Transcript 4.1 - schsgeometry

5.8
What If I Know the Hypotenuse?
Pg. 23
Sine and Cosine Ratios
5.8 – What If I Know the Hypotenuse?
Sine and Cosine Ratios
In the previous section, you used the idea
of similarity in right triangles to find a
relationship between the acute angles and
the lengths of the legs of a right triangle.
However, we do not always work just with
the legs of a right triangle – sometimes we
only know the length of the hypotenuse.
By the end of this lesson, you will be able
to use two new trigonometric ratios that
involve the hypotenuse of right triangles.
5.43 – MISSING THE LEGS
a. What is the relationship between the two
triangles below? Explain your reasoning.
Similar, AA~
b. Find the rise for the slope triangles shown at
right.
x2 + 2.32 = 32
2
x + 5.29 = 9
x2 = 3.71
x = 1.93
3.86
1.93
adjacent
c. Find the ratio
for each triangle.
hypotenuse
Why must these ratios be equal?
2.3
3
=
23
30
4.6 = 23
6 30
Ratios are
equal because
triangles are
similar
H
H
3.86
O
A
1.93
O
A
5.44 – MISSING THE LEGS In the last problem,
you used a ratio that included the hypotenuse of
∆ABC. One of these ratios is known as the sine
ratio (pronounced "sign"). This is the ratio of
the length of the side opposite the acute angle
to the length of the hypotenuse.
opposite
sin 
hypotenuse
H
O
A
adjacent
cos 
hypotenuse
H
O
A
5.45 – CALCLUATOR TIME!
a. Like the tangent, your calculator can give you
both the sine and cosine ratios for any angle.
Locate the "sin" and "cos" buttons on your
calculator and use them to find the sine and
cosine of 40°. Make sure you get the correct
answers and are in degree mode.
0.64
sin 40° = ____________________
0.77
cos 40° = _____________________
Shows each ratio
in relation to the
sides
b. Use a trig ratio to write an equation and solve
for a in the diagram at right. Does this require
the sine ratio or the cosine ratio? Begin by
labeling the opposite, adjacent, and hypotenuse.
What side do you know? What side are you
solving for?
a
sin23  15
1
a  15  sin23
a  5.86
H
A
O
c. Likewise, write an equation and solve for
b in the triangle at right.
b
cos37 
1
8
b  8  cos37
b  6.39
H
O
A
5.46 – THE STREETS OF SAN FRANCISCO
While traveling around the beautiful city of
San Francisco, Julia climbed several steep
streets. One of the steepest, Filbert Street,
has a slope angle of 31.5° according to her
guide book. Once Julia finished walking 100
feet up the hill, she decided to figure out how
high she had climbed. Julia drew the
diagram below to represent the situation.
a. Can the Pythagorean theorem be used to
find the opposite and adjacent side? Why or why
not?
No, only know
one side
b. Can special triangles be used to find the
opposite and adjacent side? Why or why not?
No, 31.5  isn’t
special
c. Can we use sine, cosine, or tangent to find
the opposite and adjacent side? Why or why
not?
H
Sine and
cosine
because we
know the
hypotenuse
and one angle
O
A
d. Julia still wants to know how many feet she
climbed vertically and horizontally when she
walked up Filbert Street. Use one of your new
trig ratios to find both parts of the missing
triangle.
H
O
x
cos31.5 
1
100
x  100  cos31.5
x  85.26 ft
A
H
O
y
sin31.5 
100
1
y  100  sin 31.5
y  52.25 ft
A
5.47 – TRIGONOMETRY
For each triangle below, decide which side is
opposite, adjacent, or the hypotenuse to the
given angle. Then determine which of the three
trig ratios will help you find x. Write and sole an
equation. SOH-CAH-TOA might help.
H
A
O
x
sin25 
9
1
x  9  sin25
x  3.80
H
A
O
cos17 
1
3
x
x  cos17  3
3
x
cos17
x  3.14
5
H
A
O
5
tan62 
x
1
x  tan62  5
5
x
tan 62
x  2.66
O
A
H
13
sin34 
x
1
x  sin34  13
13
x
sin34
x  23.25
21 3
21
H
O
A
x
sin28 
19
1
x  19  sin28
x  8.92
H
A
O
6
sin25 
x
1
x  sin25  6
6
x
sin 25
x  14.2
H
A
O
18
cos61 
1
x
x  cos61  18
18
x
cos61
x  37.13
16
8 3
A
H
O
x
cos20 
1
10
x  10  cos20
x  9.4
8
5.48 – EXACTLY!
Martha arrived for her geometry test only to find
that she forgot her calculator. She decided to
complete as much of each problem as possible.
a. In the first
problem on the
test, Martha was
asked to find the
length of x in the
triangle shown at
right. Using her
algebra skills,
she wrote and
solved an
equation. Her
work is shown
below. Explain
what she did in
each step.
Set up equation
multiplied
divided
b. Martha's answer in part (a) is called an exact
answer. Now use your calculator to help Martha
find the approximate length of x.
x = 68.62
c. In the next problem, Martha was asked to find
y in the triangle at right. Find the exact answer
for y without using a calculator. Then use a
calculator to find an approximate value for y.
H
A
O
y
tan53 
1
5
x  5  tan53
x  6.64
5.49 – CONCLUSIONS
Describe how to determine when to use
the sine, cosine, and tangent ratios. What
are the steps in setting up an equation?
How can you write an exact answer?