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1-1 Using Trigonometry to Find Lengths. You have been hired to refurbish the Weslyville Tower… (copy the diagram, 10 lines high, the width of your page.). In order to bring enough gear, you need to know the height of the tower……. How would you determine the tower’s height?.

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1-1 Using Trigonometry to Find Lengths

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1 1 using trigonometry to find lengths

1-1Using Trigonometry to Find Lengths


1 1 using trigonometry to find lengths

You have been hired to refurbish the Weslyville Tower…(copy the diagram, 10 lines high, the width of your page.)

In order to bring enough gear, you need to know the height of the tower……

How would you determine the tower’s height?


1 1 using trigonometry to find lengths

  • When it is too difficult to obtain the measurements directly, we can operate on a model instead.

  • A model is a larger or smaller version of the original object.


1 1 using trigonometry to find lengths

  • A model must have similar proportions as the initial object to be useful.

  • Trigonometry uses TRIANGLES for models.

    We construct a similar triangle to represent the situation being examined.


Imagine the sun casting a shadow on the ground

Imagine the sun casting a shadow on the ground.

Turn this situation into a right angled triangle


The length of the shadow can be measured directly

The length of the shadow can be measured directly

The primary angle can also be measured directly

X

The Height?

Sooo…

40O

200 m


1 1 using trigonometry to find lengths

Make a model!!

Draw a right angled triangle with a base of 20 cm and a primary angle of 40O, then just measure the height!


1 1 using trigonometry to find lengths

X cm

=

20 000 cm

We can generate an equation using equivalent fractions to determine the actual height!

General Model Real

X cm

Height

17 cm

=

=

Base

20 000 cm

20 cm

0.85

20 000 (0.85) = X

170 m = X


In the interest of efficiency

In the interest of efficiency..

  • Drawing triangles every time is too time consuming.

  • Someone has already done it for us, taken all the measurements, and loaded them into your calculator

  • Examine the following diagram


1 1 using trigonometry to find lengths

O

O

O

O

As the angle changes, so

shall all the sides

of the triangle.

Recall the Trig names for different sides of a triangle…


Geometry

Geometry

hypotenuse

height

O

base

Trigonometry

hypotenuse

opposite

“theta”

adjacent


1 1 using trigonometry to find lengths

Trig was first studied by Hipparchus (Greek), in 140 BC.

Aryabhata (Hindu) began to study specific ratios.

For the ratio OPP/HYP, the word “Jya” was used


1 1 using trigonometry to find lengths

Brahmagupta, in 628, continued studying the same relationship and“Jya” became “Jiba”

“Jiba became Jaib”which means “fold” in arabic


1 1 using trigonometry to find lengths

European Mathmeticians translated “jaib” into latin:

SINUS

(later compressed to SIN by Edmund gunter in 1624)


1 1 using trigonometry to find lengths

Given a right triangle, the 2 remaining angles must total 90O.

A = 10O, then B = 80O

A = 30O, then B = 60O

A

A “compliments” B

C

B


1 1 using trigonometry to find lengths

The ratio ADJ/HYP compliments the ratio OPP/HYP in the similar mathematical way.

Therefore, ADJ/HYP is called “Complimentary Sinus”

COSINE


The 3 primary trig ratios

The 3 Primary Trig Ratios

SINO = opp

O

hyp

COSO = adj

hyp

hyp

opp

TANO = opp

adj

adj


Soh cah toa

soh cah toa

1

A

X 17

FIND A:

17 X

COS25O =

17

1

A = 17 X cos25O

17m

A = 15.4 m

25O

A


Soh cah toa1

soh cah toa

1

A

X 12

FIND A:

12 X

SIN32O =

12

1

A = 12 X SIN32O

12 m

A = 6.4 m

A

32O


Soh cah toa2

soh cah toa

1

A

X 10

FIND A:

10 X

TAN63O =

10

1

A = 10 X TAN63O

63O

A = 19.6 m

10 m

A


Tan 40 o

X

Tan 40O =

200

200 (Tan40O) = X

168 m = X

X

40O

200 m


Remember equivalent fractions can be inverted

Remember: Equivalent fractions can be inverted

2

5

=

4

10

4

10

=

2

5


1 1 using trigonometry to find lengths

Page 8

[1,2] a,c

3-7


Find the height of the building

TAN 50 = H

150

1

150 X

X 150

Find the height of the building

1

(150) TAN 50 = H

HYP

OPP

H

150 m

ADJ

50O


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