The Height ProportionBase

of right Triangles

Imagine 2 similar right triangles

3m

4m

6m

8m

HeightBase

=34

=68

= 0.75decimal

The height of the larger triangle will be the key unknown…..

h

4.35m 7.61m

6.38m

50O

50O

h

4.35=

7.61

6.38

h

4.3= 1.19 X 4.3

4.3 X

h = 5.12m

h

7.24m

15.81m

26.31m

31O

31O

h

7.24=

15.81

26.31

h

7.24= 0.60 X 7.24 7.24 X

h = 4.35m

Height

Base=

H

4.1=

1

1.7

tree meter stick

1.7H = 4.1

H = 2.4 m

Let’s try some for real, as a group…outside… yes……..

KEY

TRIANGLE

Imagine trying to find the height of a tree…

1m

1.7m 4.1m

? m

1.Take advantage of the fact that we can model these 2 situations with similar triangles

2. Create a proportion

3. Solve for an unknown that we can not physically obtain!!!

A more efficient way…Imagine, as the sun moves across the sky, that it

creates many different angles for our triangles…

Technically, there is an infinite number of triangles that could be used in our

proportions….

We need to be more efficient than that….

We will limit our triangles to under 100 000

The good news….• Instead of use having to go out and measure

each possible triangle individually to get our KEY triangle….someone has done that for us……

• Not only have they done that for us, they have also given cool names to 3 of the most useful ratios….

• For example……

We need a better name for the height / base ratio…

Since both sides involved are touching the right angle… height

base

The latin word “tangens” was used…

Tangens was eventually converted to Tangent, or TAN.

O “theta”adjacent

oppositehypotenuse

Consider the ratio: hypotenuse

opposite

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

Aryabhata (Hindu) continued his work.

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

Brahmagupta, in 628, continued studying the same relationship and

“Jya” became “Jiba”

later,

Jiba became Jaib, which means “fold” in Arabic

European Mathmeticians eventually translated “jaib” into latin:

SINUS

Later compressed to the singular “SINE” by Edmund Gunter in 1624

Compressed again by calculator manufactorers into..

SIN

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

O “theta”adjacent

oppositehypotenuse

The last ratio will be…hypotenuse

adjacent

The adjacent/hyp ratio compliments the opposite/hyp ratio (called SIN)….therefore

Therefore, ADJ/HYP is called “Complimentary Sinus”

COSINE

COS

The 3 Primary Trig Ratios

O

SINO = opp

opp

adj

hyp

hyp

COSO = adj hyp

TANO = opp adj

Your calculator probably has hundreds of thousands of KEY

triangles already loaded into the memory…..

O O OO

Now we have 3 working ratios for every possible right sided triangle at our fingertips….

Solve for the following heights:

Finding the height of a building (H = ?)

150 m50O

H

TAN 50 = H150

150 X TAN 50 = H

X 150 150 X

1

1

178.76m = H

43O1000 m

H

Tan 43O =1000

H X 1000 1000 X

H= 1000 X Tan43O

H =932.52 m

1

1

25O1000 m

H

Tan 25O =1000

H X 1000 1000 X

H= 1000 X Tan25O

H =466.30 m

1

1

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:

37O

A10 m

SIN37O = A10

X 1010 X

1

1

A = 10 X SIN37O

A = 6.02 m

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