using various trigonometric ratios to find lengths of inaccessible distances
TRANSCRIPT
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
=
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