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Study Guide and Intervention
(continued)
Geometric Mean
Attitude of a Triangle
In the diagram, AABC AADB ABDC.
An altitude to the hypotenuse of a right triangle forms two right
triangles. The two triangles are similar and each is similar to the
original triangle.
Exarnpie 1
Use right ISABC with
BD L AC. Describe two geometric
means.
a. &4DB
-
/.BDC so
so
AC
—
=
MDB and EABC
AB
and
-
IxBDC,
BC
AC
=
In AABC, each leg is the geometric
mean between the hypotenuse and the
segment of the hypotenuse adjacent to
that leg.
Find x, y, and z.
PR_PQ
PQPS
15
=
In AABC, the altitude is the geometric
mean between the two segments of the
hypotenuse.
b. tABC
Exarn pie 2
A4C
PR=25,PQ=15PS=x
x
Cross multiply.
25x = 225
Divide each side by 25.
x 9
Then
y=PR—SP
25 9
= 16
PR _QR
QRRS
—
z
25
z
2
z
y
z
=
400
20
PR=25,QR=z,RS=y
y=16
Cross multiply.
Take the square root of each side.
6xerti5e5
Find x, y, and z to the nearest tenth.
2.
/
6.
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Study Guide and Intervention
3
The Pythagorean Theorem and Its Converse
The Pythagorean Theorem In a right triangle, the sum of the
squares of the measures of the legs equals the square of the measure of
the hypotenuse.
2 + b
2 = c
.
2
AABC is a right triangle, so a
AC
Exam pie 1
Prove the Pythagorean Theorem.
With altitude CD, each leg a and b is a geometric mean between
hypotenuse c and the segment of the hypotenuse adjacent to that leg.
b
c
c
a
2 =cyandb 2 =cx.
—=—and—=—,soa
x
b
a
y
Add the two equations and substitute c = y + x to get
b
y+cxc(y+x)c
+
a
c
.
2
Exwnpie 2
b. Find c.
a. Find a.
CA
.
2 +
a
2
b
Pythagorean Theorem
2 + 122
a
=
144
=
2
a
=
2 +
a
132
b
=
12, c
=
2 +
a
13
202
169 Simplify.
25
Subtract.
a=5
400
2
b
900
1300
+
viãöii
Take the square root of each side.
Pythagorean Theorem
=
+ 3Q2
36.1
a
=
2
c
=
20, b
30
Simplify.
=
Add.
=
Take the square root of each side.
C
Use a calculator.
Find x.
2.
3.
x
t;Nc
.4
.
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NAME
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PERIOD
7-) Skills Practice
Special Right Triangles
Find x and y.
I
2
6.
1313
For Exercises 7—9, use the figure at the right.
7. If a
=
B
11, find b and c.
AC
8.Ifb=15,findaandc.
9. If c
=
9, find a and b.
For Exercises 10 and 11, use the figure at the right.
A
I
B
10. The perimeter of the square is 30 inches. Find the length of BC.
c
D
11. Find the length of the diagonal RD.
12. The perimeter of the equilateral triangle is 60 meters. Find the
length of an altitude.
E
°
6
D
0
G
F
13. AGEC is a 30°-60°-90° triangle with right angle at E, and .EC is
the longer leg. Find the coordinates of G in Quadrant I for E( 1, 1)
and C(4, 1).
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Study Guide and Intervention
Trigonometry
(continued)
Use Trigo
nometric Ratios In
or if you know the measures of one a right triangle, if you know the measures of two sides
ratios to find the measures of the side and an acute angle, then you can use trigonometric
missing sides or angles of the
triangle.
Exam pie
whole number.
a. Find x.
x +
58
x
=
Find x, y, and z. Round each
measure to the nearest
b. Find y.
90
32
c. Find z.
tanA=18
tan 58°
=
18
y 18 tan 58°
y29
cosA=z
cos 58°
=
z cos 58°
=
z
z
18
18
cos 58°
34
Find x. Round to the nearest
tenth.
2.
3.
1
6.
© Glencoe/McGraw-H1II
370
Glencoe Geometry
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PERIOD
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Intervention
7-5. Study Guide and
d Depression
an
on
ti
va
le
E
of
s
le
ng
A
ms that involve
ble
n Many real-world proms
of an angle of
Angles of Elevatioca
ter
in
d
n be describe
e of sight
looking up to an object
between an observer’s lin
gle
an
the
is
ich
wh
n,
io
elevat
and a horizontal line.
top
n from point A to the
io
at
ev
el
of
e
gl
an
e
Th
base of the cliff,
is 1000 feet from the
A
t
in
po
If
°.
34
is
iff
of a cl
how high is the cliff?
cliff
Let x = the height of the
fxampie
tan 34
0
tan
=
1000 ft
opposite
adjacent
1000(tan 34°)
x
0.
Multiply each side by 100
674.5
x
Use a calculator.
is about 674.5 feet.
The height of the cliff
mber
the nearest whole nu
to
ts
en
gm
se
of
s
re
su
Round mea
Solve each problem.
arest degree.
and angles to the ne
.
to the top of a hill is 49°
A
int
po
m
fro
n
tio
va
ele
w high is
1. The angle of
9
m the base of the hill, ho
If point A is 400 feet fro
the hill?
meter-tall
of the sun when a 12.5n
tio
va
ele
of
gle
an
2. Find the
18-meter-long shadow.
telephone pole casts an
12.5 m
18 m
gle of 78°
a building makes an an
st
ain
ag
g
nin
lea
r
de
from the
3. A lad
t of the ladder is 5 feet
with the ground. The foo
ladder?
building. How long is the
standing
feet above the ground is
5
are
es
ey
e
os
wh
n
4. A perso
ntrol tower.
port 100 feet from the co
window
on the runway of an air
the
at
an air traffic controller
That person observes
n?
tio
hat is the angle of eleva
of the 132-foot tower. W
ft
5
375
Glencoe Geometry
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7-5á Study Guide and Intervention
(continued)
I
Angles of Elevation and Depression
I
angle of
Angles of Depression When an observer is looking down, the
angle of depression is the angle between the observer’s line of sight
and a horizontal line.
6xampIc
The angle of depression from the top of an
80-foot building to point A on the ground is 42°. How far
is the foot of the building from point A?
Let x = the distance from point A to the foot of the building. Since
the horizontal line is parallel to the ground, the angle of depression
LDBA is congruent to LBAC.
tan42°
80
x
x(tan 42°)
80
X
x
tan=
horizontal 0
B
angle of
depression
_—
A
80
°f
42
oppOsite
adjacent
—
Multiply each side by x.
80
tan 42°
88.8
Divide each side by tan 42°.
Use a calculator.
Point A is about 89 feet from the base of the building.
Solve each problem. Round measures of segments to the nearest whole number
and angles to the nearest degree.
1. The angle of depression from the top of a sheer cliff to
point A on the ground is 35°. If point A is 280 feet from
the base of the cliff, how tall is the cliff’?
A
280ft
2. The angle of depression from a balloon on a 75-foot
string to a person on the ground is 36°. How high is
the balloon?
3. A ski run is 1000 yards long with a vertical drop of
208 yards. Find the angle of depression from the top
of the ski run to the bottom.
4. From the top of a 120-foot-high tower, an air traffic
controller observes an airplane on the runway at an
angle of depression of 19°. How far from the base of the
tower is the airplane?
©
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4
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DATE
NAME
PERIOD
Skills Practice
The Law of Sines
. Round angle measures
Find each measure using the given measures from ISABC
nearest tenth.
to the nearest tenth degree and side measures to the
1. If rnLA
=
35, rnLB
2. If rnLB
=
17, mLC
86, mLA
3. IfmLC
4. If a
=
17, b
5. If c
=
38, b
6.If a
=
=
28, find a.
46, and c
=
18, find b.
=
38, find c.
51, and a
=
8, and rn/A
73, find mLB.
=
36, find rnLC.
=
20,andmLC= 83,flndmLA.
18, and mLB= 104, find b.
22, a
=
=
34, and rnLB
12,c
7. If rn/A
48, and b
the nearest tenth.
Solve each Es.PQR described below. Round measures to
8.p
33
27,q
=
40,mLP
=
11, mCi?
16
34,rnLQ
111
9. q
=
12, r
1O.p
=
29,q
89,p
11. IfrnLP
=
=
16, r
=
12
13
63,p
12. If mLQ
=
103, mLP
13. If mLP
=
96, mLR
14. If rnLR
=
49, mLQ
76, r
=
26
31, mLP
52,p
=
20
15. If rnLQ
16. If q
=
8, mLQ
17. Ifr
=
l5,p
©
=
=
82, r
28, mLR
21, mLP
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=
35
=
=
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Glencoe Geometry
NAME
DATE
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Skills Practice
,
The Law of Cosines
In ARST, given the following measures, find
the measure of the missing side.
1. r 5,s
8, rnZT 39
2. r
6, t
3.r
9,t
=
11, mLS
15,m/S
=
87
103
4.s=12,t10,inLR58
In iS.HIJ, given the lengths of the sides, find
the measure of the stated angle to the
nearest tenth
5 Ii
12,
l8,j
t
7, mLH
=
4
6 h
=
15,
7. h
=
23, i
8. h
=
37, i
i
=
l6,j
=
22, mU
27,j
=
29; mU
21,j
=
30; mUll
Determine whether the Law of Sines or
the Law of Cosines should be used firs
t to
solve each triangle. Then solve each triangl
e. Round angle measures to the nea
rest
degree and side measures to the nearest
tenth.
10.
L2
5
2
A
11.a
N
=
10,b
14,c =19
12.a
12,b
=
10,mLC
=
27
Solve each IxRST described below. Round
measures to the nearest tenth.
13.r 12,s = 32,t = 34
14.r
=
30,s
=
25,mLT= 42
15.r
=
15,s
=
11,mLR
16. r
=
21, s
=
28, t
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=
=
67
30
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Glencoe Geometry