# Trigonometry - PowerPoint PPT Presentation

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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. h. β. 30 º. q. r. α. 60 º.

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Trigonometry

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## 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.

h

β

30º

q

r

α

60º

(i)If |α|= 60º and |β|= 30º, express |pr| and |qr| in terms of h.

p

s

h

p

r

s

OPP

h

60º

30º

q

r

60º

p

s

h

OPP

30º

q

r

60º

p

(ii)Find|pq| in terms of h, if tan qrp=

A

s

Pythagoras’ Theorem

h

30º

q

r

60º

p

a2 = b2+c2– 2bccosA

q

r

p

a2 = b2+c2– 2bccosA

slanted edge

Sommets 5

1 90 72 -17.00787

2 270 72 -17.00787

3 270 252 -17.00787

4 90 252 -17.00787

5 180 162 68.03149

Faces 5

1 4 4 3 2 1

2 3 1 2 5

3 3 2 3 5

4 3 3 4 5

5 3 4 1 5

BSommets 5

1 70.4549365234253 18.5097949350555 -90.9217083697797

2 245.53709547791 30.1395555679914 -131.05880197977

3 204.102094663231 100.878766349906 -291.305813700551

4 29.0199357087471 89.2490057169698 -251.168720090559

5 139.83922903418 137.698138286049 -157.341972681488

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.

Pythagoras’ theorem

x

230 m

2

= 105800

x

162·6

146

x=

325·269..

=

162·6

162·6

230 m

2

Sommets 5

1 99 63 -17.00787

2 225 63 -17.00787

3 225 189 -17.00787

4 99 189 -17.00787

5 162 126 68.03149

Faces 5

1 4 4 3 2 1

2 3 1 2 5

3 3 2 3 5

4 3 3 4 5

5 3 4 1 5

BSommets 5

1 99 63 -17.00787

2 225 63 -17.00787

3 225 189 -17.00787

4 99 189 -17.00787

5 162 126 68.03149

2006 Paper 2 Q5 (b)

slanted edge

Sommets 5

1 90 72 -17.00787

2 270 72 -17.00787

3 270 252 -17.00787

4 90 252 -17.00787

5 180 162 68.03149

Faces 5

1 4 4 3 2 1

2 3 1 2 5

3 3 2 3 5

4 3 3 4 5

5 3 4 1 5

BSommets 5

1 70.4549365234253 18.5097949350555 -90.9217083697797

2 245.53709547791 30.1395555679914 -131.05880197977

3 204.102094663231 100.878766349906 -291.305813700551

4 29.0199357087471 89.2490057169698 -251.168720090559

5 139.83922903418 137.698138286049 -157.341972681488

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.

Pythagoras’ theorem

l

146 m

2

= 47754·76

l

146

l= 218·528..

162·6

162·6 m

= 219 m

Sommets 5

1 99 63 -17.00787

2 225 63 -17.00787

3 225 189 -17.00787

4 99 189 -17.00787

5 162 126 68.03149

Faces 5

1 4 4 3 2 1

2 3 1 2 5

3 3 2 3 5

4 3 3 4 5

5 3 4 1 5

BSommets 5

1 99 63 -17.00787

2 225 63 -17.00787

3 225 189 -17.00787

4 99 189 -17.00787

5 162 126 68.03149

2006 Paper 2 Q5 (b)

slanted edge

1

2

1

2

Area of triangle = base × height

= (230)(186·4)

Sommets 5

1 90 72 -17.00787

2 270 72 -17.00787

3 270 252 -17.00787

4 90 252 -17.00787

5 180 162 68.03149

Faces 5

1 4 4 3 2 1

2 3 1 2 5

3 3 2 3 5

4 3 3 4 5

5 3 4 1 5

BSommets 5

1 70.4549365234253 18.5097949350555 -90.9217083697797

2 245.53709547791 30.1395555679914 -131.05880197977

3 204.102094663231 100.878766349906 -291.305813700551

4 29.0199357087471 89.2490057169698 -251.168720090559

5 139.83922903418 137.698138286049 -157.341972681488

(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 m

h

=34736

= 186·375..

= 186·4 m

115 m

230 m

2

h

= 21436 m2

Sommets 5

1 99 63 -17.00787

2 225 63 -17.00787

3 225 189 -17.00787

4 99 189 -17.00787

5 162 126 68.03149

Faces 5

1 4 4 3 2 1

2 3 1 2 5

3 3 2 3 5

4 3 3 4 5

5 3 4 1 5

BSommets 5

1 99 63 -17.00787

2 225 63 -17.00787

3 225 189 -17.00787

4 99 189 -17.00787

5 162 126 68.03149

2006 Paper 2 Q5 (b)

slanted edge

Sommets 5

1 90 72 -17.00787

2 270 72 -17.00787

3 270 252 -17.00787

4 90 252 -17.00787

5 180 162 68.03149

Faces 5

1 4 4 3 2 1

2 3 1 2 5

3 3 2 3 5

4 3 3 4 5

5 3 4 1 5

BSommets 5

1 70.4549365234253 18.5097949350555 -90.9217083697797

2 245.53709547791 30.1395555679914 -131.05880197977

3 204.102094663231 100.878766349906 -291.305813700551

4 29.0199357087471 89.2490057169698 -251.168720090559

5 139.83922903418 137.698138286049 -157.341972681488

(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 m

h

=34736

= 186·375..

= 186·4 m

115 m

2

h

Total area

= 21436  4

= 85744 m2

= 86000 m2

Sommets 5

1 99 63 -17.00787

2 225 63 -17.00787

3 225 189 -17.00787

4 99 189 -17.00787

5 162 126 68.03149

Faces 5

1 4 4 3 2 1

2 3 1 2 5

3 3 2 3 5

4 3 3 4 5

5 3 4 1 5

BSommets 5

1 99 63 -17.00787

2 225 63 -17.00787

3 225 189 -17.00787

4 99 189 -17.00787

5 162 126 68.03149

2006 Paper 2 Q5 (b)

r

h

θ

p

q

3x

s

r

t

x

h

θ

p

q

3x

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)

s

h

p

t

x

s

r

t

x

h

θ

p

q

3x

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)

s

r

t

x

h

θ

3

x

tanθ

x

tan2θ

=

p

q

3x

Let t = tan θ

2005 Paper 2 Q5 (c)

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

A

(i)Find |bac| to the nearest degree.

5

m

4

c

d

21

2

Ðamd =

tan

21

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

(ii)A vertical pole [ad] is erected

at a such that |ad| =2, find

|amd| to the nearest degree.

2

5

m

c

2

=

am

21

amd