Right Triangle Trigonometry - Pascack Valley Regional High

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Transcript Right Triangle Trigonometry - Pascack Valley Regional High

Right Triangle
Trigonometry
1
Right Triangle Trigonometry


Trigonometry is based upon ratios of the
sides of right triangles.
The ratio of sides in triangles with the same
angles is consistent. The size of the triangle
does not matter because the triangles are
similar (same shape different size).
2
Right Triangle Trigonometry



Previously, we worked with 3 trig functions of
a right triangle: sine, cosine, and tangent.
This year, we will work with the reciprocals of
these 3 functions (cosecant, secant, and
cotangent) for a total of 6 trig functions.
In Geometry, angles were measured in
degrees. Now we will measure angles in
degrees and radians.
3
The six trigonometric functions of a right triangle,
with an acute angle , are defined by ratios of two sides
of the triangle.
hyp
opp
The sides of the right triangle are:
θ
adj
 the side opposite the acute angle ,
 the side adjacent to the acute angle ,
 and the hypotenuse of the right triangle.
4
hyp
The trigonometric functions are
opp
θ
adj
sine, cosine, tangent, cotangent, secant, and cosecant.
opp
sin  =
cos  = adj
tan  = opp
hyp
hyp
adj
csc  =
hyp
opp
sec  = hyp
adj
cot  = adj
opp
Note: sine and cosecant are reciprocals, cosine and secant are reciprocals,
and tangent and cotangent are reciprocals.
5
Reciprocal Functions
Another way to look at it…
sin  = 1/csc 
cos  = 1/sec 
tan  = 1/cot 
csc  = 1/sin 
sec  = 1/cos 
cot  = 1/tan 
6


Given 2 sides of a right triangle you should
be able to find the value of all 6 trigonometric
functions.
Example:
5

12
7
Calculate the trigonometric functions for  .
Calculate the trigonometric functions for .
5
The six trig ratios are
4
sin  =
5
3
cos  =
5
4
tan  =
3
3
cot  =
4
5
sec  =
3
5
csc  =
4
3
sin α =
5
4
cos α =
5
3
tan α =
4
4
cot α =
3
5
sec α =
4
5
csc α =
3

4

3
What is the
relationship of
α and θ?
They are
complementary
(α = 90 – θ)
8
Note sin  = cos(90  ), for 0 <  < 90
Note that  and 90  are complementary
angles.
Side a is opposite θ and also
adjacent to 90○– θ .
sin  = a and cos (90  ) = a .
b
b
So, sin  = cos (90  ).
hyp
90○– θ a
θ
b
Note : These functions of the complements are called cofunctions.
9
Cofunctions
sin  = cos (90  )
cos  = sin (90  )
tan  = cot (90  )
cot  = tan (90  )
sec  = csc (90  ) csc  = sec (90  )
10
Quotient Identities
hyp
opp
θ
adj
sin  = opp
hyp
cos  = adj
hyp
tan  =
opp
adj
opp
sin  hyp opp hyp opp




 tan
cos adj hyp adj adj
hyp
The same argument can be made for cot… since it is the
reciprocal function of tan.
11
Calculate the trigonometric functions for a 45 angle.
2
1
45
1
sin 45 =
opp
1
2
=
=
hyp
2
2
adj 1
cot 45 =
= = 1
opp 1
opp 1
tan 45 =
= = 1
1
adj
sec 45 =
2
hyp
=
=
1
adj
1
2
adj
cos 45 =
=
=
2
hyp
2
2
csc 45 =
2
hyp
=
= 2
opp
1
12
Geometry of the 30-60-90 triangle
Consider an equilateral triangle with
each side of length 2.
30○30○
The three sides are equal, so the
angles are equal; each is 60.
2
The perpendicular bisector
of the base bisects the
opposite angle.
60○
2
3
1
60○
2
1
Use the Pythagorean Theorem to
find the length of the altitude, 3 .
13
Calculate the trigonometric functions for a 30 angle.
2
1
30
3
opp 1
sin 30 =
=
hyp
2
cos 30 =
3
1
opp
tan 30 =
=
=
adj
3
3
3
adj
cot 30 =
=
= 3
1
opp
2
2 3
hyp
sec 30 =
=
=
3
3
adj
hyp 2
csc 30 =
=
= 2
opp
1
3
adj
=
2
hyp
14
Calculate the trigonometric functions for a 60 angle.
2
3
60○
1
1
adj
=
2
hyp
sin 60 =
opp
3
=
hyp
2
cos 60 =
tan 60 =
3
opp
=
= 3
1
adj
3
1
cot 60 = adj =
=
opp
3
3
hyp 2
sec 60 =
= = 2
adj 1
2
2 3
hyp
csc 60 =
=
=
opp
3
3
15
Some basic trig values
Sine
300
450
600
Cosine
Tangent
1
2
3
2
3
3
2
2
2
2
1
3
2
1
2
3
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Example: Given sec  = 4, find the values of the
other five trigonometric functions of  .
Draw a right triangle with an angle  such
4
4
hyp
that 4 = sec  =
= .
adj 1
Use the Pythagorean Theorem to solve
for the third side of the triangle.
sin  =
15
4
cos  = 1
4
tan  = 15 = 15
1
15
θ
1
1 = 4
4 15
=
sin 
15
15
1
sec  =
=4
cos 
cot  = 1 = 15
15
15
17
csc  =
Using the calculator
Function Keys
http://mathbits.com/MathBits/TISection/Algeb
ra1/TrigRatios.htm
Reciprocal Key
http://mathbits.com/MathBits/TISection/Trig/r
eciprocal.htm
Inverse Keys
http://mathbits.com/MathBits/TISection/Trig/in
versetrig.htm
18
Using Trigonometry to Solve a Right
Triangle
A surveyor is standing 115 feet from the base of the
Washington Monument. The surveyor measures the
angle of elevation to the top of the monument as 78.3.
How tall is the Washington Monument?
Figure 4.33
Applications Involving Right
Triangles
The angle you are given
is the angle of elevation,
which represents the
angle from the horizontal
upward to an object.
For objects that lie below
the horizontal, it is
common to use the term
angle of depression.
Solution
where x = 115 and y is the height of the
monument. So, the height of the Washington
Monument is
y = x tan 78.3
 115(4.82882)  555 feet.