0.809 cm
b. 22!
The missing side is the opposite side and we have the adjacent and angle hence opp a tanθ = ⇒ tan34! = adj 1.2
b
The missing side is the adjacent side and we have the hypotenuse and angle hence adj b cosθ = ⇒ cos 22! = hyp 7.5
adj
7.5 cm hyp
⇒ b = 7.5 × cos 22! = 6.95 cm opp
c.
opp 4.6 cm hyp 7.3 cm
4.6 ! ⇒ c = sin−1 = 39.1 7.3
adj c
d.
d
The missing side is the hypotenuse side and we have the adjacent side and angle hence adj 3.45 cosθ = ⇒ cos17.5! = d hyp
hyp
17.5!
3.45 cm adj
The angle is and we know the opposite side and hypotenuse hence 4.6 opp ⇒ sin c = sinθ = 7.3 hyp
opp
⇒d =
Click here to read the question again Click here to return to the index
3.45 = 3.62cm cos17.5!
Question 3 In the diagram below find the following lengths a. AC b. AD c. CD. A
! 27! 32
5.6 cm
D
B
C
Click here to read the solution to this question Click here to return to the index
Solution to question 3 A
! 27! 32
5.6 cm
D
B
C A
a.
First draw out triangle ABC separately adj 5.6 cosθ = ⇒ cos 27! = hyp AC 5.6 AC = = 6.29 cm cos 27!
b.
hyp
adj 5.6 cm
First draw out triangle ACD separately
cosθ =
adj ⇒ cos32! = hyp
⇒ AD = c.
27!
AD 5.6 cos27!
5.6 × cos32 = 5.33 cm cos 27!
opp ⇒ sin32! = hyp
⇒ AD =
CD 5.6 cos 27!
5.6 × sin32! = 3.33 cm cos 27!
Click here to read the question again Click here to return to the index
opp
C
A
32!
!
Considering triangle ACD. sinθ =
B
adj
hyp D
5.6 cos 27! opp C
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. N
120!
A
P N
60 km
Bearing to be measured
C
150!
D
B
Distance to be measured
80 km
Distance = 135 km, measured with a ruler. Bearing is 137! measured with a protractor.
L
b.
Drawing out triangles PAB and BDL separately we have P
adj 30!
A
adj
B 60
hyp
D
!
opp 60 km
hyp
opp
B
80 km
L
cosθ =
PA adj ⇒ cos30! = hyp 60
cosθ =
PA adj ⇒ cos 60! = hyp 80
⇒ PA = 60 × cos30! (East) opp AB sinθ = ⇒ sin30! = hyp 60
⇒ PA = 80 × cos60! (East) opp AB sinθ = ⇒ sin60! = hyp 80
⇒ AB = 60 × sin30! (South)
⇒ AB = 80 × sin 60! (South)
Click here to continue with solution or go to next page
The total distance east is = 60 cos30! + 80 cos 60! = 91.961… = 92.0km The total distance south is = 60 sin30! + 80 sin60! = 99.282… = 99.3km c.
The distance PL is found by Pythagoras’ theorem
PL =
( 60cos30
!
+ 80 cos 60!
) + ( 60 sin30 2
!
+ 80 sin60!
)
2
= 135.328… = 135 km
ˆ so we use triangle L We need to find angle L N
P
adj 60 cos 30! + 80 sin 60!
C
opp 60 sin30! + 80 sin60!
hyp
L
tanθ =
! ! opp ˆ = tan−1 60 sin30 + 80 sin60 ⇒ L ! ! adj 60 cos30 + 80cos 60
= 47.2!
But we need to find the bearing of Littlehampton from Portshampton, which is given by 90! + 47.2! = 137.2!
Click here to read the question again Click here to return to the index
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
D
C
28!
A
23!
B
1.75 m 20 m
Consider triangle ABC
a.
C
tanθ =
opp BC ⇒ tan23! = adj 20
BC = 20 × tan 23! = 8.49m
hyp opp A
23!
20 m adj
B
The height of the wall is = 1.75 + 8.49 = 10.239… = 10.2m
Consider triangle ABD
b.
D
opp
28!
20 m
opp BC ⇒ tan28! = adj 20
BC = 20 × tan 28! = 10.63m
hyp
A
tanθ =
B
adj
Click here to read the question again Click here to return to the index
The height of the flagpole from the ground is = 1.75 + 10.63 = 12.384m Height of flagpole is CD = 12.38 − 10.24 = 2.14 m
Question 6
A rectangular box ABCDEFGH has AB = 5 cm , BC = 6 cm and AE = 3cm Calculate a.
AC,
b.
AG,
c.
ˆ . The angle CAG
H E
3 cm
A
F
D
5 cm
Click here to read the solution to this question Click here to return to the index
G
C
B
6 cm
Solution to question 6 H
G
E
F
3 cm
D
A
C
6 cm
B
5 cm
Consider triangle ABC
a.
C
By Pythagoras’ theorem AC = 52 + 62
6 cm
A
b.
5 cm
= 61 = 7.81cm
B
Consider triangle ACG.
G
By Pythagoras’ theorem hyp
AG = 61 + 32 Note
(
61
)
2
= 61
= 70 = 8.37 cm
3 cm
A
61 m adj
c.
Consider triangle ACG. tanθ =
opp ˆ = tan−1 3 = 21.0! ⇒ CAG adj 61
Click here to read the question again Click here to return to the index
opp
C
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
Click here to read the solution to this question Click here to return to the index
12 m 9m
Solution to question 1 Using Pythagoras’ theorem we have
a. x
3 cm
x 2 = 3 2 + 52 x = 32 + 52
5 cm
= 9 + 25 = 34 = 5.83cm
b.
Using Pythagoras’ theorem, ing that 12 m is the hypotenuse. 122 = y 2 + 92
y 12 m 9m
y 2 = 122 − 92 y = 144 − 81 = 63 = 7.94m
Click here to read the question again Click here to return to the index
Question 2 For the triangles below find the missing letters. a.
b. 22!
a
7.5 cm
34!
1.2 cm
c.
d. 4.6 cm d 17.5
7.3 cm c
3.45 cm
Click here to read the solution to this question Click here to return to the index
!
b
Solution to question 2 First label the sides of each triangle, adjacent, opposite and hypotenuse. a. hyp
a
opp
34!
1.2 cm adj
⇒ a = 1.2 × tan 34! = 0.809 cm
b. 22!
The missing side is the opposite side and we have the adjacent and angle hence opp a tanθ = ⇒ tan34! = adj 1.2
b
The missing side is the adjacent side and we have the hypotenuse and angle hence adj b cosθ = ⇒ cos 22! = hyp 7.5
adj
7.5 cm hyp
⇒ b = 7.5 × cos 22! = 6.95 cm opp
c.
opp 4.6 cm hyp 7.3 cm
4.6 ! ⇒ c = sin−1 = 39.1 7.3
adj c
d.
d
The missing side is the hypotenuse side and we have the adjacent side and angle hence adj 3.45 cosθ = ⇒ cos17.5! = d hyp
hyp
17.5!
3.45 cm adj
The angle is and we know the opposite side and hypotenuse hence 4.6 opp ⇒ sin c = sinθ = 7.3 hyp
opp
⇒d =
Click here to read the question again Click here to return to the index
3.45 = 3.62cm cos17.5!
Question 3 In the diagram below find the following lengths a. AC b. AD c. CD. A
! 27! 32
5.6 cm
D
B
C
Click here to read the solution to this question Click here to return to the index
Solution to question 3 A
! 27! 32
5.6 cm
D
B
C A
a.
First draw out triangle ABC separately adj 5.6 cosθ = ⇒ cos 27! = hyp AC 5.6 AC = = 6.29 cm cos 27!
b.
hyp
adj 5.6 cm
First draw out triangle ACD separately
cosθ =
adj ⇒ cos32! = hyp
⇒ AD = c.
27!
AD 5.6 cos27!
5.6 × cos32 = 5.33 cm cos 27!
opp ⇒ sin32! = hyp
⇒ AD =
CD 5.6 cos 27!
5.6 × sin32! = 3.33 cm cos 27!
Click here to read the question again Click here to return to the index
opp
C
A
32!
!
Considering triangle ACD. sinθ =
B
adj
hyp D
5.6 cos 27! opp C
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. N
120!
A
P N
60 km
Bearing to be measured
C
150!
D
B
Distance to be measured
80 km
Distance = 135 km, measured with a ruler. Bearing is 137! measured with a protractor.
L
b.
Drawing out triangles PAB and BDL separately we have P
adj 30!
A
adj
B 60
hyp
D
!
opp 60 km
hyp
opp
B
80 km
L
cosθ =
PA adj ⇒ cos30! = hyp 60
cosθ =
PA adj ⇒ cos 60! = hyp 80
⇒ PA = 60 × cos30! (East) opp AB sinθ = ⇒ sin30! = hyp 60
⇒ PA = 80 × cos60! (East) opp AB sinθ = ⇒ sin60! = hyp 80
⇒ AB = 60 × sin30! (South)
⇒ AB = 80 × sin 60! (South)
Click here to continue with solution or go to next page
The total distance east is = 60 cos30! + 80 cos 60! = 91.961… = 92.0km The total distance south is = 60 sin30! + 80 sin60! = 99.282… = 99.3km c.
The distance PL is found by Pythagoras’ theorem
PL =
( 60cos30
!
+ 80 cos 60!
) + ( 60 sin30 2
!
+ 80 sin60!
)
2
= 135.328… = 135 km
ˆ so we use triangle L We need to find angle L N
P
adj 60 cos 30! + 80 sin 60!
C
opp 60 sin30! + 80 sin60!
hyp
L
tanθ =
! ! opp ˆ = tan−1 60 sin30 + 80 sin60 ⇒ L ! ! adj 60 cos30 + 80cos 60
= 47.2!
But we need to find the bearing of Littlehampton from Portshampton, which is given by 90! + 47.2! = 137.2!
Click here to read the question again Click here to return to the index
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
D
C
28!
A
23!
B
1.75 m 20 m
Consider triangle ABC
a.
C
tanθ =
opp BC ⇒ tan23! = adj 20
BC = 20 × tan 23! = 8.49m
hyp opp A
23!
20 m adj
B
The height of the wall is = 1.75 + 8.49 = 10.239… = 10.2m
Consider triangle ABD
b.
D
opp
28!
20 m
opp BC ⇒ tan28! = adj 20
BC = 20 × tan 28! = 10.63m
hyp
A
tanθ =
B
adj
Click here to read the question again Click here to return to the index
The height of the flagpole from the ground is = 1.75 + 10.63 = 12.384m Height of flagpole is CD = 12.38 − 10.24 = 2.14 m
Question 6
A rectangular box ABCDEFGH has AB = 5 cm , BC = 6 cm and AE = 3cm Calculate a.
AC,
b.
AG,
c.
ˆ . The angle CAG
H E
3 cm
A
F
D
5 cm
Click here to read the solution to this question Click here to return to the index
G
C
B
6 cm
Solution to question 6 H
G
E
F
3 cm
D
A
C
6 cm
B
5 cm
Consider triangle ABC
a.
C
By Pythagoras’ theorem AC = 52 + 62
6 cm
A
b.
5 cm
= 61 = 7.81cm
B
Consider triangle ACG.
G
By Pythagoras’ theorem hyp
AG = 61 + 32 Note
(
61
)
2
= 61
= 70 = 8.37 cm
3 cm
A
61 m adj
c.
Consider triangle ACG. tanθ =
opp ˆ = tan−1 3 = 21.0! ⇒ CAG adj 61
Click here to read the question again Click here to return to the index
opp
C
Related Documents c2h70
Igcse Trigonometry 705b14
November 2019 108
Trigonometry 4p3d1r
December 2019 52
Trigonometry 101 203r
December 2019 69
C3 Trigonometry 2o1i6n
December 2019 42
Trigonometry Gelfand 493v5p
July 2021 0
Trigonometry Ii 1y3z3q
August 2021 0More Documents from "Dienta Alias" l2al
Pbs Band 6- Science 204f3j
October 2019 53
Powerpoint Presentation - Sejarah Tingkatan 2 Bab 7 4o5wh
November 2021 0
Igcse Trigonometry 705b14
November 2019 108
Medusa Play 5f6u2l
October 2019 84
Free Fall Physics Lab Report 3f335h
April 2021 0