Practice 8-5 : Angles of Elevation & Depression DNG

p. 405

Describe each angle as it relates to the diagram.

ANSWERS:

a. of depression from the birds to the ship

b. of elevation from the ship to the birds

c. of depression from the ship to the submarine

d. of elevation from the submarine to the ship

Practice 8-5 : Angles of Elevation & Depression DNG

p. 405

Describe each angle as it relates to the diagram.

d. of elevation from the sailboat to the person

ANSWERS:

a. of depression from the plane to the person

b. of elevation from the person to the plane

c. of depression from the person to the sailboat

Practice 8-5 : Angles of Elevation & Depression DNG

p. 405

Find the value of x. Round the lengths to the nearest tenth.

Opposite

Adjacent

Hypotenuse

Adjacent

tan 25° =𝒙

𝟐𝟓𝟎

cos 32° =𝒙

𝟏𝟎𝟎

x = 250 tan 25°

x ≈ 116.6 ft

x = 100 cos 32°

x ≈ 𝟖𝟒.8 ft

Practice 8-5 : Angles of Elevation & Depression DNG

p. 405

Find the value of x. Round the lengths to the nearest tenth.

Opposite

Opposite

Hypotenuse

Adjacent

sin 40° =𝟑𝟎

𝒙

tan 58° =𝟓𝟎

𝒙

x =𝟑𝟎

𝐬𝐢𝐧 𝟒𝟎°

x ≈ 46.7 ft

x =𝟓𝟎

𝐭𝐚𝐧 𝟓𝟖°

x ≈ 𝟑𝟏.2 yd

Practice 8-5 : Angles of Elevation & Depression DNG

p. 405

Find the value of x. Round the lengths to the nearest tenth.

Opposite

Hypotenuse

Hypotenuse

Adjacent

sin 28° =𝟔𝟎

𝒙

cos 22° =𝟑𝟎𝟎

𝒙

x =𝟔𝟎

𝐬𝐢𝐧 𝟐𝟖°

x =𝟑𝟎𝟎

𝐜𝐨𝐬 𝟐𝟐°

x ≈ 127.8 m

x ≈ 323.6 m

Practice 8-5 : Angles of Elevation & Depression DNG

p. 405

9. A person standing 30 ft from a flagpole can see the top

of the pole at a 35° angle of elevation.

a. Draw a diagram.

b. The person’s eye level is 5 ft from the ground.

Find the height of the flagpole to the nearest foot.

30 ft 5 ft

30 ft

35

∟

h

5 ft

x

Adjacent

Opposite

tan 35° =𝒙

𝟑𝟎

x = 30 tan 35°

x ≈ 21 ft

𝒉 ≈ 𝟐𝟏 𝒇𝒕 + 𝟓𝒇𝒕 ≈ 𝟐𝟔 𝒇𝒕

The flagpole is about 26 feet tall.

QUIZ… 2/17/2015 1:59 PM

Two hills are 2 mi apart. The taller hill is 2707 ft. high. The angle

of depression from the top of the taller hill to the top of the shorter

hill is 7°. Find the height of the shorter hill to the nearest foot.

(1 mi= 5280 ft.) Let x = height of shorter hill

x ≈ 1410 ft

x x

y

x = 2707 − 𝒚

Solve for y: 7) tan 7° =

𝒚

𝟏𝟎𝟓𝟔𝟎

y = 𝟏𝟎𝟓𝟔𝟎 tan 𝟕° ≈ 𝟏𝟐𝟗𝟕

= 𝟐 𝟓𝟐𝟖𝟎 =10,560 ft

10,560 ft

x = 2707 −1297