# trigonometry

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```					Trigonometry
Say that you wanted to find out the height of a pyramid in Egypt. Impressed by the heights at
which these were built, with all that human labour that went into their construction, this
particular pyramid seems most impressive.

You ask your tour guide about its height. You are given a response, but you want to calculate
height yourself, given your state of awe. Would you be able to do so?

Absolutely, yes! How so?

Assuming that you have verified the size of the area of the pyramid, and all the implications
thereof, with the calculated distance you are from it and the angle from your eye, from which the
top of the pyramid appears, you may calculate the actual height of the pyramid.

All such a possibility is found in the gamut of trigonometry.

Trigonometry involves the relationship between the angles and the length of sides of a
triangle. In other words, the lengths of the sides of a triangle have a bearing on each angle
within the triangle. To understand more fully, we examine firstly, the simplest triangle ─ the
right-angled triangle:

The Right-angled Triangles in Trigonometry
The longest side of a right angled triangle is called the hypotenuse. Unlike Pythagoras Theorem,
the other two sides are neither horizontal nor vertical. They actually have names; adjacent or
opposite.

The next identifiable side of a right angle triangle in trigonometry is called the opposite. This
side, as it were faces the acute angle in question, or for which the acute angle being considered
opens up to it. By virtue of having identified the opposite, we unwittingly identify the

A
Hypotenuse
The opposite of  is AB

B                             C

Notice the C = . Since this acute angle is given (as it were), then AB becomes the opposite.
AB faces the angle . The angle  opens up to AB. Naturally then, BC becomes the adjacent.
We further illustrate:
The opposite of  is JK
C
J                              K



The opposite of  is DE
D                        E                                                         L

In triangle CDE, the acute angle being highlighted is . Since this is the case, the side which is
called the opposite is DE. In triangle JKL, the highlighted acute angle is . Hence the opposite
is the side JK.

Trigonometrical Ratios
Trig ratios (the shortened form of trigonometrical ratios) only apply to right-angled triangles.
These ratios tell that there is a direct bearing of any two given sides of the triangle and its acute
angle. As hinted on the outset, the 90° angle in this triangle ensures that those ratios are
established, just about the same as it does for Pythagoras Theorem.

There are three such ratios: Sine (shortened as sin), cosine (shortened as cos) and tangent
(shortened as tan).

Sine
Sine (sin) is the ratio of the opposite divided by the hypotenuse. Sine of an angle  (denoted sin
), is the ratio that corresponds with .

Sine   =           opposite                   (SOH is the acronym for sin= opp/hyp)
hypotenuse

The sine of each acute angle of a r.a.t is specific and conversely, the ratio that corresponds to
each acute angle is always specific. For instance:

Sin 30° = 0.5                                 Sin 35° ≈ 0.573
Sin 40° ≈ 0.643                               Sin 45° ≈ 0.707

So, no matter what the size of the right angled triangle, once the angle at one of the vertex is ,
the ratio of the opposite divided by the hypotenuse is always going to be the same. Thus, for any
:
Sin  =         opposite
hypotenuse

Naturally, once a ratio of an angle is given, we can find the desired degree. In the case of sine, it
is equal to the inverse of the calculated ratio (i.e. the opposite divided by the hypotenuse). That
is:

      =       sin-1    opposite
hypotenuse

No doubt, since the sine ratio involves an acute angle, its opposite and the hypotenuse, we use
sine ratios only with respect to these measurements. Even in saying so, a math problem dictates
that one of three pieces of information is missing.

Problem:
QRS is a triangle with S = 90°, QR = 35 cm and RS = 18cm. Determine the angle at Q.

Solution:
R

18cm         35cm

S                Q

Sin Q =        RS              =       18             =       0.514 (3dp)
QR                      35

Q =           sin-1 0.514     =       30.9497° (4 dp) ≈      31°

Cosine
Cosine represents the ratio of the adjacent divided by the hypotenuse of a r.a.t. Like sine, the
cosine ratio is specific for each angle. Therefore, the size of any r.a.t. is irrelevant to the ratio
that each angle produces – the adjacent vís–a–vís the hypotenuse. Ideally:

hypotenuse
Naturally, cosine of an angle is the corresponding ratio of its adjacent divided by the hypotenuse.
Therefore, for any :
hypotenuse

Once a math problem calls for the use of the adjacent and the hypotenuse, it calls for the
application of the cosine ratio. Ideally, when using this ratio in any math problem, one of three
things (the angle, the size of the adjacent, or the size of the hypotenuse) is missing. Our aim
thereafter is to ascertain the value of the missing measurement.

We now look at one example.

Problem:
FGH is a triangle such that G = 90°, F = 40° and FG = 23cm. Evaluate the length of FH.

Solution:
H

Let FG = h, FH = g and GH = f

Cos F = h             g=       h                                         40°
g                      cos F                                  F         23cm    G

g=        23          =         23          =       30. 024 (3 dp)
Cos 40°               0.766

Tangent
In considering sine and cosine, we noted that their ratios of the opposite and adjacent were
relative to the hypotenuse, respectively. With regards to tangent, however, the ratio represents
the value of the opposite divided by the adjacent. That is:

Tangent       =       opposite                      (tan = opposite/adjacent – TOA)

Like sine and cosine, the tangent ratio is specific for each angle. Thus, for any , of an acute
angle of a right-angle triangle:

Tan          =       opposite

Of course, to use this ratio, we consider only the adjacent, hypotenuse and opposite of a right
angled triangle, with one of these measurements being outstanding. We now look at a problem:
Problem:
The triangle DGN is such that DG = 2cm, GN = 3.3cm and G marks the angle of 90°.

(i)          Evaluate the size of D
(ii)         Hence, or otherwise, calculate the size of N to the nearest whole

Solution:
(i)
G           3.3cm         N                      Tan D = GN =             3.3cm
DG                2cm
2cm

Tan D = 1.65
D    = tan-1 1.65       =       58.78°

D

Problem:
An electronic compass is placed on the ground, some 40m away from a vertical building. The
angle of elevation (the angular measurement from the compass to the height of the building, in
this instance) is computed at 80°. Given that the compass and the building are at the same
ground level, calculate the height of the building.

Solution:
(To get a gist of the nature of the question, we frame the components of the question
graphically.)

Let us assume that the point at which the compass is at is C, and the height of the building is GB.

B

80°
C    40m   G
GB = tan 80°         GB = tan 80°      GB = tan 80°  40m ≈ 5.671  40
CG                   40m

GB ≈ 2,026.84m

```
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 views: 2 posted: 9/12/2012 language: English pages: 6