### Trigonometry

```Trigonometry
3D Trigonometry
s
p, q and r are points on level ground, [sr] is
a vertical flagpole of height h. The angles
of elevation of the top of the flagpole
from p and q are α and β, respectively.
q
h
β
30º
(i) If | α | = 60º and | β | = 30º,
express | pr | and | qr | in terms of h.
60º
α
p
r
s
s
h
OPP
60º
p
h
tan 60 
pr
h
3
pr
r
q
h
30º
60º
h
pr 
3
p
r
s
h
tan 30 
qr
1
h

3 qr
q
30º
qr  3h
p
60º
h
OPP
r
(ii) Find | pq | in terms of h, if tan qrp = 8.
s
Pythagoras’ Theorem
x 2  12  ( 8 )2
 1 8
9
q
30º
3h
3x
8
A
1
p
1h
cos A 
3
r
60º
h
3
a2 = b2 + c2 – 2bccosA
 
2
 
 h 
 h  1 
qp  3h  
  2 3h 
 3 
 3
 3  
2
2
2
2
9
h

h

2
h
h
2
 3h 2 
 h2 
3
3 3
1
2
cos A 
8h

3
3
q
r
2
qp 
2
2
8h

3
8
h
3
p
h
3
a2 = b2 + c2 – 2bccosA
The great pyramid at Giza in
Egypt has a square base and
four triangular faces.
The base of the pyramid is
of side 230 metres and the
pyramid is 146 metres high.
slanted edge
The top of the pyramid is directly above the centre of the base.
(i) Calculate the length of one of the slanted edges, correct to the
nearest metre.
Pythagoras’ theorem
x 2  2302  2302
x  105800
x
2
146
162·6
x  325·269..
 162·6
2
2006 Paper 2 Q5 (b)
162·6
230 m
230 m
The great pyramid at Giza in
Egypt has a square base and
four triangular faces.
The base of the pyramid is
of side 230 metres and the
pyramid is 146 metres high.
slanted edge
The top of the pyramid is directly above the centre of the base.
(i) Calculate the length of one of the slanted edges, correct to the
nearest metre.
Pythagoras’ theorem
l 2  1462  162·62
l
l 
47754·76
l  218·528..
2
146
162·6
 219 m
2006 Paper 2 Q5 (b)
162·6 m
146 m
(ii) Calculate, correct to
two significant numbers,
the total area of the four
triangular faces of the
pyramid (assuming they
are smooth flat surfaces)
Pythagoras’ theorem
219  h  115
2
2
2
slanted edge
219 m
h
h2  2192  1152
115 m230 m
h 2  34736  186·375..  186·4
m
1
Area of triangle  base × height
2
1
 (230)(186·4)
2
2

21436
m
2006 Paper 2 Q5 (b)
(ii) Calculate, correct to
two significant numbers,
the total area of the four
triangular faces of the
pyramid (assuming they
are smooth flat surfaces)
Pythagoras’ theorem
219  h  115
2
2
2
slanted edge
219 m
h2  2192  1152
115 m
h 2  34736  186·375..  186·4
m
Total area  21436  4
 85744 m2
 86000 m2
2006 Paper 2 Q5 (b)
h
s
r
h
θ
p
q
3x
2θ
p
h
tan  
3x
r
t
x
h
θ
3x
h  3x tan 
pqrs is a vertical wall of height h on level ground. p is a point
on the ground in front of the wall. The angles of elevation of r
from p is θ and the angle of elevation of s from p is 2θ.
| pq | = 3| pt |.
Find θ.
2005 Paper 2 Q5 (c)
q
s
s
h
p
2θ
x
2θ
t
p
h
tan 2 
x
r
t
x
h
θ
3x
h  x tan 2
pqrs is a vertical wall of height h on level ground. p is a point
on the ground in front of the wall. The angles of elevation of r
from p is θ and the angle of elevation of s from p is 2θ.
| pq | = 3| pt |.
Find θ.
2005 Paper 2 Q5 (c)
q
s
2 tan 
tan 2 
1  tan 2 
2 tan 
3xtan θ  xtan 2θ
1  tan 2 
2θ
p
3t 1  t
t  3t  0
t 1  3t
1
t 
3
2
x
h
θ
q
3x
Let t = tan θ t  0
2t
3t 
2
1 t
3
r
t
2
2
  2t
3t  3t  2t
0
1  3t  0
1
t  tan  
3
2005 Paper 2 Q5 (c)
3
2


6
d
abc is an isosceles triangle on a horizontal
plane, such that |ab|  |ac|  5 and |bc|  4.
m is the midpoint of [bc].
2
5
b
A
5
(i) Find | bac | to the nearest degree.
a 2  b2  c 2  2bc cos A
m4
42  52  52  2(5)(5)cos A
16  25  25  50cos A
16  50  50cos A
c
50cos A  50  16
34
cos A 
50
A  47·156....
A  47
a
d
abc is an isosceles triangle on a horizontal
plane, such that |ab|  |ac|  5 and |bc|  4.
m is the midpoint of [bc].
(ii) A vertical pole [ad] is erected
at a such that |ad |  2, find
|amd | to the nearest degree.
am  2  5
2
2
5
b
2
m
am  25  4
am  21
amd
2
21
a
21 5
2
2
2
tan amd 
2
c
amd  23·578..  24
```