Transcript 8-5 Angles of Elevation and Depression

8-5 Angles of Elevation and
Depression
The Student will be able to:
1. Solve problems involving angles of elevation and
depression.
2. Use angles of elevation and depression to find the distance
between two objects.
Angles of Elevation and Depression
Angle of elevation - the angle formed by a horizontal line and an
observer’s line of sight to an object above the horizontal line.
Angle of depression – the angle formed by a horizontal line and an
observer’s line of sight to an object below the horizontal line.
The two horizontal lines are parallel so the angle of elevation
and the angle of depression are equal.
Example 1:
The cross bar of a goalpost is 10 feet high. If a goal attempt is
made 25 yards from the base of the goalpost that clears the goal
by 1 foot, what is the smallest angle of elevation at which the
ball could have been kicked to the nearest degree?
1st
– Draw a picture.
2nd – Which angle are
you looking for?
Where he kicked it from.
1 ft
11 ft
10 ft
3rd – Which trig ratio applies and do
you use it or its inverse?
opp
= tan x
adj
You’re given two sides.
Use the inverse operation.
x°
25 yds 75 ft
æ opp ö
tan ç
÷ = mÐx
è adj ø
-1 æ 11 ö
ta n ç ÷ = mÐx
è 75 ø
-1
Example 2:
A lifeguard is watching a beach from a line of sight 6 feet above
the ground. She sees a swimmer at an angle of depression of 8°.
How far away from the tower is the swimmer?
1st – Which side are we looking for?
From the bottom of the tower
to the swimmer.
6 ft
2nd – Which trig ratio applies and
do you use it or its inverse?
opp
= tan x You’re given one side and
adj
one angle.
x°
8°
Use the trig ratio.
x » 43 ft
You Try It:
1. At the circus, a person in the audience at ground level
watches the high-wire routine. A 5-foot-6 inch tall acrobat is
standing on a platform that is 25 feet off the ground. How far is
the audience member from the base of the platform, if the angle
of elevation from the audience member’s line of sight to the top
of the acrobat’s head is 27°? 60 ft 5.5 ft
tan =
opp
adj
366in
tan 27 =
x
366in
x=
tan 27
718.31 in or 59.86 ft ≈ x
25 ft
27°
x
2. Maria is at the top of a cliff and sees a seal in the water. If the
cliff is 40 feet above the water and the angle of depression is
52°, what is the horizontal distance from the seal to the cliff, to
the nearest foot? 31 ft
52°
opp
tan =
adj
40
x
40
x=
tan 52
tan 52 =
31.3 ft ≈ x
40 ft
52°
x
Two Angles of Elevation or Depression
Angles of elevation or depression to two different objects can be
used to estimate the distance between those objects. Similarly, the
angles from two different positions of observation to the same
object can be used to estimate the object’s height.
Example 3:
Two buildings are sited from atop a 200-meter skyscraper.
Building A is sited at a 35° angle of depression, while Building B is
sighted at a 36° angle of depression. How far apart are the two
buildings to the nearest meter?
36 35
1st – Draw a picture.
2nd – Which side are we
200
looking for?
The distance between the
two buildings.
36
35
Hint: Find the length of the
base of both triangles first.
A
xB
3rd – Which trig ratio applies and
do you use it or its inverse?
36
200
You’re given one side and one
angle. You must solve for the
base angle of each triangle. Find
the base length of building b first.
opp
tan x =
adj
35
A
35
x
36
B
275
4th – Find the distance between
building a & building b.
building a – building b = x
286 m – 275 m = x
11 m = x
b = 275 m
a = 286 m
You Try It:
Miko and Tyler are visiting the Great Pyramid in Egypt. From
where Miko is standing, the angle of elevation to the top of the
pyramid is 48.6°. From Tyler’s position, the angle of elevation is
50°. If they are standing 20 feet apart, how tall is the pyramid?
1st – Which side are we
looking for?
The height of the pyramid.
Hint: Find the base length
of the smaller triangle first.
3rd – Which trig ratio applies and
do you use it or its inverse?
You’re given only one angle. You
must solve for x.
y
y + 20
opp
tan 50 =
adj
1.1918y = x
4th – Solve for the bigger
triangle.
x
1.1343 =
y + 20
1.1343(y + 20) = x
Hint: Substitute for x from
step 1.
y
y + 20
5th – Substitute y back in original
1.1343y + 22.6855 =1.1918y
equation (step 1) and solve for x.
22.6855 =1.1918y -1.1343y
1.1918y = x
22.6855 =.0575y
1.1918(394.5304) = x
22.6855
=y
470.2 = x
.0575
394.5304 = y