Using Various Trigonometric Ratios

to Find Lengths of inaccessible distances

• 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.

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

•Trigonometry uses TRIANGLES for models.

We use similar triangles to represent the situation being examined.

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

O O OO

As the angle changes, so

shall all the sides

of the triangle.

Recall the Trig names for different sides of a triangle…

Geometry

O “theta”adjacent

oppositehypotenuse

Trigonometrybase

heighthypotenuse

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

Brahmagupta, in 628, continued studying the

same relationship and “Jya” became “Jiba”

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

European Mathmeticians translated “jaib” into latin:

SINUS(later compressed to SIN by

Edmund gunter in 1624)

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

A = 10O, then B = 80O

A = 30O, then B = 60O

A

BC

A “compliments” B

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

O

SINO = opp

opp

adj

hyp

hyp

COSO = adj hyp

TANO = opp adj

soh cah toaFIND A:

25O

A

17m

COS25O = A17

X 1717 X

1

1

A = 17 X cos25O

A = 15.4 m

soh cah toaFIND A:

32O

A12 m

SIN32O = A12

X 1212 X

1

1

A = 12 X SIN32O

A = 6.4 m

soh cah toaFIND A:

63O

A

10 m

TAN63O = A10

X 1010 X

1

1

A = 10 X TAN63O

A = 19.6 m

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?

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 primary angle can also be measured

directly

The Height?

200 m 40O

X

Sooo…

Tan 40O = X

200 m 40O

X

200200 (Tan40O) = X

168 m = X

Remember: Equivalent fractions can be inverted

24

510

=

42

105

=

Page 338

Page 344