Trigonometry
3D Trigonometry
r
s
h
p
qβ
α
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 flagpolefrom p and q are α and β, respectively.
(i) If | α | = 60º and | β | = 30º, express | pr | and | qr | in terms of h.
30º
60º
s
h
rp ADJ
OPP
tan 60h
pr
3h
pr
3
hpr
s
h
r
p
q30º
60º
60º
tan30h
qr
1
3
h
qr 3qr h
s
h
r
p
qOPP
ADJ
30º
60º
r
s
h
p
q3h
(ii) Find | pq | in terms of h, if tan qrp = 8.
3
h
a2 = b2 + c2 – 2bccosAA
8
1
2 2 21 81 89
( )x
x
Pythagoras’ Theorem
3
1cos
3A
30º
60º
r
p
q
3
h
a2 = b2 + c2 – 2bccosA
2
22 13 2 3
33 3
h hqp h h
2
2 223
3 3
hh h
2 2 29 2
3
h h h
28
3
h
28
3
hqp
8
3h
1cos
3A
slanted edge
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.
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.
230 m
230 m
Pythagoras’ theoremx
2 2 2230 230x
105800
2
325·269..
x
2
x 162·6
162·6
162·6
146
2006 Paper 2 Q5 (b)
slanted edge
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.
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.
146 m
162·6 m
Pythagoras’ theoreml
47754·76
2
l 218·528..
l
219 m162·6
146
2 2 2146 162·6l
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)
slanted edge
219 m
230 m 347362 186·375..
h 186·4 m
h
115 m
Pythagoras’ theorem2 2 2219 115h
2 2 2219 115h
Area of triangle base × height
12
(230)(186·4)
12
21436 m2
2006 Paper 2 Q5 (b)
slanted edge
(ii) Calculate, correct to two significant numbers, the total area of the four triangular faces of the pyramid (assuming they are smooth flat surfaces)
219 m
347362 186·375..
h 186·4 m
h
115 m
Pythagoras’ theorem2 2 2219 115h
2 2 2219 115h
Total area 21436 4
85744 m2 86000 m2
2006 Paper 2 Q5 (b)
θ
2θ
3x
x
q
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.
p
s
r
h
tθ3x qp
r
h
tan3
h
x 3 tanh x
2005 Paper 2 Q5 (c)
θ
2θ
3x
x
q
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.
p
s
r
h
t2θ
x tp
s
h
tan 2h
x tan 2h x
2005 Paper 2 Q5 (c)
2
2tan
1 tan
tan θx θ
2θ
3x
x
qp
s
r
h
t2
2tantan 2
1 tan
tan 2θx3
Let t = tan θ
2
23
1
tt
t
23 1 2t t t
33 0t t 21 3 0t t
0t
2 1
3t
1tan
3t
33 3 2t t t
21 3 0t
6
2005 Paper 2 Q5 (c)
a
c
d
(i) Find | bac | to the nearest degree.b 5
m 5
2
4
A
2 2 24 5 5 2(5)(5)cos A
2 2 2 2 cosa b c bc A
16 25 25 50cos A
16 50 50cos A 50cos 50 16A
34cos
50A 47·156....A
abc is an isosceles triangle on a horizontal plane, such that |ab| |ac| 5 and |bc| 4.m is the midpoint of [bc].
47A
abc is an isosceles triangle on a horizontal plane, such that |ab| |ac| 5 and |bc| 4.m is the midpoint of [bc]. a
c
d
b 5
m 5
2
(ii) A vertical pole [ad] is erectedat a such that |ad | 2, find|amd | to the nearest degree.
2
2 2 22 5am 225 4am
23·578..amd 24
221am
21
amd
amd 2tan
21