Title: Example 1 Geometric Mean
Author: Tim Heppner
Last saved by: MHE
Company: MCGRAW-HILL

Geometry	Chapter 8

Lesson 8-5

Example 1  Angle of Elevation
SURVEYING  A surveyor is 130 feet from a tower; the tower is 86 feet high. The surveyor's instrument is 4.75 feet above the ground. If the surveyor sights the top of the tower with his instruments, what is the angle of elevation?
Make a drawing.

Since BD is 86 feet and CD is 4.75 feet, BC is 81.25 feet. Let x represent m∠BAC.

	tan x°	= BC/AC	tan = opposite/adjacent
	tan x°	= 81.25/130	BC = 81.25, AC = 130
	x	= tan^-1 ﴾81.25/130﴿	Solve for x.
	x	≈ 32.0	Use a calculator.

The angle of elevation is about 32.0°.

Example 2  Angle of Depression
Short-Response Test Item
A fire is sighted by a ranger in a fire tower, 75 meters 	
above the ground. If the ranger is looking down at an 
angle of depression of 22°, how far from the base of 
the tower is the fire?

Read the Test Item
The angle of depression between the ranger's line of sight and the horizontal is 22°. Use trigonometry to find the distance from the base of the tower to the fire.

Solve the Test Item

Method 1

The ground and the horizontal level with the ranger's eyes are parallel. Therefore, m∠BAC = m∠ACD since they are alternate interior angles.

	tan 22°	= 75/CD
	CD tan 22°	= 75
	CD	= 75/(tan 22)
	CD	≈ 185.6

Method 2

The horizontal line level with the ranger's eyes and the segment from the ground to the ranger are perpendicular. So, ∠BAC and ∠CAD are complementary angles. Therefore, m∠CAD = 90 - 22 or 68.

	tan 68°	= CD/75
	75 tan 68°	= CD
	185.6	≈ CD

The fire is about 185.6 meters from the base of the tower.

Example 3  Indirect Measurement
A plane is flying at an altitude of 300 meters on a heading
towards both a maintenance hangar and the main runway. 	
The pilot sights the end of the main runway at
an angle of depression of 15°. She then sights the 
maintenance hangar at an angle of depression 
of 32°. What is the distance between the end of 
the main runway and the maintenance hangar?

△CDA and △CDB are right triangles, and CD = 300. The distance between the end of the runway and the hangar is AB or AD – BD.  Use the right triangles to find these two lengths.

Because  and  are horizontal lines, they are parallel. Thus, ∠ECA ≅ ∠CAD and 
∠ECB ≅ ∠CBD because they are alternate interior angles. This means that m∠CAD = 15
and m∠CBD = 32.

	tan 15°	= 300/AD	tan = opposite/adjacent; m∠CAD = 15
	AD tan 15°	= 300	Multiply each side by AD.
	AD	= 300/(tan 15)	Divide each side by tan 15.
	AD	≈ 1119.62	Use a calculator.
	
	tan 32°	= 300/BD	tan = opposite/adjacent; m∠CBD = 32
	BD tan 32°	= 300	Multiply each side by BD.
	BD	= 300/(tan 32)	Divide each side by tan 32°.
	BD 	≈ 480.10	Use a calculator.

The distance between the end of the runway and the maintenance hangar is
AD - BD  ≈ 1119.62 - 480.10 or about 639.52 meters.