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Topic 2
Periodic Functions and Applications I
Periodic Functions and Applications I
trigonometry – definition and practical applications of the sin, cos and tan ratios
simple practical applications of the sine and cosine rules
Model Calculate the value of the pronumeral in each of the following:
cos x = 12
19
x = cos-1 12 19
= 5050’
X°
19 cm
12 cm
Page 33 Exercise 2.1
Find the perimeter of the figure below.
mON
ON
ON
6740sin
43
4340sin
Perimeter = 30+43+30+51+67
= 221m
Page 36 Exercise 2.2 No. 1 – 6
Page 42 Exercise 2.4 No. 1, 3, 5 - 8
Worked solution Ex 2.4 No. 6
The Sine Rule
A
a
b
C
c
B
b
A
a
sinsinsin
B
c
C
Model Find x in the following
85
c =
x
)2(65.13
65sin85sin
1585sin
15
65sin
sinsin
dp
x
xC
c
B
b
65A
B
C
b =
15
Example 11, page 46 (Modelling and Problem Solving)
Jermaine sees the top of a sand dune at an angle of 16°. On walking 40m closer, she finds the angle of elevation increases to 22°. What is the height of the sand dune?
16° 22°
40m
h
AC
B
D
* If we knew the length of line DB, we could easily work out the
height.
16° 22°
40m
h
AC
B
D
Consider triangle ABD
16°
6°
158°
40m
d
b
a .).1(5.105
16sin6sin
406sin
40
16sin
sinsin
pdma
a
aB
b
A
a
16° 22°A
B
DC
h
mh
h
h
5.39
22sin5.1055.105
22sin
Height
105.5m
Page 46 Exercise 2.5 N0. 1, 2(b,c,d,e), 3, 4, 7-9
The Cosine Rule
Baccab cos2222
Abccba cos2222
A
a
b
B
c
C
Cabbac cos2222
Similarly
Model Find the length of the unknown side in the triangle below:
12
1732
A
B
C
c =
b=
)2(33.9
87
00.87
32cos408144289
32cos121721217
cos222
222
dp
a
Abccba
a
Model Find the size of the largest angle in the triangle below:
12
17
A
B
C
c =
b=
'57114
cos192
81
cos19281
cos192208289
cos19214464289
cos128212817
cos2222
222
B
B
B
B
B
B
Baccab
a=
8
N.B. Largest angle is always opposite the largest side
Page 50 Exercise 2.6
Examples 15 & 16, page 51 & 52 Page 53 Exercise 2.7No. 1(a,b,c,f), 2(a,b,c,f), 4, 7, 11-15
Worked solution Ex 2.7 No. 13
3D Applications
Example 17: page 56
- From point A, a mountain point due north is at an elevation of 20°. From point B, 2Km east of A and on the same level as A, the bearing of the peak is N40°W. Find the height of the peak above A and B.
- From point A, a mountain point due north is at an elevation of 20°. From point B, 2Km east of A and on the same level as A, the bearing of the peak is N40°W. Find the height of the peak above A and B.
Consider triangle ABM
Angle BAM is a right angle
- From point A, a mountain point due north is at an elevation of 20°. From point B, 2Km east of A and on the same level as A, the bearing of the peak is N40°W. Find the height of the peak above A and B.
Consider triangle ABM
A
M
B
50°
2 Km
Side AM, which is common to both triangles, can now be calculated
50tan22
50tan
AM
AM
- From point A, a mountain point due north is at an elevation of 20°. From point B, 2Km east of A and on the same level as A, the bearing of the peak is N40°W. Find the height of the peak above A and B.
Now consider triangle AMP
A
P
M
20°2 tan50
dp) (3 Km868.0
20tan50tan250tan2
20tan
PM
PM
PM
Example 18: page 57
John observes that the top of a transmission tower at bearing 038° is at an elevation of 12°. Mary is 575m due east of John and can see the tower on a bearing of 295°. What is the height of the tower?
Example 18: page 57
John observes that the top of a transmission tower at bearing 038° is at an elevation of 12°. Mary is 575m due east of John and can see the tower on a bearing of 295°. What is the height of the tower?
52° 25°
103°
FH
W
575m
We need side WH (f)
f
w =(1dp)4.249
25sin103sin
575103sin
575
25sin
sinsin
mf
f
fW
w
F
f
Example 18: page 57
John observes that the top of a transmission tower at bearing 038° is at an elevation of 12°. Mary is 575m due east of John and can see the tower on a bearing of 295°. What is the height of the tower?
12°
T
H W249.4m
We can now calculate tower height
.).1(0.53
12tan4.2494.249
12tan
PDm
TW
TW
Page 59 Exercise 2.8
Worked solution Ex 2.7 No. 13
