#### Transcript The Escalator Problem

```Problem 1
The Escalator
Click on the link below to view
the lesson starter.
Stuck on an Escalator
Problem 1 (Part 1)
In order to save our friends, the rescue team
must determine the total length of the
escalator. If the vertical rise of the escalator
has been measured at 195 ft., 9.5 in. at an
angle of elevation of 10.4°, what is the length
of the escalator to the nearest foot?
1.
Draw the figure.
2.
Convert measurement to desired
unit (feet).
3.
Calculate the length of the
escalator.
1. Draw the figure.
M
x
195 ft., 9.5 in.
10.4°
E
O
2. Convert measurement to desired unit (feet).
195 ft., 9.5 in.
9.5 in. x 1 ft. = 9.5 ft. = 0.792
ft.
12 in. 12
195 ft. + 0.792 ft. = 195.792 ft.
M
195.792 ft.
E
x
10.4°
O
3. Calculate the length of the escalator.
Sin O° = ME
MO
Sin 32.4° = 90.792 ft.
x
x (Sin 32.4°) = 90.792 ft.
x = 90.792 ft.
Sin 32.4°
x = 169.443 ft.
MO = 169.443 ≈ 169 ft.
1.
Calculate starting distance from the
bottom of the escalator.
2.
Calculate distance traveled along
the incline.
3.
Use trigonometry to calculate
horizontal distance.
Problem 1 (Part 2)
The rescue team determines our friends are
stuck a third of the length of the escalator
from the bottom when the escalator begins to
move again. If the escalator moves our
friends for just 10 seconds before stopping
again, along the incline at a speed of 3 feet
per second, what is the horizontal distance
traveled to the nearest tenth?
1. Calculate starting distance from the
bottom of the escalator.
…our friends are stuck a third of the length of
the escalator from the bottom…
Current location = (1/3) (169 ft.) = 56.33 ft.
M
E
10.4°
O
2. Calculate distance traveled along
the incline.
…the escalator moves our friends for 10
seconds along the incline at a speed of 3
feet per second…
d = r (t)
d = 3 (10) = 30 ft.
M
E
10.4°
O
3. Use trigonometry to calculate
horizontal distance.
M
J
E
T
10.4°
y
Cos O° = TO
OJ
Cos (10.4°) =
y
86.33 ft.
y = (Cos 10.4°) (86.33) ≈ 84.9
ft.
O
Problem 1 (Part 3)
After this 10 seconds of movement, would it
be shorter for the stranded riders to walk the
rest of the way up or revert back down to the
bottom of the escalator? Explain why.
Problem 2
Jessica observed a mountain climber
reaching the summit, which is
known to be at 2,358 ft. If she
sighted the climber standing 1500 ft.
from the base, at what angle did
Jessica sight the mountain climber to
the nearest degree?
1.
Draw a figure to represent the
problem.
2.
Determine which trigonometric
ratio to use.
3.
Calculate the angle of
elevation.
1. Draw a figure to represent
the problem.
M
2,358 ft.
x°
S
1,500 ft.
C
2. Determine which trigonometric
ratio to use.
M
2,358 ft.
x°
S
1,500 ft.
Tangent
C
3. Calculate the Angle of Elevation.
Tan C° = MS
SC
Tan x° = 2,358 ft.
1,500 ft.
x = Tan-1 2,358 ft.
1,500 ft.
x = 57.538°
C≈ 58°
Problem 3
A rescue helicopter pilot sights
a life raft at an angle of
o
depression of 26 . The
helicopter is 3 km above the
water. What is the pilot’s
surface distance from the raft
to the nearest km?
1.
Draw a figure to represent the
problem.
2.
Determine which trigonometric ratio
to use.
3.
Calculate the surface distance.
1. Draw a figure to represent
the problem.
H
26°
3 km
26°
W
x km
R
2. Determine which trigonometric
ratio to use.
H
26°
3 km
26°
W
x km
Tangent
R
3. Calculate the surface distance.
Tan R° = HW
WR
Tan 26° = 3 km
x km
x (Tan 26°) = 3 km
x = 3 km
Tan 26°
x = WR ≈ 6.1509 km
Problem 4
Kevin is standing at the back of the cruise ship and
observes two sea turtles following each other,
swimming in a straight line in the opposite direction
of the ship. Kevin’s position is 206 meters above sea
level and the angles of depression to the two sea
turtles are 43° and 47°. Calculate the distance
between the two sea turtles to the nearest meter.
K
43° 47°
206m
S
T
O
1.
Separate and re-draw the two
triangles.
2.
Calculate individual horizontal
distances.
3.
Calculate the difference between
the two horizontal distances.
1. Separate and re-draw the two triangles.
43°
K
47°
206m
43°
S
K
206m
47°
x
O
T
y
O
2. Calculate individual horizontal distances.
Tan S° = KO
SO
Tan T° = KO
TO
Tan 43° = 206 m
x
Tan 47° = 206 m
y
x (Tan 43°) = 206 m
x (Tan 47°) = 206 m
x = 206 m
Tan 43°
x = 206 m
Tan 47°
x = 220.90795
x = 193.03062
SO = 220.90795 m
SO = 193.03062 m
3. Calculate the differences between the
two horizontal distances.
ST = SO – TO
ST = 220.90795 m – 193.03062 m
ST = 27.8773 m
ST ≈ 28 m
```