Download Section 10.6 – Right Triangle Trigonometry – pg 153

153
Section 10.6 – Right Triangle Trigonometry
Objective #1:
Understanding Adjacent, Hypotenuse, and Opposite sides
of an acute angle in a right triangle.
In a right triangle, the hypotenuse is always the longest side; it is the side
opposite of (or "across from") the right angle. The hypotenuse of the
following triangles is highlighted in gray.
The adjacent side of an acute angle is the side of the triangle that forms the
acute angle with the hypotenuse. The adjacent side is the side of the angle
that is "adjacent" to or next to the hypotenuse. The adjacent side depends
on which of the acute angles one is using. The adjacent side of the
following angles (labeled A) is highlighted in gray.
A
A
A
A
The opposite side of an acute angle is the side of the triangle that is not a
side of the acute angle. It is "across from" or opposite of the acute angle.
The opposite side depends on which of the acute angles one is using. The
opposite side of the following angles (labeled A) is highlighted in gray.
A
A
A
A
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For each acute angle in the following right triangles, identify the
adjacent, hypotenuse, and opposite side:
R
Ex. 1
B
Ex. 2
Q
A
C
X
Solution:
For ∠R,
the adjacent side is QR,
the opposite side is QX,
and the hypotenuse is RX.
For ∠X
the adjacent side is QX,
the opposite side is QR,
and the hypotenuse is RX.
Solution:
For ∠A,
the adjacent side is AC,
the opposite side is BC,
and the hypotenuse is AB.
For ∠B,
the adjacent side is BC,
the opposite side is AC,
and the hypotenuse is AB.
Objective #2:
Understanding Trigonometric Ratios or Functions.
Recall that two triangles are similar if they have the same shape, but are
not necessarily the same size. In similar triangles, the corresponding
angles are equal. Thus, if ∆ABC ∼ ∆HET, then m∠A = m∠H, m∠B = m∠E,
and m∠C = m∠T. The corresponding sides are not equal. However, the
ratios of corresponding sides are equal since one triangle is in proportion to
the other triangle. Thus, if ∆ABC ∼ ∆HET, then
B
E
AB BC AC
=
=
HE ET HT
Now, let's consider the proportion
AB
BC
=
. If we cross multiply,
HE
ET
A
C
we get (AB)•(ET) = (HE)•(BC). Now,
H
divide both sides by (BC)•(ET) and simplifying, we get:
(AB)•(ET) = (HE)•(BC)
AB = HE
(BC)•(ET)
(BC)•(ET)
BC
ET
T
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Now consider angles ∠C and ∠T.
If ∠A and ∠H are right angles, then
the proportion
AB
BC
=
HE
ET
B
E
implies that
the ratio of the opposite side to the
hypotenuse is the same for any right
triangle if the corresponding acute
A
C
angles have the same measure.
H
T
We will define this ratio as the Sine Ratio or Sine Function. The
abbreviation for the sine function is "sin", but it is pronounced as "sign".
Sine Ratio or Function
Given an acute angle A in a right triangle, then
the sine ratio of A, is the ratio of the opposite
side of A to the hypotenuse.
oppositeside
opp
sin A =
=
hypotenuse
hyp
A
opp
adj
hyp
For each acute angle in the following right triangles, find the sine of
the angle:
16.1 in
Ex. 3
24.1 cm
Ex. 4
55˚
50˚
42.0 cm
19.2 in
25.0 in
35˚
34.4 cm
Solution:
opp
24.1
sin 35˚ =
=
Solution:
sin 40˚ =
= 0.5738… ≈ 0.574
opp
34.4
sin 55˚ =
=
= 0.644
sin 50˚ =
= 0.8190… ≈ 0.819
= 0.768
hyp
hyp
42.0
42.0
40˚
opp
hyp
=
16.1
25.0
opp
hyp
=
19.2
25.0
Since the opposite is never greater than the hypotenuse, then the value of
the sine function cannot exceed 1. If the opposite side is equal to the
hypotenuse, then the measure of the angle is 90˚ and the triangle
degenerates into a vertical line. This means that sin 90˚ = 1. Similarly, if the
opposite side has zero length meaning the sine ratio is 0, then the measure
of the angle is 0˚ and the triangle degenerates into a horizontal line. Hence,
sin 0˚ = 0.
156
We have seen that if the corresponding acute angles of rights triangles
have the same measure, the ratio of the opposite side to the hypotenuse is
fixed. We can make the same argument that the ratio of the adjacent side
to the hypotenuse is also fixed. We will define this ratio as the Cosine
Ratio or Cosine Function. The abbreviation for the cosine function is
"cos", but it is pronounced as "co-sign".
Cosine Ratio or Function
Given an acute angle A in a right triangle, then
the cosine ratio of A, is the ratio of the adjacent
side of A to the hypotenuse.
cos A =
adjacentside
hypotenuse
=
hyp
A
adj
hyp
opp
adj
For each acute angle in the following right triangles, find the cosine of
the angle:
16.1 in
Ex. 5
24.1 cm
Ex. 6
55˚
50˚
42.0 cm
19.2 in
25.0 in
35˚
34.4 cm
Solution:
adj
34.4
cos 35˚ =
=
Solution:
cos 40˚ =
= 0.8190… ≈ 0.819
= 0.768
hyp
cos 55˚ =
adj
hyp
42.0
=
24.1
42.0
= 0.5738… ≈ 0.574
cos 50˚ =
40˚
adj
hyp
=
19.2
25.0
adj
hyp
=
16.1
25.0
= 0.644
Since the adjacent is never greater than the hypotenuse, then the value of
the cosine function cannot exceed 1. If the adjacent side is equal to the
hypotenuse, then the measure of the angle is 0˚ and the triangle
degenerates into a horizontal line. This means that cos 0˚ = 1. Similarly, if
the adjacent side has zero length meaning the cosine ratio is 0, then the
measure of the angle is 90˚ and the triangle degenerates into a vertical line.
Hence, cos 90˚ = 0.
We have seen that if the corresponding acute angles of rights triangles
have the same measure, the ratio of the opposite side to the hypotenuse is
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fixed and the ratio of the adjacent side to the hypotenuse is fixed. We can
make the same argument that the ratio of the opposite side to the adjacent
side is also fixed. We will define this ratio as the Tangent Ratio or Tangent
Function. The abbreviation for the tangent function is "tan", but it is
pronounced as "tangent".
Tangent Ratio or Function
Given an acute angle A in a right triangle, then
the tangent ratio of A, is the ratio of the opposite
side of A to the adjacent side of A.
oppositeside
opp
tan A =
=
adjacentside
hyp
A
opp
adj
adj
For each acute angle in the following right triangles, find the tangent
of the angle:
16.1 in
Ex. 7
24.1 cm
Ex. 8
55˚
42.0 cm
34.4 cm
Solution:
opp
tan 35˚ =
=
adj
24.1
34.4
= 0.7005… ≈ 0.701
opp
adj
=
= 1.427… ≈ 1.43
19.2 in
25.0 in
35˚
tan 55˚ =
50˚
34.4
24.1
Solution:
tan 40˚ =
opp
adj
40˚
=
16.1
19.2
≈ 0.8385… = 0.839
tan 50˚ =
opp
adj
=
19.2
16.1
= 1.192… ≈ 1.19
When the opposite side has zero length, then the measure of the angle is
zero. The tangent ratio is the ratio of 0 to the adjacent side, which is equal
to zero. Hence, tan 0˚ = 0. Similarly, if the adjacent side has zero length,
then the measure of the angle is 90˚. The tangent ratio is the ratio of the
opposite side to 0, which is undefined. Thus, tan 90˚ is undefined. Unlike
the sine and cosine function, the values of the tangent function can exceed
one.
There are some memories aids that you can use to remember the
trigonometric ratios. If you remember the order sine, cosine, and tangent,
then you might find one of these phrases helpful:
158
Trigonometric
Ratio
Ratio of
sides
Rated G
Phrase
Rated PG-13
Phrase
opp
hyp
adj
hyp
opp
adj
oscar
had
a
heap
of
apples
oh
h∗ll
another
hour
of
algebra
sin A =
cos A =
tan A =
If you want to include the trigonometric functions themselves, you can use
the following phrases:
s-o-h-c-a-h-t-o-a
s
i
n
e
o
p
p
o
s
i
t
e
h
y
p
o
t
e
n
u
s
e
c
o
s
i
n
e
a
d
j
a
c
e
n
t
h
y
p
o
t
e
n
u
s
e
t
a
n
g
e
n
t
o
p
p
o
s
i
t
e
a
d
j
a
c
e
n
t
sinful oscar had consumed a heaping taste of apples
sinful
i
n
e
€
€
€
oscar
p
p
o
s
i
t
e
had
y
p
o
t
e
n
u
s
e
consumed
o
s
i
n
e
a
d
j
a
c
e
n
t
heaping
y
p
o
t
e
n
u
s
e
taste
a
n
g
e
n
t
of
p
p
o
s
i
t
e
apples.
d
j
a
c
e
n
t
159
Objective #3:
Evaluating trigonometric ratios on a scientific calculator.
On a scientific calculator, SIN key is for the sine ratio, COS key is for the
cosine ratio, and TAN key is for the tangent ratio. To evaluate a
trigonometric ratio for a specific angle, you first want to be in the correct
mode. If the angle is in degree, then the calculator needs to be in degree
mode. Hit the appropriate trigonometric ratio, type the angle and hit enter.
Evaluate the following:
Ex. 9a
Ex. 9c
cos 26˚
2
sin 16 ˚
Ex. 9b
tan 78.3˚
3
Solution:
a)
We put our calculator in degree mode, hit the COS key, type 26
and hit enter:
cos 26˚ = 0.8987… ≈ 0.899
b)
We put our calculator in degree mode, hit the TAN key, type
78.3 and hit enter:
tan 78.3˚ = 4.828… ≈ 4.83
c)
We put our calculator in degree mode, hit the SIN key, type
2
2
16 and hit enter:
sin 16 ˚ = 0.2868… ≈ 0.287
3
Objective #4:
3
Finding Angles using Inverse Trigonometric Ratio.
If we know the ratio of two sides of a right triangle, we can use inverse
trigonometric ratios or functions to find the angle. The inverse trigonometric
function produces an angle for the answer.
Inverse Trigonometric Functions:
Given a right triangle, we can find the acute
angle A using any of the following:
1) Inverse sine (arcsine)
opp
A = sin – 1(
)
hyp
A
opp
adj
hyp
2) Inverse cosine (arccosine)
adj
A = cos – 1(
)
hyp
3) Inverse tangent (arctangent)
A = tan – 1(
opp
adj
)
On a scientific calculator, these functions are labeled SIN – 1, COS – 1 and
TAN – 1. They are usually directly above the SIN, COS, and TAN keys. To
160
find the angle on the calculator, first set your calculator in degree mode.
Second, hit the 2nd (or INV) and then the appropriate trigonometric ratio
key followed by the ratio of the two sides you are given and hit enter.
Solve the following. Round to the nearest tenth:
Ex. 10a
sin x = 0.833
Ex. 10c
tan x = 1.82
Ex. 10b
cos x = 0.528
Solution:
a)
Since sin x = 0.833, then x = sin – 1(0.833).
Hit 2nd SIN, type 0.833 and hit enter: sin – 1(0.833) ≈ 56.4˚
b)
Since cos x = 0.528, then x = cos – 1(0.528).
Hit 2nd COS, type 0.528 and hit enter: cos – 1(0.528) ≈ 58.1˚
c)
Since tan x = 1.82, then x = tan – 1(1.82).
Hit 2nd TAN, type 1.82 and hit enter: tan – 1(1.82) ≈ 61.2˚
Objective #5:
Finding the missing sides and angles of a right triangle.
Given either two sides or a side and one of the acute angles of a right
triangle, we can use our trigonometric ratios and the Pythagorean Theorem
to solve for the missing sides and angles of a right triangle.
Finding the missing sides and angles of the following. Round all sides
and angles to the nearest tenth:
E
Ex. 11
Ex.
12
B
18 ft
71.5˚
23˚
C
A
Solution:
Since ∠A and ∠B are
complimentary, then
m∠B = 90˚ – 23˚ = 67˚.
We have the hypotenuse
and the acute angle A. We
can use the sine of 23˚ to
find BC and the cosine of
23˚ to find AC.
Solution: F 13 in G
Since ∠E and ∠G are
complimentary, then
m∠E = 90˚ – 71.5˚ = 18.5˚ = 18˚.
We have the adjacent side and
and the acute angle G. We can
use the tangent of 71.5˚ to find
the EF and the cosine of 71.5˚
to find EG.
€
161
sin 23˚ =
opp
hyp
=
BC
18
opp
adj
tan 71.5˚ =
EF
13
=
To solve, multiply by 18:
BC = 18•sin 23˚
= 18•0.3907… = 7.03…
€
≈€7.0 ft
adj
AC
cos 23˚ =
=
To solve, multiply by 13:
EF = 13•tan 71.5˚
= 13•2.98… = 38.85…
€ in €
≈ 38.9
adj
13
cos 71.5˚ =
=
To solve for BC, multiply
by 18:
AC = 18•cos 23˚
€
=€18•0.9205… = 16.56…
≈ 16.6 ft
Thus, BC ≈ 7.0 ft. AC ≈ 16.6 ft
and m∠B = 67˚.
To solve for EG, multiply by
EG and then divide by cos 71.5˚:
EG•cos 71.5˚ = 13
EG = 13/cos 71.5˚ =13/0.317…
= 40.97… ≈ 41.0 in
Hence, EF ≈ 38.9 in, EG ≈ 41.0 in,
and m∠E = 18.5˚.
hyp
Ex. 13
C
hyp
18
7 cm
B
Ex. 14
F
10.1 m
5 cm
E
A
Solution:
We have the two legs of a
right triangle. We can use
the inverse tangent function
to find ∠B:
.
tan B =
EG
opp
adj
=
5
7
Thus, B = tan – 1(
Solution:
We have the leg and the
hypotenuse of a right triangle.
we can use the inverse cosine to
to find ∠G.
cos G =
5
7
) ≈ 35.5˚
Since ∠A and ∠B are
€
€complimentary,
then
m∠A = 90˚ – 35.5˚ = 54.5˚
€
To find the length AB, we can
use the Pythagorean Theorem:
(AB)2 = (7)2 + (5)2
(AB)2 = 49 + 25
(AB)2 = 49 + 25
G
15 m
adj
hyp
=
10.1
15
Hence, G = cos – 1(
10.1
)
15
≈ 47.7˚
Since ∠E and ∠G are
€
€complimentary,
then
m∠E = 90˚ – 47.7˚ = 42.3˚
€
To find the length EF, we can
use the Pythagorean Theorem:
(10.1)2 + (EF)2 = (15)2
102.01 + (EF)2 = 225
subtract 102.01 from both sides
162
(AB)2 = 74
AB = 74 = 8.60… ≈ 8.6 cm
Thus, m∠B ≈ 35.5˚,
m∠A ≈ 54.5˚, and AB ≈ 8.6 cm.
€
Objective #6:
(EF)2 = 122.99
EF = 122.99 =11.09… ≈ 11.1 m
Hence, m∠G ≈ 47.7˚,
m∠E = 42.3˚, and EF ≈ 11.1 m.
€
Solving applications involving right triangles.
The key to solving applications involving right triangles is drawing a
reasonable diagram to represent the situation. Next, we need determine
what information we are given and what information we need to find.
Finally, we then use the appropriate trigonometric function to solve the
problem.
Solve the following:
Ex. 15
A surveyor measures the angle between her instruments and
the top of a tree and finds the angle of elevation to be 65.7˚. If her
instruments are 4 ft high off of the ground and are 25 feet away from
the center of the base of the tree and the ground is leveled, how high
is the tree?
Solution:
We begin by drawing the surveyor's
instruments and the tree. Since her
instruments are 4 ft above the ground,
we can draw a horizontal line from her
her instruments to four feet above the
center of the base of the tree. Drawing
x
a line from the top of the tree to her
instruments and a vertical line down
65.7˚
from the top of the tree four feet above
25 feet
the center of the base of the tree, we
get a right triangle. The adjacent side
is 25 feet and we are looking for the opposite of the triangle. Once we
get the opposite side, we will need to add 4 ft to the answer to find
the height of the tree. The trigonometric ratio that uses the opposite
and adjacent sides is the tangent function. Hence,
opp
x
tan 65.7˚ =
=
(multiply by 25 to solve for x)
adj
25
x = 25•tan 65.7˚ = 25•2.214… = 55.36… ft
Now, add 4 ft to find the height of the tree:
55.36… + 4 = 59.36 ≈ 59.4 ft
So, the tree is approximately 59.4 ft tall.
163
Ex. 16
A helicopter flying directly above an ocean liner at a height of
1400 ft, locates another boat at a 23˚ angle of depression. How far
apart are the two ships? Round to the nearest hundred feet.
Solution:
The angle of depression is the angle between a horizontal line and
the line of observation. In other words, it is the angle one has to look
down to see the observed object, which is a ship in this case. The
helicopter is flying directly above the ocean liner. We can draw a
vertical line between the helicopter and the ocean liner. Next we can
draw a line from the helicopter to the other ship. The angle between
this two lines and the angle of
depression are complimentary
23˚
angles. Thus, the angle between
67˚
the two lines is 90˚ – 23˚ = 67˚.
1400 ft
Now, drawing a line between the
two ships creates a right triangle.
The vertical height is the adjacent
side of 67˚ and we are looking for
the opposite side of 67˚. So, we will
need to use the tangent function.
x
opp
x
tan 67˚ =
=
(multiply by 1400 to solve for x)
adj
1400
x = 1400•tan 67˚= 1400•2.35… = 3298.1… ft ≈ 3300 ft
So, the ships are approximately 3300 ft apart.
€
€ a wheel
Ex. 17
For
chair accessible ramp to be ADA compliant, the
ratio of the rise to the run of the ramp must be 1 to (12 or greater). A
29 ft 9 in long ramp leading into a building climbs 2 ft 5 in.
a) Is this ramp ADA compliant?
b) If a new city ordinance states that the incline of a handicapped
access can be no more than 4.5˚, does this ramp comply with the
new ordinance? If not, how long does the ramp need to be? Round to
the nearest inch.
Solution:
a) We will begin by drawing a picture:
Length
← Angle of Elevation
Run
Rise
164
The Length of the ramp corresponds to the hypotenuse, and the Rise
and Run correspond to the two legs. Thus, we will need convert the
measurements into inches and use the Pythagorean theorem to
calculate the run of the ramp:
( 12in
) + 5 in = 24 in + 5 in = 29 in
1ft
29 ft 12in
Length = 29 ft + 9 in =
( 1ft ) + 9 in = 348 in + 9 in = 357 in
1
2 ft
1
Rise = 2 ft + 5 in =
(Run)2 = (Length)2 – (Rise)2 = (357)2 – (29)2 = 127449 – 841 = 126608
€ €
Run = 126608 = 355.52… in
The ratio of the
€ Rise
€ to the Run is 29 to 355.52… Dividing by 29, we
get: 1 to 12.269… Since 12.269… ≥ 12, the ramp is ADA compliant.
€ b) Since the ramp climbs 29 in, then
357 in
x
the height of the ramp is 29 in. This
29 in
will correspond to the opposite side
of the incline. The length of the ramp is 357 in. This will correspond to
the hypotenuse. This means we will need to use the inverse sine
function to find the angle.
opp
29
sin x˚ =
=
(use the inverse sine)
x=
hyp
357
29
sin – 1 (
) = 4.659…˚
357
which is greater than 4.5˚
Thus, the current ramp does not comply with the ordinance.
€ climbs 29 in, then
Since the ramp
y
the height of the ramp is 29 in. This
29 in
4.5˚
€
will correspond to the opposite side
of the incline. The length of the ramp is
length of the ramp which is the hypotenuse. This means we will need
to use the sine function.
opp
29
sin 4.5˚ =
=
(multiply by y)
hyp
y
y sin 4.5˚ = 29
(divide by sin 4.5˚)
29
= 369.619… ≈ 370 in
y=
sin 4.5
=
29
0.07845...
€
€
But, 370 ÷ 12 = 30 with a remainder 10, so
370 in = 30 ft 10 in
The
€ ramp€should be 30 ft 10 in to comply with the ordinance. Note:
Another requirement for a ramp to ADA compliant is that is one 5 ft by
5 ft rest platform is required after a run of 30 feet in a ramp.