Download Sect 9.1– Applications Involving Right Triangles

142
Sect 9.1– Applications Involving Right Triangles
Objective 1:
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
to the nearest tenth and all angles to the nearest minute:
E
B
Ex. 1
Ex. 2
18 ft
71.5˚
23˚
C
A
F 13 in G
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.
sin 23˚ =
=
Solution:
Since ∠E and ∠G are
complimentary, then
m∠E = 90˚ – 71.5˚ = 18.5˚ = 18˚30'.
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.
tan 71.5˚ =
=
To solve, multiply by 18:
BC = 18•sin 23˚
= 18•0.3907… = 7.03…
≈ 7.0 ft
cos 23˚ =
=
To solve, multiply by 13:
EF = 13•tan 71.5˚
= 13•2.98… = 38.85…
≈ 38.9 in
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˚30'.
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Ex. 3
C
7 cm
B
Ex. 4
F
10.1 m
5 cm
E
A
15 m
G
Solution:
We have the two legs of a
right triangle. We can use
the inverse tangent function
Solution:
We have the leg and the
hypotenuse of a right triangle.
we can use the inverse cosine to
to find ∠B: tan B =
to find ∠G: cos G =
Thus, B = tan – 1(
=
)
= 35.53… ˚
Convert into DMS and
round:
35˚32'15"6 ≈ 35˚32'
Thus, m∠B = 35˚32'
Since ∠A and ∠B are
complimentary, then
m∠A = 90˚ – 35˚32' = 54˚28'
To find the length AB, we can
use the Pythagorean Theorem:
(AB)2 = (7)2 + (5)2
(AB)2 = 49 + 25
(AB)2 = 74
AB =
= 8.60… ≈ 8.6 cm
Thus, m∠B ≈ 35˚32',
m∠A ≈ 54˚28', and AB ≈ 8.6 cm.
Objective 2:
Hence, G = cos – 1(
=
)
= 47.67… ˚
Convert into DMS and round:
47˚40'30"5 ≈ 47˚41'
Therefore, m∠G = 47˚41'
Since ∠E and ∠G are
complimentary, then
m∠E = 90˚ – 47˚41' = 42˚19'
To find the length EF, we can
use the Pythagorean Theorem:
(10.1)2 + (EF)2 = (15)2
102.01 + (EF)2 = 225
– 102.01
= – 102.01
2
(EF) = 122.99
EF =
=11.09… ≈ 11.1 m
Hence, m∠G ≈ 47˚41',
m∠E = 42˚19', and EF ≈ 11.1 m.
Solving applications involving right triangles.
In solving right triangles, we will use the fact that the two acute angles
are complementary and we will use the Pythagorean Theorem.
Right Triangle Theorem
Let A, B, and C be the angles of the triangle and a, b, and c be the
lengths of the corresponding sides. If C is a right angle, then
A + B = 90˚
and
c2 = a2 + b2
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Solve the following:
Ex. 5
A road rises 223 ft over a horizontal distance of 2500 ft. What is
the angle of elevation? (round to the nearest tenth).
Solution:
First, draw a diagram:
223 ft
θ
2500 ft
We have the opposite and adjacent sides of a right triangle, so we
want to use the inverse tangent function to find the angle
opp
223
tan θ =
=
(use the inverse tangent)
adj
θ = tan – 1(
Ex. 6
2500
223
)
2500
= 5.097… ≈ 5.1˚
€For a €
wheel 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.
c) 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 a maximum of 30 feet in
a ramp. Does the new ramp require a rest platform?
Solution:
a) We will begin by drawing a picture:
Length
Rise
← Angle of Elevation
Run
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:
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Rise =
€
(
) + 5 in = 29 in
Length =
29 ft
1
(
) + 9 in = 357 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… is greater than 12, the ramp is
ADA compliant.
b) Since the ramp climbs 29 in, then
357 in
the height of the ramp is 29 in. This
29 in
x
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.
29
sin x =
=
(use the inverse sine)
357
x = sin – 1 (
29
)
357
= 4.659…˚ 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.
sin 4.5˚ =
=
y sin 4.5˚ = 29
29
y=
=

sin 4.5
29
y
29
0.07845...
(multiply by y)
(divide by sin 4.5˚)
= 369.619… ≈ 370 in
€
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.
€
€
c) To determine if the new ramp needs a rest platform, we will need
to calculate the Run:
(Run)2 = (Length)2 – (Rise)2 = (370)2 – (29)2 = 136900 – 841 = 136059
Run = 136059 = 368.86… in
Converting to feet, we get: 368.86…/12 = 30.738… feet which is
greater than 30 feet so the new ramp will need a rest platform.
€
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In navigation, the bearing or direction is measured as the acute angle
either east or west of north or south. Thus, if a ship is traveling 35˚ east of
north, it's bearing would be N35˚E while a airplane flying 22˚ west of south
would have a bearing of S22˚W.
Identify the bearing for each point:
Ex. 7
N
A
B 77˚ 38˚
W
E
C
65˚
D
5˚
S
Solution:
A)
The bearing is N38˚E.
B)
The bearing is N77˚W.
C)
The bearing is S65˚E
D)
The bearing is S5˚W.
Solve the following:
Ex. 8 A ship leaves port in the Florida Keys with a bearing of S85˚E with a
speed of 20 knots. After two hours, the ship turns 90˚ toward the
north. After 3 additional hours of maintaining the same speed, what is
the bearing of the ship from the port?
Solution:
We begin by drawing a diagram of
the situation. The ship travels 20(2)
or 40 nautical miles S85˚E from the
Florida Keys. it then turns 90˚ toward
the north and travels 20(3) = 60
nautical miles. If we connect a line from
the Florida Keys to the ending point,
we will form a right triangle. We will
label the angle formed by the 40 nautical
θ
40
85˚
60
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mile side and the line connecting the Florida Keys to the ending point
θ. Since we have the opposite side and the adjacent side, we will use
the tangent function to find θ.
tan(θ) =
opp
adj
=
60
40
= 1.5.
Thus, θ = tan – 1(1.5) = 56.3099…˚ ≈ 56.3˚
Since the 40 nautical mile side had a bearing of S85˚E, the angle
between that side and the positive x-axis is 90˚ – 85˚ = 5˚.
€
€The angle
between the positive x-axis and the line connecting the
Florida Keys to the ending point is θ – 5˚ = 56.3˚ – 5˚ = 51.3˚.
Finally, the angle between the line connecting the Florida Keys to the
ending point and the positive y-axis is 90˚ – 51.3˚ = 38.7˚
This means the bearing from the Florida Keys to the ship is N38.7˚E.